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	<title>EjerciciosFyQ</title>
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	<description>Ejercicios Resueltos, Situaciones de aprendizaje y V&#205;DEOS de F&#237;sica y Qu&#237;mica para Secundaria y Bachillerato</description>
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<item xml:lang="es">
		<title>Vectores de posici&#243;n y desplazamiento entre dos puntos en una trayectoria curvil&#237;nea (8324)</title>
		<link>https://ejercicios-fyq.com/Vectores-de-posicion-y-desplazamiento-entre-dos-puntos-en-una-trayectoria</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Vectores-de-posicion-y-desplazamiento-entre-dos-puntos-en-una-trayectoria</guid>
		<dc:date>2024-10-07T03:24:18Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Posici&#243;n</dc:subject>
		<dc:subject>Desplazamiento</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>

		<description>
&lt;p&gt;A partir del gr&#225;fico siguiente, dibuja los vectores de posici&#243;n de cada uno de los puntos marcados y el desplazamiento entre los puntos B y D.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores" rel="directory"&gt;Algebra de vectores&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/Posicion" rel="tag"&gt;Posici&#243;n&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Desplazamiento" rel="tag"&gt;Desplazamiento&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A partir del gr&#225;fico siguiente, dibuja los vectores de posici&#243;n de cada uno de los puntos marcados y el desplazamiento entre los puntos B y D.&lt;/p&gt;
&lt;div class='spip_document_2012 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;a href='https://ejercicios-fyq.com/IMG/png/ej_8324.png' class=&#034;spip_doc_lien mediabox&#034; type=&#034;image/png&#034;&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L500xH435/ej_8324-99861.png?1758425397' width='500' height='435' alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Los vectores de posici&#243;n los puedes dibujar si unes el origen de coordenadas (A) con cada uno de los puntos marcados. El gr&#225;fico queda como: &lt;br/&gt;&lt;/p&gt;
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&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;a href='https://ejercicios-fyq.com/IMG/png/ej_8324_2.png' class=&#034;spip_doc_lien mediabox&#034; type=&#034;image/png&#034;&gt; &lt;img src='https://ejercicios-fyq.com/IMG/png/ej_8324_2.png' width=&#034;3000&#034; height=&#034;2611&#034; alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt; &lt;p&gt;&lt;br/&gt; El desplazamiento es el vector resultante de la diferencia entre la posici&#243;n final y la posici&#243;n inicial, es decir: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/af6488ac2f95efe0a07d8a23c68812a0.png' style=&#034;vertical-align:middle;&#034; width=&#034;171&#034; height=&#034;20&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\Delta \vec{r}_{BD} = \vec{r}_D - \vec{r}_B}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\Delta \vec{r}_{BD} = \vec{r}_D - \vec{r}_B}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt;&lt;/math&gt; El gr&#225;fico final es: &lt;br/&gt;&lt;/p&gt;
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&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;a href='https://ejercicios-fyq.com/IMG/png/ej_8324_3.png' class=&#034;spip_doc_lien mediabox&#034; type=&#034;image/png&#034;&gt; &lt;img src='https://ejercicios-fyq.com/IMG/png/ej_8324_3.png' width=&#034;3000&#034; height=&#034;2611&#034; alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/div&gt;
		
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<item xml:lang="es">
		<title>Opuesto al vector resultante de dos vectores (7970)</title>
		<link>https://ejercicios-fyq.com/Opuesto-al-vector-resultante-de-dos-vectores-7970</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Opuesto-al-vector-resultante-de-dos-vectores-7970</guid>
		<dc:date>2023-06-25T13:45:15Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>
		<dc:subject>Componentes</dc:subject>

		<description>
&lt;p&gt;Calcula el vector necesario para que la resultante sea nula, si ya tengo dos vectores que son: y .&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores" rel="directory"&gt;Algebra de vectores&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Componentes" rel="tag"&gt;Componentes&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calcula el vector necesario para que la resultante sea nula, si ya tengo dos vectores que son: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L118xH22/913e987ffdc89c46797ab191ee5a2b33-29739.png?1732964416' style='vertical-align:middle;' width='118' height='22' alt=&#034;\vec{A} = 40\ N\ (185^o)&#034; title=&#034;\vec{A} = 40\ N\ (185^o)&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L119xH22/dcd14509a322efff906036c5c62a883d-dce90.png?1732964416' style='vertical-align:middle;' width='119' height='22' alt=&#034;\vec{B} = 80\ N\ (275^o)&#034; title=&#034;\vec{B} = 80\ N\ (275^o)&#034; /&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;La resoluci&#243;n anal&#237;tica del problema es la m&#225;s c&#243;moda. En primer lugar, calculas las componentes de cada vector dado: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/59abc9dbcb50713f94811c1f44673598.png' style=&#034;vertical-align:middle;&#034; width=&#034;495&#034; height=&#034;50&#034; alt=&#034;\left \vec{A} = A\cdot cos\ 185^o\ \vec{i} + A\cdot sen\ 185^o\ \vec{j} \atop \vec{B} = B\cdot cos\ 275^o\ \vec{i} + B\cdot sen\ 275^o\ \vec{j} \right \}\ \to\ {\color[RGB]{0,112,192}{\bm{\left \vec{A} = -39.85\ \vec{i} - 3.49\ \vec{j} \atop \vec{B} = 6.97\ \vec{i} - 79.7\ \vec{j} \right \}}}}&#034; title=&#034;\left \vec{A} = A\cdot cos\ 185^o\ \vec{i} + A\cdot sen\ 185^o\ \vec{j} \atop \vec{B} = B\cdot cos\ 275^o\ \vec{i} + B\cdot sen\ 275^o\ \vec{j} \right \}\ \to\ {\color[RGB]{0,112,192}{\bm{\left \vec{A} = -39.85\ \vec{i} - 3.49\ \vec{j} \atop \vec{B} = 6.97\ \vec{i} - 79.7\ \vec{j} \right \}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Ahora sumas ambos vectores, componente a componente, para obtener el vector resultante: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9b1b3ce82606b940ab2f4d4372c78193.png' style=&#034;vertical-align:middle;&#034; width=&#034;199&#034; height=&#034;21&#034; alt=&#034;\vec{A} + \vec{B} = \color[RGB]{0,112,192}{\bm{-32.9\ \vec{i} - 83.2\ \vec{j}}}&#034; title=&#034;\vec{A} + \vec{B} = \color[RGB]{0,112,192}{\bm{-32.9\ \vec{i} - 83.2\ \vec{j}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El vector necesario es el opuesto al vector resultante calculado: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0ba067cdc6841dfe878988c575674c61.png' style=&#034;vertical-align:middle;&#034; width=&#034;163&#034; height=&#034;29&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\vec{C} = 39.2\ \vec{i} + 83.2\ \vec{j}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\vec{C} = 39.2\ \vec{i} + 83.2\ \vec{j}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Algebra de vectores referido a sus m&#243;dulos (7856)</title>
		<link>https://ejercicios-fyq.com/Algebra-de-vectores-referido-a-sus-modulos-7856</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Algebra-de-vectores-referido-a-sus-modulos-7856</guid>
		<dc:date>2023-02-13T05:02:50Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>

		<description>
&lt;p&gt;Demuestra que la desigualdad , se cumple para los vectores y . &#191;Qu&#233; tienen que cumplir dos vectores y para que se verifique la igualdad&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Vectores-Cinematica-Dinamica-y-Energia-2-o-Bach" rel="directory"&gt;Vectores, Cinem&#225;tica, Din&#225;mica y Energ&#237;a (2.&#186; Bach)&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Demuestra que la desigualdad &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L117xH22/19551b9c016ae4f59d634999da6c0d37-59424.png?1733062919' style='vertical-align:middle;' width='117' height='22' alt=&#034;|\vec{a} + \vec{b}| \leqslant |\vec{a}| + |\vec{b}|&#034; title=&#034;|\vec{a} + \vec{b}| \leqslant |\vec{a}| + |\vec{b}|&#034; /&gt; , se cumple para los vectores &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L95xH21/538ce977da310fee170559495ed807f0-c9df7.png?1733062919' style='vertical-align:middle;' width='95' height='21' alt=&#034;\vec{a} = \vec{i} - \vec{j} + \vec{k}&#034; title=&#034;\vec{a} = \vec{i} - \vec{j} + \vec{k}&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L130xH21/165f5a807bf6d74504d997808534227a-8b817.png?1733062919' style='vertical-align:middle;' width='130' height='21' alt=&#034;\vec{b} = -2\vec{i} + 5\vec{j} - 3\vec{k}&#034; title=&#034;\vec{b} = -2\vec{i} + 5\vec{j} - 3\vec{k}&#034; /&gt; . &#191;Qu&#233; tienen que cumplir dos vectores &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L10xH13/8fd082536a0a420385519d1473c9d27e-788ae.png?1733062919' style='vertical-align:middle;' width='10' height='13' alt=&#034;\vec{a}&#034; title=&#034;\vec{a}&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L9xH18/0f4c4ce0863d100a12c90c114fd9abeb-e7813.png?1733062919' style='vertical-align:middle;' width='9' height='18' alt=&#034;\vec{b}&#034; title=&#034;\vec{b}&#034; /&gt; para que se verifique la igualdad &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L125xH22/d8d89717d55da7d151c03e43bcb11932-66460.png?1733062919' style='vertical-align:middle;' width='125' height='22' alt=&#034;|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| ?&#034; title=&#034;|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| ?&#034; /&gt;&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Lo primero que debes hacer el calcular los m&#243;dulos que tienes que relacionar: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6cccec07a1e1bda61e10549468b654c0.png' style=&#034;vertical-align:middle;&#034; width=&#034;213&#034; height=&#034;31&#034; alt=&#034;\left |\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \color[RGB]{0,112,192}{\bm{\sqrt{3}}}&#034; title=&#034;\left |\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \color[RGB]{0,112,192}{\bm{\sqrt{3}}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6324e1a4266777ccb704681764c50502.png' style=&#034;vertical-align:middle;&#034; width=&#034;240&#034; height=&#034;21&#034; alt=&#034;|\vec{b} = \sqrt{(-2)^2 + 5^2 + (-3)^2} = \color[RGB]{0,112,192}{\bm{\sqrt{38}}}&#034; title=&#034;|\vec{b} = \sqrt{(-2)^2 + 5^2 + (-3)^2} = \color[RGB]{0,112,192}{\bm{\sqrt{38}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Para hacer el vector de la suma primero debes sumar los vectores: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/106a81bcdf8adc02597dfc0234c6b44d.png' style=&#034;vertical-align:middle;&#034; width=&#034;422&#034; height=&#034;22&#034; alt=&#034;\vec{a} + \vec{b} = (1 - 2)\ \vec{i} + (-1 + 5)\ \vec{j} + (1 - 3)\ \vec{k} = -\vec{i} + 4\ \vec{j} - 2\ \vec{k}&#034; title=&#034;\vec{a} + \vec{b} = (1 - 2)\ \vec{i} + (-1 + 5)\ \vec{j} + (1 - 3)\ \vec{k} = -\vec{i} + 4\ \vec{j} - 2\ \vec{k}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Ahora haces el m&#243;dulo del vector que has obtenido: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6b9926a779168ae8c998e637b3848ee6.png' style=&#034;vertical-align:middle;&#034; width=&#034;273&#034; height=&#034;21&#034; alt=&#034;|\vec{a} + \vec{b}| = \sqrt{(-1)^2 + 4^2 + (-2)^2} = \color[RGB]{0,112,192}{\bm{\sqrt{21}}}&#034; title=&#034;|\vec{a} + \vec{b}| = \sqrt{(-1)^2 + 4^2 + (-2)^2} = \color[RGB]{0,112,192}{\bm{\sqrt{21}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Como puedes ver, se cumple al desigualdad: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3475efd64bcf0e7c7e0d48817167c311.png' style=&#034;vertical-align:middle;&#034; width=&#034;150&#034; height=&#034;27&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\sqrt{21} &lt; \sqrt{3} + \sqrt{38}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\sqrt{21} &lt; \sqrt{3} + \sqrt{38}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; El m&#243;dulo de la suma de vectores lo puedes expresar tambi&#233;n como: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c86e6281aa703825e3bd4b48243070a9.png' style=&#034;vertical-align:middle;&#034; width=&#034;221&#034; height=&#034;22&#034; alt=&#034;|\vec{a} + \vec{b}| = \sqrt{a^2 + b^2 + 2ab\cdot cos\ \alpha}&#034; title=&#034;|\vec{a} + \vec{b}| = \sqrt{a^2 + b^2 + 2ab\cdot cos\ \alpha}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Para saber cuando se verifica la igualdad en la ecuaci&#243;n de partida solo tienes que hacer el cuadrado en ambos miembros e igualar: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b212bcf6dc119ec0ccb08a062b510dd3.png' style=&#034;vertical-align:middle;&#034; width=&#034;476&#034; height=&#034;22&#034; alt=&#034;|\vec{a} + \vec{b}|^2 = (|\vec{a}| + |\vec{b}|)^2\ \to\ \color[RGB]{2,112,20}{\bm{a^2 + b^2 + 2ab\cdot cos\ \alpha = a^2 + b^2 + 2ab}}&#034; title=&#034;|\vec{a} + \vec{b}|^2 = (|\vec{a}| + |\vec{b}|)^2\ \to\ \color[RGB]{2,112,20}{\bm{a^2 + b^2 + 2ab\cdot cos\ \alpha = a^2 + b^2 + 2ab}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Esta igualdad se verifica cuando &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/40d0b09143c6d0402f795304d5aed7e6.png' style=&#034;vertical-align:middle;&#034; width=&#034;67&#034; height=&#034;12&#034; alt=&#034;cos\ \alpha = 1&#034; title=&#034;cos\ \alpha = 1&#034; /&gt;, es decir, si &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/060f5d74c4a7f2a503421b8d84a9d463.png' style=&#034;vertical-align:middle;&#034; width=&#034;45&#034; height=&#034;12&#034; alt=&#034;\color[RGB]{192,0,0}{\bm{\alpha = 0}}&#034; title=&#034;\color[RGB]{192,0,0}{\bm{\alpha = 0}}&#034; /&gt;. &lt;b&gt;Para que se cumpla la igualdad, los vectores tienen que ser paralelos&lt;/b&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Suma de vectores en coordenadas polares (7840)</title>
		<link>https://ejercicios-fyq.com/Suma-de-vectores-en-coordenadas-polares-7840</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Suma-de-vectores-en-coordenadas-polares-7840</guid>
		<dc:date>2023-01-24T07:17:36Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>
		<dc:subject>Coordenadas polares</dc:subject>

		<description>
&lt;p&gt;Calcula el desplazamiento resultante de la suma de los vectores y y expr&#233;sala en coordenadas polares.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Vectores-Cinematica-Dinamica-y-Energia-2-o-Bach" rel="directory"&gt;Vectores, Cinem&#225;tica, Din&#225;mica y Energ&#237;a (2.&#186; Bach)&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Coordenadas-polares" rel="tag"&gt;Coordenadas polares&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calcula el desplazamiento resultante de la suma de los vectores &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L103xH18/d061141ad2c9ea9f424fc9d07926c454-07d53.png?1732989495' style='vertical-align:middle;' width='103' height='18' alt=&#034;A = (5\ m, 30^o)&#034; title=&#034;A = (5\ m, 30^o)&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L113xH18/c76712e5915fa53076595e95ebcfbe85-10258.png?1732989495' style='vertical-align:middle;' width='113' height='18' alt=&#034;B = (3\ m, 220^o)&#034; title=&#034;B = (3\ m, 220^o)&#034; /&gt; y expr&#233;sala en coordenadas polares.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;El planteamiento que voy a desarrollar para resolver el ejercicio es hacer la suma en coordenadas cartesianas y la conversi&#243;n a coordenadas polares. &lt;br/&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;Lo primero que debes hacer es expresar los vectores en coordenadas cartesianas. Para ello aplicas las f&#243;rmulas: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9e210e8c311bd5e9e819f6110837918c.png' style=&#034;vertical-align:middle;&#034; width=&#034;104&#034; height=&#034;40&#034; alt=&#034;\left x = r\cdot cos\ \alpha \atop y = r\cdot sen\ \alpha \right \}&#034; title=&#034;\left x = r\cdot cos\ \alpha \atop y = r\cdot sen\ \alpha \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes los datos de cada vector, pero teniendo cuidado con el modo de la calculadora: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/402be173a2babf4490cb35f8b60cfcd5.png' style=&#034;vertical-align:middle;&#034; width=&#034;435&#034; height=&#034;50&#034; alt=&#034;\left \vec{A} = 5\cdot cos\ 30\ \vec{i} + 5\cdot sen\ 30\ \vec{j}\ \to\ {\color[RGB]{0,112,192}{\bm{\vec{A} = 4.33\ \vec{i} + 2.5\ \vec{j}}}} \atop \vec{B} = 3\cdot cos\ 220\ \vec{i} + 3\cdot sen\ 220\ \vec{j}\ \to\ {\color[RGB]{0,112,192}{\bm{\vec{B} = -2.3\ \vec{i} - 1.93\ \vec{j}}}} \right \}&#034; title=&#034;\left \vec{A} = 5\cdot cos\ 30\ \vec{i} + 5\cdot sen\ 30\ \vec{j}\ \to\ {\color[RGB]{0,112,192}{\bm{\vec{A} = 4.33\ \vec{i} + 2.5\ \vec{j}}}} \atop \vec{B} = 3\cdot cos\ 220\ \vec{i} + 3\cdot sen\ 220\ \vec{j}\ \to\ {\color[RGB]{0,112,192}{\bm{\vec{B} = -2.3\ \vec{i} - 1.93\ \vec{j}}}} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El desplazamiento es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8ce6bc012cfa377bad345528d3e64e6d.png' style=&#034;vertical-align:middle;&#034; width=&#034;101&#034; height=&#034;16&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\Delta \vec{r} = \vec{B} - \vec{A}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\Delta \vec{r} = \vec{B} - \vec{A}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes y haces la operaci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/afc2b9c0d315f6372fd699bc61e56e11.png' style=&#034;vertical-align:middle;&#034; width=&#034;495&#034; height=&#034;21&#034; alt=&#034;\Delta \vec{r} = (-2.3 - 4.33)\ \vec{i} + (-1.93 - 2.5)\ \vec{j}\ \to\ \color[RGB]{0,112,192}{\bm{\Delta \vec{r} = -6.63\ \vec{i} - 4.43\ \vec{j}}}&#034; title=&#034;\Delta \vec{r} = (-2.3 - 4.33)\ \vec{i} + (-1.93 - 2.5)\ \vec{j}\ \to\ \color[RGB]{0,112,192}{\bm{\Delta \vec{r} = -6.63\ \vec{i} - 4.43\ \vec{j}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Ambas coordenadas son negativas, por lo que el desplazamiento est&#225; en el tercer cuadrante. La conversi&#243;n la haces a partir de las coordenadas. Puedes despejar el valor de &lt;i&gt;r&lt;/i&gt; de la coordenada &lt;i&gt;x&lt;/i&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d7d1a7b7fd45b1bf2cf0c5305184ce2f.png' style=&#034;vertical-align:middle;&#034; width=&#034;317&#034; height=&#034;36&#034; alt=&#034;x = r\cdot cos\ \alpha\ \to\ r = \frac{x}{cos\ \alpha}\ \to\ \color[RGB]{2,112,20}{\bm{r = \frac{-6.63}{cos\ \alpha}}}&#034; title=&#034;x = r\cdot cos\ \alpha\ \to\ r = \frac{x}{cos\ \alpha}\ \to\ \color[RGB]{2,112,20}{\bm{r = \frac{-6.63}{cos\ \alpha}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes en la coordenada &lt;i&gt;y&lt;/i&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bf1de01bd61b615a628e8cb316250c6c.png' style=&#034;vertical-align:middle;&#034; width=&#034;395&#034; height=&#034;36&#034; alt=&#034;-4.43 = \frac{-6.63}{cos\ \alpha}\cdot sen\ \alpha\ \to\ tg\ \alpha = \frac{4.43}{6.63}\ \to\ \color[RGB]{0,112,192}{\bm{\alpha = 33.7^o}}&#034; title=&#034;-4.43 = \frac{-6.63}{cos\ \alpha}\cdot sen\ \alpha\ \to\ tg\ \alpha = \frac{4.43}{6.63}\ \to\ \color[RGB]{0,112,192}{\bm{\alpha = 33.7^o}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Debes sumar 180 al &#225;ngulo obtenido porque el desplazamiento est&#225; en el tercer cuadrante. Ahora solo te queda calcular el m&#243;dulo, &lt;i&gt;r&lt;/i&gt;, y lo haces con la primera ecuaci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/405d989515bed00566cea28dad1cbffc.png' style=&#034;vertical-align:middle;&#034; width=&#034;223&#034; height=&#034;35&#034; alt=&#034;r = \frac{-6.63}{cos\ 213.7^o}\ \to\ \color[RGB]{0,112,192}{\bf r = 7.97\ m}&#034; title=&#034;r = \frac{-6.63}{cos\ 213.7^o}\ \to\ \color[RGB]{0,112,192}{\bf r = 7.97\ m}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El desplazamiento, en coordenadas polares, es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/83290176a8d077c3356b18ab19e2f8c4.png' style=&#034;vertical-align:middle;&#034; width=&#034;187&#034; height=&#034;26&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\Delta \vec{r}= (7.97\ m, 213.7^o)}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\Delta \vec{r}= (7.97\ m, 213.7^o)}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Suma de vectores y producto escalar (7405)</title>
		<link>https://ejercicios-fyq.com/Suma-de-vectores-y-producto-escalar-7405</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Suma-de-vectores-y-producto-escalar-7405</guid>
		<dc:date>2021-11-26T06:46:41Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>EDICO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>
		<dc:subject>Producto escalar</dc:subject>

		<description>
&lt;p&gt;Para los siguientes vectores: , y . Calcula .&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores" rel="directory"&gt;Algebra de vectores&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EDICO" rel="tag"&gt;EDICO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Producto-escalar" rel="tag"&gt;Producto escalar&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Para los siguientes vectores: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L111xH21/60fb39e39271b2d83a97f34b8fe97bf5-670ce.png?1732951300' style='vertical-align:middle;' width='111' height='21' alt=&#034;\vec A = 3\ \vec i + \vec j - \vec k&#034; title=&#034;\vec A = 3\ \vec i + \vec j - \vec k&#034; /&gt; , &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L139xH21/3c05c84ee9193aeac84c0f3403e67397-4e521.png?1732951300' style='vertical-align:middle;' width='139' height='21' alt=&#034;\vec B = - \vec i + 2\ \vec j + 5\ \vec k&#034; title=&#034;\vec B = - \vec i + 2\ \vec j + 5\ \vec k&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L100xH21/6e103e13cc085613790c2f55a1a1a93b-6eee8.png?1732951300' style='vertical-align:middle;' width='100' height='21' alt=&#034;\vec C = 2\ \vec j - 3\ \vec k&#034; title=&#034;\vec C = 2\ \vec j - 3\ \vec k&#034; /&gt; . Calcula &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L81xH22/7680ab7e46c6edfa6fee304cb7dbb063-856f8.png?1732951300' style='vertical-align:middle;' width='81' height='22' alt=&#034;\vec B\cdot (\vec A + \vec C)&#034; title=&#034;\vec B\cdot (\vec A + \vec C)&#034; /&gt; .&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;El orden en el que hacer la operaci&#243;n es importante. En primer lugar debes sumar los vectores &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9ac9a5e9881810996e08e1226f561427.png' style=&#034;vertical-align:middle;&#034; width=&#034;14&#034; height=&#034;17&#034; alt=&#034;\vec{A}&#034; title=&#034;\vec{A}&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8cf943f35f95da0c266ec28738154362.png' style=&#034;vertical-align:middle;&#034; width=&#034;13&#034; height=&#034;18&#034; alt=&#034;\vec{C}&#034; title=&#034;\vec{C}&#034; /&gt; y luego hacer el producto escalar del vector &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/69e3966668f4dabe833bedf0903ccb0c.png' style=&#034;vertical-align:middle;&#034; width=&#034;13&#034; height=&#034;17&#034; alt=&#034;\vec{B}&#034; title=&#034;\vec{B}&#034; /&gt; con el vector resultante de la suma anterior. &lt;br/&gt; &lt;br/&gt; La soluci&#243;n que debes obtener es: &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5627fd3bd88e391634d952083d4efeb5.png' style=&#034;vertical-align:middle;&#034; width=&#034;161&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\vec{B}\cdot (\vec{A} + \vec{C}) = -17}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\vec{B}\cdot (\vec{A} + \vec{C}) = -17}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt; &lt;br/&gt; &lt;i&gt;Puedes ver la resoluci&#243;n si haces clic en la siguiente imagen.&lt;/i&gt; &lt;br/&gt;&lt;/p&gt;
&lt;div class='spip_document_1528 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;a href='https://ejercicios-fyq.com/IMG/jpg/ej_7405.jpg' class=&#034;spip_doc_lien mediabox&#034; type=&#034;image/jpeg&#034;&gt; &lt;img src='https://ejercicios-fyq.com/IMG/jpg/ej_7405.jpg' width=&#034;1236&#034; height=&#034;346&#034; alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Descarga el enunciado y la resoluci&#243;n del problema en formato EDICO si lo necesitas&lt;/b&gt;.&lt;/p&gt;
&lt;div class='spip_document_1534 spip_document spip_documents spip_document_file spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt;
&lt;a href=&#034;https://ejercicios-fyq.com/apuntes/descarga.php?file=Ej_7405.edi&#034; class=&#034; spip_doc_lien&#034; title='Zip - ' type=&#034;application/zip&#034;&gt;&lt;img src='https://ejercicios-fyq.com/plugins-dist/medias/prive/vignettes/zip.svg?1772792240' width='64' height='64' alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Vector unitario en la direcci&#243;n de un vector resultante (7207)</title>
		<link>https://ejercicios-fyq.com/Vector-unitario-en-la-direccion-de-un-vector-resultante-7207</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Vector-unitario-en-la-direccion-de-un-vector-resultante-7207</guid>
		<dc:date>2021-06-01T06:12:01Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>
		<dc:subject>M&#243;dulo</dc:subject>

		<description>
&lt;p&gt;Se dan los siguientes vectores , y . Halla un vector unitario en la direcci&#243;n del vector .&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Vectores-dimensiones-y-unidades" rel="directory"&gt;Vectores, dimensiones y unidades&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Modulo" rel="tag"&gt;M&#243;dulo&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Se dan los siguientes vectores &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L120xH21/58ff9ed37ef3822d7635bca376b74589-feff8.png?1732984645' style='vertical-align:middle;' width='120' height='21' alt=&#034;\vec A = 3\vec i - \vec j - 4\ \vec k&#034; title=&#034;\vec A = 3\vec i - \vec j - 4\ \vec k&#034; /&gt; , &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L152xH21/825930aaab03c2f1bc65293a51bb0db9-60eab.png?1732984645' style='vertical-align:middle;' width='152' height='21' alt=&#034;\vec B = -2\ \vec i + 4\ \vec j - 3\ \vec k&#034; title=&#034;\vec B = -2\ \vec i + 4\ \vec j - 3\ \vec k&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L112xH21/4aa0f701a3eb775175b24271f9543f95-29137.png?1732984645' style='vertical-align:middle;' width='112' height='21' alt=&#034;C = \vec i + 2\ \vec j - \vec k&#034; title=&#034;C = \vec i + 2\ \vec j - \vec k&#034; /&gt; . Halla un vector unitario en la direcci&#243;n del vector &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L103xH19/94e59eba04782a7f692f77d1d7c58912-364f4.png?1732984645' style='vertical-align:middle;' width='103' height='19' alt=&#034;3\vec A - 2\vec B +4\vec C&#034; title=&#034;3\vec A - 2\vec B +4\vec C&#034; /&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Puedes empezar por calcular los vectores que luego tienes que sumar: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5eeffddcc9da7517f27eea47bc5a84f6.png' style=&#034;vertical-align:middle;&#034; width=&#034;155&#034; height=&#034;21&#034; alt=&#034;3\vec A = 9\ \vec i - 3\ \vec j - 12\ \vec k&#034; title=&#034;3\vec A = 9\ \vec i - 3\ \vec j - 12\ \vec k&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/203d29e52a54ce288e0902bd7588b4da.png' style=&#034;vertical-align:middle;&#034; width=&#034;160&#034; height=&#034;21&#034; alt=&#034;-2\vec B = 4\ \vec i - 8\ \vec j + 6\ \vec k&#034; title=&#034;-2\vec B = 4\ \vec i - 8\ \vec j + 6\ \vec k&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0a58eaf1f709d51323087d68fd7e08e1.png' style=&#034;vertical-align:middle;&#034; width=&#034;147&#034; height=&#034;21&#034; alt=&#034;4\vec C = 4\ \vec i + 8\ \vec j - 4\ \vec k&#034; title=&#034;4\vec C = 4\ \vec i + 8\ \vec j - 4\ \vec k&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si sumas los tres vectores anteriores obtienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/374ac439d5db61f0800f56b00a0965c5.png' style=&#034;vertical-align:middle;&#034; width=&#034;173&#034; height=&#034;20&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{\vec R = 17\ \vec i - 3\ \vec j - 10\ \vec k}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{\vec R = 17\ \vec i - 3\ \vec j - 10\ \vec k}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Calcula el m&#243;dulo del vector resultante: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7d95aaf73e1b707fbff1dce69be2177b.png' style=&#034;vertical-align:middle;&#034; width=&#034;212&#034; height=&#034;17&#034; alt=&#034;R = \sqrt{17^2 + 3^2 + 10^2} = \color[RGB]{0,112,192}{\bf 19.95}}&#034; title=&#034;R = \sqrt{17^2 + 3^2 + 10^2} = \color[RGB]{0,112,192}{\bf 19.95}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El vector unitario pedido es el cociente entre el vector resultante y su m&#243;dulo: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e4b8b6f6646f56a31ebfb37261167ee2.png' style=&#034;vertical-align:middle;&#034; width=&#034;558&#034; height=&#034;40&#034; alt=&#034;\vec{u}_R = \frac{\vec R}{R} = \frac{17}{19.95}\ \vec i - \frac{3}{19.95}\ \vec j - \frac{10}{19.95}\ \vec k\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{\vec{u}_R = 0.85\ \vec i - 0.15\ \vec j - 0.50\ \vec k}}}&#034; title=&#034;\vec{u}_R = \frac{\vec R}{R} = \frac{17}{19.95}\ \vec i - \frac{3}{19.95}\ \vec j - \frac{10}{19.95}\ \vec k\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{\vec{u}_R = 0.85\ \vec i - 0.15\ \vec j - 0.50\ \vec k}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Fuerza resultante de cuatro fuerzas concurrentes (6855)</title>
		<link>https://ejercicios-fyq.com/Fuerza-resultante-de-cuatro-fuerzas-concurrentes-6855</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Fuerza-resultante-de-cuatro-fuerzas-concurrentes-6855</guid>
		<dc:date>2020-11-02T07:08:15Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Fuerza resultante</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>

		<description>
&lt;p&gt;Halla la fuerza que equilibra el sistema formado por las fuerzas; en el primer cuadrante, formando un &#225;ngulo de con la direcci&#243;n positiva del eje X, hacia el sureste, formando por debajo del eje X, en la direcci&#243;n sur, en el tercer cuadrante, formando con la direcci&#243;n negativa del eje X.&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Dinamica-1-o-Bach" rel="directory"&gt;Din&#225;mica (1.&#186; Bach)&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Fuerza-resultante" rel="tag"&gt;Fuerza resultante&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Halla la fuerza que equilibra el sistema formado por las fuerzas; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L68xH15/71a3481cad63c2e70fa81de5d82306b9-509ad.png?1732990868' style='vertical-align:middle;' width='68' height='15' alt=&#034;\ce{F_1 = 50 N}&#034; title=&#034;\ce{F_1 = 50 N}&#034; /&gt; en el primer cuadrante, formando un &#225;ngulo de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L22xH13/88b8cbd61c84b1df8acec397df92fc5e-69dd9.png?1732990868' style='vertical-align:middle;' width='22' height='13' alt=&#034;62^o&#034; title=&#034;62^o&#034; /&gt; con la direcci&#243;n positiva del eje X, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L76xH15/3c4356f1b96df716421ae88c51ba88f9-a3b39.png?1732990868' style='vertical-align:middle;' width='76' height='15' alt=&#034;\ce{F_2 = 180 N}&#034; title=&#034;\ce{F_2 = 180 N}&#034; /&gt; hacia el sureste, formando &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L22xH13/1ae214a7bf42fa3205839ff84e5ddc1c-e5d3e.png?1732990868' style='vertical-align:middle;' width='22' height='13' alt=&#034;23^o&#034; title=&#034;23^o&#034; /&gt; por debajo del eje X, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L76xH16/4a80abb744e6a01529e18661a7c67202-d472f.png?1732990868' style='vertical-align:middle;' width='76' height='16' alt=&#034;\ce{F_3 = 130 N}&#034; title=&#034;\ce{F_3 = 130 N}&#034; /&gt; en la direcci&#243;n sur, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L76xH15/fec05aeac7d3853412752eb59d32f52b-84421.png?1732990868' style='vertical-align:middle;' width='76' height='15' alt=&#034;\ce{F_4 = 125 N}&#034; title=&#034;\ce{F_4 = 125 N}&#034; /&gt; en el tercer cuadrante, formando &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L22xH13/f13e28ce829b262174de9bc35f8a8b7d-c9043.png?1732956198' style='vertical-align:middle;' width='22' height='13' alt=&#034;25^o&#034; title=&#034;25^o&#034; /&gt; con la direcci&#243;n negativa del eje X.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Es muy aconsejable hacer un esquema de la situaci&#243;n descrita en el enunciado para hacerte una idea de c&#243;mo est&#225;n situadas las fuerzas: &lt;br/&gt;&lt;/p&gt;
&lt;div class='spip_document_1221 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;img src='https://ejercicios-fyq.com/IMG/png/ej_6855.png' width=&#034;640&#034; height=&#034;435&#034; alt='' /&gt;
&lt;/figure&gt;
&lt;/div&gt; &lt;p&gt;&lt;i&gt;(Si clicas en la miniatura podr&#225;s ver el esquema con m&#225;s detalle)&lt;/i&gt; &lt;br/&gt; &lt;br/&gt; Ahora debes descomponer los vectores en sus componentes sobre los ejes marcados, para poder hacer la suma de esas componentes y obtener el vector resultante: &lt;br/&gt;&lt;/p&gt;
&lt;div class='spip_document_1222 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;img src='https://ejercicios-fyq.com/IMG/png/ej_6855_2.png' width=&#034;635&#034; height=&#034;430&#034; alt='' /&gt;
&lt;/figure&gt;
&lt;/div&gt; &lt;p&gt;&lt;i&gt;(Si clicas en la miniatura podr&#225;s ver el esquema con m&#225;s detalle)&lt;/i&gt; &lt;br/&gt; Para obtener las componentes dibujadas debes tener cuidado con los &#225;ngulos. Observa c&#243;mo est&#225;n considerados en las siguientes ecuaciones: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/01ed8f5639064c70bb4b52fa35614844.png' style=&#034;vertical-align:middle;&#034; width=&#034;197&#034; height=&#034;19&#034; alt=&#034;\vec{F}_{1x} = 50\cdot cos\ 62 = \color[RGB]{0,112,192}{\bm{23.47\ \vec i}}&#034; title=&#034;\vec{F}_{1x} = 50\cdot cos\ 62 = \color[RGB]{0,112,192}{\bm{23.47\ \vec i}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/47f8b658e1c728c68932600997a64ab3.png' style=&#034;vertical-align:middle;&#034; width=&#034;200&#034; height=&#034;22&#034; alt=&#034;\vec{F}_{1y} = 50\cdot sen\ 62 = \color[RGB]{0,112,192}{\bm{44.15\ \vec j}}&#034; title=&#034;\vec{F}_{1y} = 50\cdot sen\ 62 = \color[RGB]{0,112,192}{\bm{44.15\ \vec j}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ac73bcd25a03cbc7fa5701e110bd1c48.png' style=&#034;vertical-align:middle;&#034; width=&#034;263&#034; height=&#034;22&#034; alt=&#034;\vec{F}_{2x} = 180\cdot cos\ (360 - 23) = \color[RGB]{0,112,192}{\bm{165.7\ \vec i}}&#034; title=&#034;\vec{F}_{2x} = 180\cdot cos\ (360 - 23) = \color[RGB]{0,112,192}{\bm{165.7\ \vec i}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/99f386d0e77aa35de7ed7fb2b12327a1.png' style=&#034;vertical-align:middle;&#034; width=&#034;280&#034; height=&#034;22&#034; alt=&#034;\vec{F}_{2y} = 180\cdot sen\ (360 - 23) = \color[RGB]{0,112,192}{\bm{- 70.33\ \vec j}}&#034; title=&#034;\vec{F}_{2y} = 180\cdot sen\ (360 - 23) = \color[RGB]{0,112,192}{\bm{- 70.33\ \vec j}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a57545fc8d3d3b861aeae4b8ddb4052c.png' style=&#034;vertical-align:middle;&#034; width=&#034;277&#034; height=&#034;22&#034; alt=&#034;\vec{F}_{4x} = 125\cdot cos\ (180 + 25) = \color[RGB]{0,112,192}{\bm{-113.3\ \vec i}}&#034; title=&#034;\vec{F}_{4x} = 125\cdot cos\ (180 + 25) = \color[RGB]{0,112,192}{\bm{-113.3\ \vec i}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d24130dd3d12f8fdf73718579f30666b.png' style=&#034;vertical-align:middle;&#034; width=&#034;280&#034; height=&#034;22&#034; alt=&#034;\vec{F}_{4y} = 125\cdot sen\ (180 + 25) = \color[RGB]{0,112,192}{\bm{- 52.83\ \vec j}}&#034; title=&#034;\vec{F}_{4y} = 125\cdot sen\ (180 + 25) = \color[RGB]{0,112,192}{\bm{- 52.83\ \vec j}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Ya solo tienes que sumar todas las componentes, sin olvidar la fuerza &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/46613afe81f59ab853fd630b80f29c59.png' style=&#034;vertical-align:middle;&#034; width=&#034;16&#034; height=&#034;20&#034; alt=&#034;\vec {F}_3&#034; title=&#034;\vec {F}_3&#034; /&gt;, y obtienes la fuerza resultante: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/47259066a2ad5d4999f48dec82d4db26.png' style=&#034;vertical-align:middle;&#034; width=&#034;593&#034; height=&#034;21&#034; alt=&#034;F_R = (23.47 + 165.7 - 113.3)\ \vec i + (44.15 - 70.33 - 130 - 52.83)\ \vec j = \color[RGB]{2,112,20}{\bm{75.9\ \vec{i} - 209\ \vec{j}}}&#034; title=&#034;F_R = (23.47 + 165.7 - 113.3)\ \vec i + (44.15 - 70.33 - 130 - 52.83)\ \vec j = \color[RGB]{2,112,20}{\bm{75.9\ \vec{i} - 209\ \vec{j}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La fuerza que equilibra la fuerza resultante obtenida es justo la fuerza opuesta a la calculada: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/58dc50036f89b8670cbf2c02dbbb944c.png' style=&#034;vertical-align:middle;&#034; width=&#034;172&#034; height=&#034;29&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\vec{F} = -75.9\ \vec i + 209\ \vec j}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\vec{F} = -75.9\ \vec i + 209\ \vec j}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>&#193;ngulos de dos fuerzas para que la resultante solo tenga componente x (6660)</title>
		<link>https://ejercicios-fyq.com/Angulos-de-dos-fuerzas-para-que-la-resultante-solo-tenga-componente-x-6660</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Angulos-de-dos-fuerzas-para-que-la-resultante-solo-tenga-componente-x-6660</guid>
		<dc:date>2020-06-18T12:34:10Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Fuerza resultante</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>

		<description>
&lt;p&gt;Si , determina los &#225;ngulos y de tal forma que la fuerza resultante est&#233; dirigida a lo largo del eje positivo x y tenga una magnitud de 88.98 N.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Dinamica-1-o-Bach" rel="directory"&gt;Din&#225;mica (1.&#186; Bach)&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Fuerza-resultante" rel="tag"&gt;Fuerza resultante&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Si &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L121xH16/e88ddf12cf0f672189db526aedd46e32-6aebd.png?1733018048' style='vertical-align:middle;' width='121' height='16' alt=&#034;F_1 = F_2 = 30\ lbf&#034; title=&#034;F_1 = F_2 = 30\ lbf&#034; /&gt; , determina los &#225;ngulos &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L17xH40/2554a2bb846cffd697389e5dc8912759-c6c87.png?1732958881' style='vertical-align:middle;' width='17' height='40' alt=&#034;\theta&#034; title=&#034;\theta&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L10xH16/1ed346930917426bc46d41e22cc525ec-ef9cf.png?1733014986' style='vertical-align:middle;' width='10' height='16' alt=&#034;\phi&#034; title=&#034;\phi&#034; /&gt; de tal forma que la fuerza resultante est&#233; dirigida a lo largo del eje positivo &lt;i&gt;x&lt;/i&gt; y tenga una magnitud de 88.98 N.&lt;/p&gt;
&lt;div class='spip_document_1132 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L360xH326/ej_6660-2d6e4.jpg?1758396239' width='360' height='326' alt='' /&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/math&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Al tener el mismo valor ambas fuerzas puedes denotarlas con la misma letra. Si las expresas en forma vectorial y teniendo en cuenta las componentes obtienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8731c796e046d23b8a59dcc30bfbb701.png' style=&#034;vertical-align:middle;&#034; width=&#034;208&#034; height=&#034;21&#034; alt=&#034;\vec F_1 = F\cdot cos\ \theta\ \vec i + F\cdot sen\ \theta\ \vec j&#034; title=&#034;\vec F_1 = F\cdot cos\ \theta\ \vec i + F\cdot sen\ \theta\ \vec j&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/75eef5109fafc83fb9767958f04bfe98.png' style=&#034;vertical-align:middle;&#034; width=&#034;263&#034; height=&#034;22&#034; alt=&#034;\vec F_2 = F\cdot cos\ (-\phi)\ \vec i + F\cdot sen\ (-\phi)\ \vec j&#034; title=&#034;\vec F_2 = F\cdot cos\ (-\phi)\ \vec i + F\cdot sen\ (-\phi)\ \vec j&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La primera condici&#243;n puede ser que la suma de las componentes del eje &lt;i&gt;y&lt;/i&gt; tiene que ser cero o, lo que es lo mismo, que ambas componentes tienen que tener el mismo m&#243;dulo: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6ad952127af08970603635585a18f3f2.png' style=&#034;vertical-align:middle;&#034; width=&#034;446&#034; height=&#034;25&#034; alt=&#034;\cancel{F}\cdot sen\ \theta = \cancel{F}\cdot sen\ (-\phi)\ \to\ sen\ \theta = sen\ (-\phi})\ \to\ \fbox{\color[RGB]{2,112,20}{\bm{\theta = -\phi}}}&#034; title=&#034;\cancel{F}\cdot sen\ \theta = \cancel{F}\cdot sen\ (-\phi)\ \to\ sen\ \theta = sen\ (-\phi})\ \to\ \fbox{\color[RGB]{2,112,20}{\bm{\theta = -\phi}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La segunda condici&#243;n ser&#225; que la suma de las componentes en la direcci&#243;n &lt;i&gt;x&lt;/i&gt; tiene que ser igual al valor indicado en el enunciado: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0b31a8615a4f224041bbc3ba5a7ad238.png' style=&#034;vertical-align:middle;&#034; width=&#034;555&#034; height=&#034;36&#034; alt=&#034;F\cdot cos\ \theta + F\cdot cos\ (-\phi) = 88.98\ to\ 2F\cdot cos\ \theta = 88.98\ \to\ \color[RGB]{0,112,192}{\bm{\theta = arccos\ \frac{88.98}{2F}}}&#034; title=&#034;F\cdot cos\ \theta + F\cdot cos\ (-\phi) = 88.98\ to\ 2F\cdot cos\ \theta = 88.98\ \to\ \color[RGB]{0,112,192}{\bm{\theta = arccos\ \frac{88.98}{2F}}}&#034; /&gt; &lt;br/&gt; Ahora debes tener mucho ciudado con las unidades porque el valor de la fuerza viene dado en lbf y no en N: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a6871a03f31bf99f745d3d91b259e9b4.png' style=&#034;vertical-align:middle;&#034; width=&#034;242&#034; height=&#034;53&#034; alt=&#034;30\ \cancel{lbf}\cdot \frac{4.45\ N}{1\ \cancel{lbf}} = 133.5\ N&#034; title=&#034;30\ \cancel{lbf}\cdot \frac{4.45\ N}{1\ \cancel{lbf}} = 133.5\ N&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Ya puedes calcular el &#225;ngulo: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c68d718f850e7de1cef5965871a55240.png' style=&#034;vertical-align:middle;&#034; width=&#034;245&#034; height=&#034;40&#034; alt=&#034;\theta = arccos\ \frac{88.98\ \cancel{N}}{(2\cdot 133.5)\ \cancel{N}} = \fbox{\color[RGB]{192,0,0}{\bf 70.5^o}}&#034; title=&#034;\theta = arccos\ \frac{88.98\ \cancel{N}}{(2\cdot 133.5)\ \cancel{N}} = \fbox{\color[RGB]{192,0,0}{\bf 70.5^o}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Ambos &#225;ngulos tienen el mismo valor pero uno por encima del eje x y el otro por debajo.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Fuerzas con las que dos tractores remolcan un tronco para que la resultante sea horizontal (6659)</title>
		<link>https://ejercicios-fyq.com/Fuerzas-con-las-que-dos-tractores-remolcan-un-tronco-para-que-la-resultante-sea</link>
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		<dc:date>2020-06-18T10:44:07Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Vectores</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>

		<description>
&lt;p&gt;Un tronco de madera es remolcado por dos tractores A y B. Determina las magnitudes de las dos fuerzas de remolque y , si se requiere que la fuerza resultante tenga una magnitud de 10 kN y est&#233; dirigida a lo largo del eje horizontal. Considera que la fuerza forma un &#225;ngulo de con la horizontal y por encima de ella y forma un &#225;ngulo de y por debajo de la horizontal.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Dinamica-1-o-Bach" rel="directory"&gt;Din&#225;mica (1.&#186; Bach)&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Vectores" rel="tag"&gt;Vectores&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Un tronco de madera es remolcado por dos tractores A y B. Determina las magnitudes de las dos fuerzas de remolque &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L20xH15/18a71fd0d3ff0cbdbee63523af19659c-25482.png?1733064707' style='vertical-align:middle;' width='20' height='15' alt=&#034;F _A&#034; title=&#034;F _A&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L19xH15/459533ae0b092252bad3a64c21aeb79d-a1675.png?1733064707' style='vertical-align:middle;' width='19' height='15' alt=&#034;F _B&#034; title=&#034;F _B&#034; /&gt;, si se requiere que la fuerza resultante tenga una magnitud de 10 kN y est&#233; dirigida a lo largo del eje horizontal. Considera que la fuerza &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L20xH15/18a71fd0d3ff0cbdbee63523af19659c-25482.png?1733064707' style='vertical-align:middle;' width='20' height='15' alt=&#034;F _A&#034; title=&#034;F _A&#034; /&gt; forma un &#225;ngulo de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L22xH13/f630d7bac0dce45f77e1c0c9e1dbf67e-1bd08.png?1732952054' style='vertical-align:middle;' width='22' height='13' alt=&#034;30 ^o&#034; title=&#034;30 ^o&#034; /&gt; con la horizontal y por encima de ella y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L19xH15/459533ae0b092252bad3a64c21aeb79d-a1675.png?1733064707' style='vertical-align:middle;' width='19' height='15' alt=&#034;F _B&#034; title=&#034;F _B&#034; /&gt; forma un &#225;ngulo de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L21xH13/fe32f3d7e20b36a2700e8e7d70a28223-7adbd.png?1733064707' style='vertical-align:middle;' width='21' height='13' alt=&#034;15 ^o&#034; title=&#034;15 ^o&#034; /&gt; y por debajo de la horizontal.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;El primer paso a dar es descomponer las fuerzas de los tractores A y B en sus componentes y escribirlas en forma vectorial: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ae831ea71c8ab3042be4726a8af02d11.png' style=&#034;vertical-align:middle;&#034; width=&#034;414&#034; height=&#034;21&#034; alt=&#034;\vec F_A = F_A\cdot cos\ 30\ \vec i + F_A\cdot sen\ 30\ \vec j = \color[RGB]{0,112,192}{\bm{0.87F_A\ \vec i + 0.5F_A\ \vec j}}&#034; title=&#034;\vec F_A = F_A\cdot cos\ 30\ \vec i + F_A\cdot sen\ 30\ \vec j = \color[RGB]{0,112,192}{\bm{0.87F_A\ \vec i + 0.5F_A\ \vec j}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/36f4f2e460c2aa397a443b65233b6f2a.png' style=&#034;vertical-align:middle;&#034; width=&#034;478&#034; height=&#034;22&#034; alt=&#034;\vec F_B = F_B\cdot cos\ (-15)\ \vec i + F_A\cdot sen\ (-15)\ \vec j = \color[RGB]{0,112,192}{\bm{0.97F_B\ \vec i - 0.26F_B\ \vec j}}&#034; title=&#034;\vec F_B = F_B\cdot cos\ (-15)\ \vec i + F_A\cdot sen\ (-15)\ \vec j = \color[RGB]{0,112,192}{\bm{0.97F_B\ \vec i - 0.26F_B\ \vec j}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La suma de las componentes verticales tiene que ser cero para cumplir con la condici&#243;n dada en el enunciado. Adem&#225;s, la suma de las componentes horizontales ha de ser igual a 10 kN para cumplir con la segunda condici&#243;n. Obtienes un sistema de dos ecuaciones con dos inc&#243;gnitas que puedes resolver por sustituci&#243;n:&lt;/p&gt;
&lt;p&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/2075ea60a1663d2d09cfa5ce17c639a0.png' style=&#034;vertical-align:middle;&#034; width=&#034;440&#034; height=&#034;40&#034; alt=&#034;\left 0.5F_A - 0.26F_B = 0\ \atop 0.87F_A + 0.95F_B = 10 \right\}\ \to\ F_A = \frac{0.26F_B}{0.5}\ \to\ \color[RGB]{0,112,192}{\bm{F_A = 0.52F_B}}&#034; title=&#034;\left 0.5F_A - 0.26F_B = 0\ \atop 0.87F_A + 0.95F_B = 10 \right\}\ \to\ F_A = \frac{0.26F_B}{0.5}\ \to\ \color[RGB]{0,112,192}{\bm{F_A = 0.52F_B}}&#034; /&gt;&lt;/p&gt;
&lt;p&gt;Ahora sustituyes en la segunda ecuaci&#243;n y calculas: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/361765947d395ef12d10e0ec3a7556a1.png' style=&#034;vertical-align:middle;&#034; width=&#034;421&#034; height=&#034;35&#034; alt=&#034;0.87\cdot 0.52F_B + 0.97F_B = 10\ \to\ F_B = \frac{10\ kN}{1.42} = \fbox{\color[RGB]{192,0,0}{\bf 7.04\ kN}}&#034; title=&#034;0.87\cdot 0.52F_B + 0.97F_B = 10\ \to\ F_B = \frac{10\ kN}{1.42} = \fbox{\color[RGB]{192,0,0}{\bf 7.04\ kN}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; El valor de la otra fuerza es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/754a810996ca2b63653688e5f38dcf3c.png' style=&#034;vertical-align:middle;&#034; width=&#034;308&#034; height=&#034;21&#034; alt=&#034;F_A = 0.52F_B = 0.52\cdot 7.04\ kN = \fbox{\color[RGB]{192,0,0}{\bf 3.66\ kN}}&#034; title=&#034;F_A = 0.52F_B = 0.52\cdot 7.04\ kN = \fbox{\color[RGB]{192,0,0}{\bf 3.66\ kN}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Operaciones con vectores (5945)</title>
		<link>https://ejercicios-fyq.com/Operaciones-con-vectores-5945</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Operaciones-con-vectores-5945</guid>
		<dc:date>2019-10-31T07:34:57Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Algebra de vectores</dc:subject>
		<dc:subject>Producto escalar</dc:subject>
		<dc:subject>Producto vectorial</dc:subject>

		<description>
&lt;p&gt;Para los siguientes vectores: ; ; y , determina: &lt;br class='autobr' /&gt;
a) ; ; ; . &lt;br class='autobr' /&gt;
b) La magnitud de cada vector y los &#225;ngulos que forman con los ejes x , y , z. &lt;br class='autobr' /&gt;
c) Los productos escalares: ; ; ; . &lt;br class='autobr' /&gt;
d) Los productos vectoriales: ; ; ;&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Vectores-dimensiones-y-unidades" rel="directory"&gt;Vectores, dimensiones y unidades&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Algebra-de-vectores-579" rel="tag"&gt;Algebra de vectores&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Producto-escalar" rel="tag"&gt;Producto escalar&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Producto-vectorial" rel="tag"&gt;Producto vectorial&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Para los siguientes vectores: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L110xH22/4d5fd0728de7b952ad6396651f52b1d4-13f40.png?1732972934' style='vertical-align:middle;' width='110' height='22' alt=&#034;\vec A = (4, -1, -6)&#034; title=&#034;\vec A = (4, -1, -6)&#034; /&gt;; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L99xH22/5d46a9543ed1495fbb6d2cfa0d5da63f-30e07.png?1732972934' style='vertical-align:middle;' width='99' height='22' alt=&#034;\vec B = (5, 7, -2)&#034; title=&#034;\vec B = (5, 7, -2)&#034; /&gt;; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L111xH22/534959000799ffe85b3c71465c929e70-56deb.png?1732972934' style='vertical-align:middle;' width='111' height='22' alt=&#034;\vec C = (-8, -5, 2)&#034; title=&#034;\vec C = (-8, -5, 2)&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L99xH22/95fce6e993b879408ddeb89f9c640140-00e63.png?1732972934' style='vertical-align:middle;' width='99' height='22' alt=&#034;\vec D = (9, -4, 0)&#034; title=&#034;\vec D = (9, -4, 0)&#034; /&gt;, determina:&lt;/p&gt;
&lt;p&gt;a) &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L56xH22/bb8f4cdd437dff8d7e784174efec6b24-8f817.png?1732972934' style='vertical-align:middle;' width='56' height='22' alt=&#034;(\vec A + \vec B)&#034; title=&#034;(\vec A + \vec B)&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L56xH22/84f2c4816ae99ec8f9deae9e86f83a1f-a0134.png?1732972934' style='vertical-align:middle;' width='56' height='22' alt=&#034;(\vec A - \vec B)&#034; title=&#034;(\vec A - \vec B)&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L57xH22/eb81786c8e324dda31f15a3f9bb3d4ca-d9cf3.png?1732972934' style='vertical-align:middle;' width='57' height='22' alt=&#034;(\vec D + \vec C)&#034; title=&#034;(\vec D + \vec C)&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L56xH22/fa6a40f67e7f1b564121cc3699fe4dfa-23014.png?1732972934' style='vertical-align:middle;' width='56' height='22' alt=&#034;(\vec A - \vec D)&#034; title=&#034;(\vec A - \vec D)&#034; /&gt;.&lt;/p&gt;
&lt;p&gt;b) La magnitud de cada vector y los &#225;ngulos que forman con los ejes x , y , z.&lt;/p&gt;
&lt;p&gt;c) Los productos escalares: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L38xH17/80066fbf13c9ca5f7e55d20e66b20272-a8efe.png?1732972934' style='vertical-align:middle;' width='38' height='17' alt=&#034;\vec A\cdot \vec B&#034; title=&#034;\vec A\cdot \vec B&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L50xH23/5c9208ed341fce66d32e3c8a73d702c7-60e1f.png?1732972934' style='vertical-align:middle;' width='50' height='23' alt=&#034;\vec D\cdot \vec C&#034; title=&#034;\vec D\cdot \vec C&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L39xH18/4569c5f3b0b86507ef880d2b287fc1ee-bdb90.png?1732972934' style='vertical-align:middle;' width='39' height='18' alt=&#034;\vec B\cdot \vec C&#034; title=&#034;\vec B\cdot \vec C&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L38xH17/80774e2d9481c886550618872f9b9de4-19a4f.png?1732972934' style='vertical-align:middle;' width='38' height='17' alt=&#034;\vec B\cdot \vec D&#034; title=&#034;\vec B\cdot \vec D&#034; /&gt;.&lt;/p&gt;
&lt;p&gt;d) Los productos vectoriales: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L46xH17/79b1af1c7b52d5833b723aa2380387fe-63eae.png?1732972934' style='vertical-align:middle;' width='46' height='17' alt=&#034;\vec A\times \vec B&#034; title=&#034;\vec A\times \vec B&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L47xH18/1fd756e4258115369ddc66bb6ebbbbb5-aa99a.png?1732972934' style='vertical-align:middle;' width='47' height='18' alt=&#034;\vec D\times \vec C&#034; title=&#034;\vec D\times \vec C&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L47xH18/08ef9278202b2f7a4729ea5bb013313e-1b91c.png?1732972934' style='vertical-align:middle;' width='47' height='18' alt=&#034;\vec B\times \vec C&#034; title=&#034;\vec B\times \vec C&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L47xH17/d37d21a98a8784d0dd55b7e4b200f0cb-63798.png?1732972934' style='vertical-align:middle;' width='47' height='17' alt=&#034;\vec B\times \vec D&#034; title=&#034;\vec B\times \vec D&#034; /&gt;&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) Para obtener las sumas y diferencias entre vectores tan solo debemos hacer esas sumas o diferencias entre sus componentes: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ac8082c0b44eaa057069071c32747506.png' style=&#034;vertical-align:middle;&#034; width=&#034;379&#034; height=&#034;26&#034; alt=&#034;(\vec A + \vec B) = [(4 + 5), (-1 + 7), (-6 - 2)] = \fbox{\color[RGB]{192,0,0}{\bf (9, 6, -8)}}&#034; title=&#034;(\vec A + \vec B) = [(4 + 5), (-1 + 7), (-6 - 2)] = \fbox{\color[RGB]{192,0,0}{\bf (9, 6, -8)}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/67cae2aed0c2c131eb3b3ab38de23cf1.png' style=&#034;vertical-align:middle;&#034; width=&#034;392&#034; height=&#034;26&#034; alt=&#034;(\vec A - \vec B) = [(4 - 5), (-1 - 7), (-6 + 2)] = \fbox{\color[RGB]{192,0,0}{\bf (-1, -8, -4)}}&#034; title=&#034;(\vec A - \vec B) = [(4 - 5), (-1 - 7), (-6 + 2)] = \fbox{\color[RGB]{192,0,0}{\bf (-1, -8, -4)}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Operando del mismo modo, las otras dos operaciones resultan: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a3106a55bbf04e55c911ecb16aa55156.png' style=&#034;vertical-align:middle;&#034; width=&#034;162&#034; height=&#034;26&#034; alt=&#034;(\vec D + \vec C) = \fbox{\color[RGB]{192,0,0}{\bf (1, -9, 2)}}&#034; title=&#034;(\vec D + \vec C) = \fbox{\color[RGB]{192,0,0}{\bf (1, -9, 2)}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8d9e045c2d6d1f952334c3cd66d31a35.png' style=&#034;vertical-align:middle;&#034; width=&#034;168&#034; height=&#034;26&#034; alt=&#034;(\vec A - \vec D) = \fbox{\color[RGB]{192,0,0}{\bf (-5, 3, -6)}}&#034; title=&#034;(\vec A - \vec D) = \fbox{\color[RGB]{192,0,0}{\bf (-5, 3, -6)}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Los cosenos directores de los vectores se obtienen al hacer el cociente entre cada una de las componentes del vector y su m&#243;dulo. Lo hacemos para el vector &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1ca591d80112a79abcdfc90b3c732d6f.png' style=&#034;vertical-align:middle;&#034; width=&#034;18&#034; height=&#034;22&#034; alt=&#034;\vec A &#034; title=&#034;\vec A &#034; /&gt; y luego ponemos los resultados para el resto de los vectores. En primer lugar calculamos el m&#243;dulo del vector: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7d85d2fc34b83ce5c053fecdd1eb90a8.png' style=&#034;vertical-align:middle;&#034; width=&#034;243&#034; height=&#034;21&#034; alt=&#034;A = \sqrt{4^2 + (-1)^2 + (-6)^2} = \color[RGB]{2,112,10}{\bm{\sqrt{53}}}&#034; title=&#034;A = \sqrt{4^2 + (-1)^2 + (-6)^2} = \color[RGB]{2,112,10}{\bm{\sqrt{53}}}&#034; /&gt;. &lt;br/&gt; &lt;br/&gt; Eje X: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/83a96ea4855bf20991b56b232febdda5.png' style=&#034;vertical-align:middle;&#034; width=&#034;309&#034; height=&#034;39&#034; alt=&#034;cos\ \alpha = \frac{A_x}{A}\ \to\ \alpha = arccos\ \frac{4}{\sqrt{53}} = \fbox{\color[RGB]{192,0,0}{\bm{56.7^o}}}&#034; title=&#034;cos\ \alpha = \frac{A_x}{A}\ \to\ \alpha = arccos\ \frac{4}{\sqrt{53}} = \fbox{\color[RGB]{192,0,0}{\bm{56.7^o}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Eje Y: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/412d4cd4befca6552babd3231db9e613.png' style=&#034;vertical-align:middle;&#034; width=&#034;309&#034; height=&#034;39&#034; alt=&#034;cos\ \beta = \frac{A_y}{A}\ \to\ \beta = arccos\ \frac{-1}{\sqrt{53}} = \fbox{\color[RGB]{192,0,0}{\bm{97.9^o}}}&#034; title=&#034;cos\ \beta = \frac{A_y}{A}\ \to\ \beta = arccos\ \frac{-1}{\sqrt{53}} = \fbox{\color[RGB]{192,0,0}{\bm{97.9^o}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Eje Z: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0e3998100457f6f7bce1eed27e83f93b.png' style=&#034;vertical-align:middle;&#034; width=&#034;316&#034; height=&#034;39&#034; alt=&#034;cos\ \gamma = \frac{A_y}{A}\ \to\ \gamma = arccos\ \frac{-6}{\sqrt{53}} = \fbox{\color[RGB]{192,0,0}{\bm{145.5^o}}}&#034; title=&#034;cos\ \gamma = \frac{A_y}{A}\ \to\ \gamma = arccos\ \frac{-6}{\sqrt{53}} = \fbox{\color[RGB]{192,0,0}{\bm{145.5^o}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Para el vector &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/2d2cfe9ec6b10171498084f20a44241e.png' style=&#034;vertical-align:middle;&#034; width=&#034;22&#034; height=&#034;52&#034; alt=&#034;\vec B &#034; title=&#034;\vec B &#034; /&gt;, su m&#243;dulo es &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a7d3575e4576cc71eb1e3e7513f86168.png' style=&#034;vertical-align:middle;&#034; width=&#034;65&#034; height=&#034;17&#034; alt=&#034;B = \sqrt{78}&#034; title=&#034;B = \sqrt{78}&#034; /&gt;: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e5d4fb14837e6ba172f55a4ed4a57dbf.png' style=&#034;vertical-align:middle;&#034; width=&#034;286&#034; height=&#034;25&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\alpha = 55.5^o\ ;\ \beta = 37.6^o\ ;\ \gamma = 103.1^o}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\alpha = 55.5^o\ ;\ \beta = 37.6^o\ ;\ \gamma = 103.1^o}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;br/&gt; Para el vector &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/386195f68d0457b24be169d3e80f9421.png' style=&#034;vertical-align:middle;&#034; width=&#034;22&#034; height=&#034;52&#034; alt=&#034;\vec C&#034; title=&#034;\vec C&#034; /&gt;, su m&#243;dulo es &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/75a681e8f032d7947ac3eacf46326f2a.png' style=&#034;vertical-align:middle;&#034; width=&#034;65&#034; height=&#034;17&#034; alt=&#034;C = \sqrt{93}&#034; title=&#034;C = \sqrt{93}&#034; /&gt;: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4f0be73dbb38564e0affa388bff42ff2.png' style=&#034;vertical-align:middle;&#034; width=&#034;267&#034; height=&#034;25&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\alpha = 146^o\ ;\ \beta = 121.2^o\ ;\ \gamma = 78^o}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\alpha = 146^o\ ;\ \beta = 121.2^o\ ;\ \gamma = 78^o}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Para el vector &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bce853cae5198d3271847372ea37de4e.png' style=&#034;vertical-align:middle;&#034; width=&#034;23&#034; height=&#034;52&#034; alt=&#034;\vec D&#034; title=&#034;\vec D&#034; /&gt;, su m&#243;dulo es &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0a648c93e5ab8ec60f50403710567850.png' style=&#034;vertical-align:middle;&#034; width=&#034;75&#034; height=&#034;17&#034; alt=&#034;D = \sqrt{117}&#034; title=&#034;D = \sqrt{117}&#034; /&gt;: &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/54173bac29bd2add1899b218c1812142.png' style=&#034;vertical-align:middle;&#034; width=&#034;271&#034; height=&#034;25&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\alpha = 33.7^o\ ;\ \beta = 111.7^o\ ;\ \gamma = 90^o}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{\alpha = 33.7^o\ ;\ \beta = 111.7^o\ ;\ \gamma = 90^o}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) El producto escalar de dos vectores es un n&#250;mero y se obtiene multiplicando las componentes entre s&#237;. Lo hacemos para el primer caso y luego ponemos el resultado para el resto de operaciones: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/fe7c49d1fd09e57703d8259cac113457.png' style=&#034;vertical-align:middle;&#034; width=&#034;287&#034; height=&#034;22&#034; alt=&#034;\vec A\cdot \vec B = (A_x\cdot B_x) + (A_y\cdot B_y) + (A_z\cdot B_z)&#034; title=&#034;\vec A\cdot \vec B = (A_x\cdot B_x) + (A_y\cdot B_y) + (A_z\cdot B_z)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3eac7efe07955216dc0dfa903e7c8112.png' style=&#034;vertical-align:middle;&#034; width=&#034;321&#034; height=&#034;22&#034; alt=&#034;\vec A\cdot \vec B = (4\cdot 5) + (-1\cdot 7) + [-6\cdot (-2)] = \fbox{\color[RGB]{192,0,0}{\bf 25}}&#034; title=&#034;\vec A\cdot \vec B = (4\cdot 5) + (-1\cdot 7) + [-6\cdot (-2)] = \fbox{\color[RGB]{192,0,0}{\bf 25}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/08a6a379b70b051dbf7deae6dabbb052.png' style=&#034;vertical-align:middle;&#034; width=&#034;349&#034; height=&#034;22&#034; alt=&#034;\vec D\cdot \vec C = [9\cdot (-8)] + [(-4)\cdot (-5)] + (0\cdot 2) = \fbox{\color[RGB]{192,0,0}{\bf -52}}&#034; title=&#034;\vec D\cdot \vec C = [9\cdot (-8)] + [(-4)\cdot (-5)] + (0\cdot 2) = \fbox{\color[RGB]{192,0,0}{\bf -52}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d4cec85fd8faaaae005d1fd41c2e93b5.png' style=&#034;vertical-align:middle;&#034; width=&#034;337&#034; height=&#034;22&#034; alt=&#034;\vec B\cdot \vec C = [5\cdot (-8)] + [7\cdot (-5)] + (-2\cdot 2) = \fbox{\color[RGB]{192,0,0}{\bf -79}}&#034; title=&#034;\vec B\cdot \vec C = [5\cdot (-8)] + [7\cdot (-5)] + (-2\cdot 2) = \fbox{\color[RGB]{192,0,0}{\bf -79}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f2c2d17978909d4db3b5bf62a6380733.png' style=&#034;vertical-align:middle;&#034; width=&#034;309&#034; height=&#034;22&#034; alt=&#034;\vec B\cdot \vec D = (5\cdot 9) + [7\cdot (-4)] + (-2\cdot 0) = \fbox{\color[RGB]{192,0,0}{\bf 17}}&#034; title=&#034;\vec B\cdot \vec D = (5\cdot 9) + [7\cdot (-4)] + (-2\cdot 0) = \fbox{\color[RGB]{192,0,0}{\bf 17}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; d) El resultado del producto vectorial de dos vectores es un vector que es perpendicular al plano que foman los vectores multiplicados. Se obtienen las componentes de este vector a partir de la resoluci&#243;n de un determinante. Los hacemos para el primer caso y escribimos las soluciones del resto: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1e62e49b091d01712eac3ee99652a5d4.png' style=&#034;vertical-align:middle;&#034; width=&#034;328&#034; height=&#034;63&#034; alt=&#034;\vec A\times \vec B = \left| \begin{array}{ccc} \vec i &amp; \vec j &amp; \vec k\\ 4 &amp; -1 &amp; -6\\ 5 &amp; 7 &amp; -2 \end{array} \right| = \fbox{\color[RGB]{192,0,0}{\bm{44\vec i - 22\vec j + 33\vec k}}}&#034; title=&#034;\vec A\times \vec B = \left| \begin{array}{ccc} \vec i &amp; \vec j &amp; \vec k\\ 4 &amp; -1 &amp; -6\\ 5 &amp; 7 &amp; -2 \end{array} \right| = \fbox{\color[RGB]{192,0,0}{\bm{44\vec i - 22\vec j + 33\vec k}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/75e98c63b00d4f62151199137f2cbf56.png' style=&#034;vertical-align:middle;&#034; width=&#034;211&#034; height=&#034;29&#034; alt=&#034;\vec D\times \vec C = \fbox{\color[RGB]{192,0,0}{\bm{-8\vec i - 18\vec j - 77\vec k}}}&#034; title=&#034;\vec D\times \vec C = \fbox{\color[RGB]{192,0,0}{\bm{-8\vec i - 18\vec j - 77\vec k}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5b32d664d28f0d49a8c99ffc9200b404.png' style=&#034;vertical-align:middle;&#034; width=&#034;185&#034; height=&#034;29&#034; alt=&#034;\vec B\times \vec C = \fbox{\color[RGB]{192,0,0}{\bm{4\vec i + 6\vec j + 31\vec k}}}&#034; title=&#034;\vec B\times \vec C = \fbox{\color[RGB]{192,0,0}{\bm{4\vec i + 6\vec j + 31\vec k}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/aa542c8f176b2d18029f9517acda5657.png' style=&#034;vertical-align:middle;&#034; width=&#034;211&#034; height=&#034;29&#034; alt=&#034;\vec B\times \vec D = \fbox{\color[RGB]{192,0,0}{\bm{-8\vec i - 18\vec j - 83\vec k}}}&#034; title=&#034;\vec B\times \vec D = \fbox{\color[RGB]{192,0,0}{\bm{-8\vec i - 18\vec j - 83\vec k}}}&#034; /&gt;&lt;/p&gt; &lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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