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Josan



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MensajePublicado: Vie Feb 24, 2012 12:53 pm    Título del mensaje: Problema de selectividad Responder citando

Calcular un punto P de la recta que pasa por el Origen y tiene vector director V(1,3,4) cuya proyección en el plano z=0 es Q y tal que el triángulo formado por los tres puntos OPQ tenga área 1.

Es urgente su resolución, gracias.[/tex]
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magvixupi



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MensajePublicado: Sab Feb 25, 2012 4:00 am    Título del mensaje: Responder citando

Hola Josan:

La ecuación de la recta es: (x,y,z)=(x _0,y _ 0,z _ 0)+t(1,3,4)

Si la recta debe pasar por el origen => (x,y,z)=t(1,3,4)

=> x=t, y=3t, z=4t (ecuaciones 1)

Las coordenadas de cada punto son: P(t,3t,4t), Q(t,3t,0), O(0,0,0)

La altura del triangulo es z=4t La base del triángulo es OQ

El área del triángulo:

\frac{\sqrt{t^{2}+(3t)^{2}}*4t}{2}=1

Calcular el valor de t y reemplazar en (ecuaciones 1) para hallar las coordenadas del punto P sobre la recta.

Saludos.
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judy9146



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MensajePublicado: Mar Sep 11, 2012 3:56 am    Título del mensaje: cheap Ralph Lauren Polo Responder citando

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judy9146



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MensajePublicado: Lun Nov 12, 2012 7:44 am    Título del mensaje: cheap designer handbags Responder citando

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MensajePublicado: Mar Nov 27, 2012 12:20 pm    Título del mensaje: Coach Outlet Online Responder citando

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