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I have the line $$D: \frac{x-7}{12} = \frac{y + 1}{-6} = \frac{z-2}{-2}$$ that crosses the plane $$P : -4x - 5y - z -3 = 0$$ on point $A=(47/4, -67/8, 41/8)$.

I must find a line from point $A$ that is perpendicular to $D$ inside $P$. How?

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Well, the direction vector of the line is $\overrightarrow{\mathbf{v}}=(1/12,-1/6,-1/2)$. With this you can write down the equation of the plane of vectors orthogonal to the line at the point $A$. Then this plane and the plane $P$ define the line you're looking for. – Brian Fitzpatrick Apr 19 '13 at 23:59
@Brian Fitzpatrick I Believe the direction vector of D is V = (12, -6, -2). But, basically your saying that I must find another plane orthogonal to the line and find the line of intersection between the 2 planes? – Machinegon Apr 20 '13 at 0:03
my bad. that's right. – Brian Fitzpatrick Apr 20 '13 at 0:05
up vote 0 down vote accepted

There are only two ways this is possible.

  1. The line is not perpendicular to the plane. In this case, there is a unique answer. Take the cross product of the direction vector of the line and the normal vector of the plane. That will give you a direction perpendicular to the original line, and also perpendicular to the normal of the plane. Then write the equation of a line through your given point in the direction of the cross product.

  2. The line is perpendicular to the plane. In this case there are infinitely many different answers. Just write down any vector perpendicular to the normal of the plane (ignore the original line) and write a new line through the given point in the direction of your new vector.

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Note that the direction vector of line $D$ is $\,\vec v = \langle 12, -6, -2 \rangle$.

Use this vector to find the equation of plane of vectors, let's call it $Q$, which is orthogonal to line $D$ at the point $A =(47/4, -67/8, 41/8)$. The line you are seeking is the line at which the planes $Q$ and $P: -4x - 5y - z -3 = 0$ intersect.

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Always wonder if it was clear to the OP? +1 – Amzoti Apr 20 '13 at 2:17

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