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If a vector line, with equation: $$l=\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} + u\begin{pmatrix}2 \\ 2 \\ 1 \end{pmatrix}$$ Intersects a plane with equation: $$\Pi=\begin{pmatrix}3 \\ 4 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}s + \begin{pmatrix}5 \\ 5 \\ 4\end{pmatrix}t$$

How can you find a vector that is parallel to the plane and also perpendicular to the vector line?

I am not sure of how to find this vector but I was considering crossing the direction vector of the vector line with the cross product of the two direction vectors of the plane (For example: $u \times (s \times t)$).

Any help will be very much appreciated.

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    Cross the direction vectors of the plane to find a normal direction vector. Cross that with the direction vector of the vector line to find a direction vector that is perpendicular to the line, and perpendicular to the normal of the plane (or parallel to the plane.) – 2017-01-05
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    Thanks.. I've been able to solve it – 2017-01-05

3 Answers 3

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Hint

$1.$ The vector $(2,4,6)\times (5,5,4)$ is perpendicular to the plane;

$2.$ The vector $(2,2,1)$ is the direction of the line.

The vector that you are looking for is $v$. Then it has to fit:

$1.$ $v\cdot [(2,4,6)\times (5,5,4)]=0$ and

$2.$ $v\cdot (2,2,1)=0$

Can you finish?

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Note that, the vector parallel to plane will be in the span of $(2,4,6)$ and $(5,5,4)$ and we want it to be perpendicular to the line, so we have following $<(2,4,6)s+(5,5,4)t, (2,2,1)>=0 \rightarrow3s+4t=0$. Choose $s=-4$ and $t=3$. The desired vector is $-4(2,4,6)+3(5,5,4)$

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it must be $$\vec{a}\cdot (2,2,1)=0$$ and $$\vec a\cdot (-14,22,-10)=0$$