1. ## Vectors: Intersection

I can't seem to work out the answer. Help would be appreciated.

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Q:

Find an equation of the plane, in the form $\bold r \bold . \bold n = p$, which contains the line $l$ and the point with the position vector $\bold a$ where $l : \bold r = t ( 2 \bold i + 3 \bold j - \bold k)$, $\bold a = \bold i + 4 \bold k$.

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P.S. The answer is $\bold r \bold . (4 \bold i - 3 \bold j - \bold k) = 0$

2. Originally Posted by Air
Q:

Find an equation of the plane, in the form $\bold r \bold . \bold n = p$, which contains the line $l$ and the point with the position vector $\bold a$ where $l : \bold r = t ( 2 \bold i + 3 \bold j - \bold k)$, $\bold a = \bold i + 4 \bold k$.
The line passes through the origin because there isn't any constant summand. The vector $\overrightarrow{OA} = \vec a = (i + 0 +4k)$ lies in the plane. Therefore the plane passes through the origin and therefore p = 0.

Calculate the normal vector $\vec n$ :

$\left|\begin{array}{ccc}i&j&k\\1&0&4\\2&3&-1\end{array} \right| = -12i + 9j + 3k = -3(4i - 3j - k)$

Thus the equation of the plan is

$r(4i - 3j - k)=0$