Find the equation of the plane containing the point (0, 7, -7) and the line
$\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$
I'm not sure how to tackle this question, since the equation of the line is given in the Cartesian form and not in the vector form.
(Not homework- I am preparing for an exam)
$\endgroup$1 Answer
$\begingroup$If we set $P(-1,+3,-2)$, $Q(0,7,-7)$ and $\vec{v}(-3,2,1)$ is the leading vector of the line(the expression multiplied by a parameter to form the locus for a line), so you can find the normal vector to the plane as follows: $$\vec{PQ}\times\vec{v}=\begin{vmatrix} i & j & k\\ -1 & -4 & 5\\ -3 &2 &1 \end{vmatrix}=-14i-14k-14j$$ So the equation of the plane would be: $$\mathscr{P}:-14(x-0)-14(y-7)-14(z+7)=0$$
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