Let the plane P: 8x+_1y+_2z+12=0 be parallel to the line L: x+22=y-33=z+45.

Question

Let the plane $P:\ 8x+\alpha_1y+\alpha_2z+12=0$ be parallel to the line $L:\ \dfrac{x+2}{2}=\dfrac{y-3}{3}=\dfrac{z+4}{5}$. If the intercept of $P$ on the $y$-axis is $1$, then the distance between $P$ and $L$ is:

Accepted Answer

Parallel to $L$: $8(2)+3\alpha_1+5\alpha_2=0\Rightarrow 3\alpha_1+5\alpha_2=-16$.

$y$-intercept of $P$ is $1$ (i.e. $P$ meets $(0,1,0)$): $\alpha_1+12=0\Rightarrow \alpha_1=-12$, giving $\alpha_2=4$.
$$P:\ 8x-12y+4z+12=0 \implies 2x-3y+z+3=0.$$
Using the point $(-2,3,-4)$ on $L$:
$$d=\frac{|2(-2)-3(3)+(-4)+3|}{\sqrt{4+9+1}}=\frac{|-4-9-4+3|}{\sqrt{14}}=\frac{14}{\sqrt{14}}=\sqrt{14}.$$

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