If the distance of the point (1,-2,3) from the plane x+2y-3z+10=0, measured parallel to th…

Question

If the distance of the point $(1,-2,3)$ from the plane $x+2y-3z+10=0$, measured parallel to the line $\dfrac{x-1}{3}=\dfrac{2-y}{m}=\dfrac{z+3}{1}$, is $\dfrac{7}{\sqrt2}$, then the value of $|m|$ is equal to:

Answer: 2

Accepted Answer

Direction $(3,-m,1)$. General point $B=(3\lambda+1,\ -m\lambda-2,\ \lambda+3)$ on this line through $(1,-2,3)$; on the plane:
$$(3\lambda+1)+2(-m\lambda-2)-3(\lambda+3)+10=0 \implies (3-2m-3)\lambda=2 \implies \lambda=-\frac1m.$$
$$|AB|^2=(3\lambda)^2+(m\lambda)^2+\lambda^2=9\lambda^2+m^2\lambda^2+\lambda^2.$$
Since $m\lambda=-1$: $|AB|^2=10\lambda^2+1=\dfrac72 \implies \lambda^2=\dfrac14 \implies \lambda=\pm\dfrac12.$
$$m=-\frac{1}{\lambda}=\mp2 \implies |m|=2.$$

Solutions · 1