If an angle between the line x+12=y-21=z-3-2 and the plane x-2y-kz=3 is ^-12√23, then a va…

Question

If an angle between the line $\dfrac{x+1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{-2}$ and the plane $x-2y-kz=3$ is $\cos^{-1}\dfrac{2\sqrt2}{3}$, then a value of $k$ is

Accepted Answer

Line direction $\vec b=(2,1,-2)$, $|\vec b|=3$; plane normal $\vec n=(1,-2,-k)$, $|\vec n|=\sqrt{5+k^2}$.
$$\sin\theta=\frac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}=\frac{|2k|}{3\sqrt{5+k^2}}.$$
Since $\cos\theta=\dfrac{2\sqrt2}{3}$, $\sin\theta=\dfrac13$, so
$$\frac{2|k|}{3\sqrt{5+k^2}}=\frac13\implies 4k^2=5+k^2\implies k^2=\frac53\implies k=\pm\sqrt{\frac53}.$$
(Note: the printed option values $\pm\frac53,\pm\frac35$ appear to be OCR-flattened square-root expressions; the source answer key marks option A as correct, corresponding to the positive root $k=\sqrt{5/3}$.)

Solutions · 1