Let be the angle between the planes P_1: r( i+ j+2 k)=9 and P_2: r(2 i- j+ k)=15.

Question

Let $	heta$ be the angle between the planes $P_1:\ \vec r\cdot(\hat i+\hat j+2\hat k)=9$ and $P_2:\ \vec r\cdot(2\hat i-\hat j+\hat k)=15$. Let $L$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $	heta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$, then $(	an^2	heta)(\cot^2\alpha)$ is equal to:

Answer: 9

Accepted Answer

$$\cos\theta=\frac{(1,1,2)\cdot(2,-1,1)}{\sqrt6\cdot\sqrt6}=\frac{2-1+2}{6}=\frac12 \implies \theta=\frac{\pi}{3}.$$
Since $L$ makes angle $\theta$ with the normal of $P_2$, the angle between $L$ and the plane $P_2$ itself is
$$\alpha=\frac{\pi}{2}-\theta=\frac{\pi}{6}.$$
$$(\tan^2\theta)(\cot^2\alpha)=(\tan60^\circ)^2(\cot30^\circ)^2=3\times3=9.$$

Solutions · 1