Equations of Planes — Question 3

Question 3 – Solution

Strategy The intersection direction is perpendicular to both normals. Obtain one point by fixing a convenient coordinate.

Intersection line With $\vec n_1=\left\langle 1,2,-1\right\rangle$ and $\vec n_2=\left\langle 2,-1,2\right\rangle$, $$\vec n_1\times\vec n_2=\left\langle 3,-4,-5\right\rangle.$$ Setting $z=0$ gives $x+2y=3$, $2x-y=1$, hence $(x,y)=(1,1)$. Therefore $$\boxed{\vec r=\left\langle 1,1,0\right\rangle+t\left\langle 3,-4,-5\right\rangle}.$$

Angle Since $\vec n_1\cdot\vec n_2=-2$, $\|\vec n_1\|=\sqrt 6$, and $\|\vec n_2\|=3$, $$\boxed{\theta=\cos^{-1}\!\left(\frac{2}{3\sqrt 6}\right)\approx 74.2^\circ}.$$