Equations of Planes — Question 6

Question 6 – Solution

Strategy Substitute the line into the plane. The line–plane angle is complementary to the angle between the direction and the normal.

Intersection Substitution gives $3(1+2t)+(2-t)-3t=5$, so $2t=0$ and $t=0$. Thus $Q=\boxed{(1,2,0)}$.

Angle With $\vec d=\left\langle 2,-1,3\right\rangle$ and $\vec n=\left\langle 3,1,-1\right\rangle$, $$\sin\alpha=\frac{|\vec d\cdot\vec n|}{\|\vec d\|\|\vec n\|}=\frac{2}{\sqrt{154}}.$$ Hence $\boxed{\alpha=\sin^{-1}(2/\sqrt{154})\approx 9.27^\circ}$.

Verification Since $\vec d\cdot\vec n\ne 0$, the line is not parallel to the plane.