Let the line : x=1-y-2=z-3, R, meet the plane P: x+2y+3z=4 at the point (,,).

Question

Let the line $\ell:\ x=\dfrac{1-y}{-2}=\dfrac{z-3}{\lambda}$, $\lambda\in\mathbb R$, meet the plane $P:\ x+2y+3z=4$ at the point $(\alpha,\beta,\gamma)$. If the angle between the line $\ell$ and the plane $P$ (measured as the angle between $\ell$ and the normal) is $\cos^{-1}\left(\dfrac{3}{\sqrt{14}}\right)$, then $\alpha+2\beta+6\gamma$ is equal to:

Answer: 11

Accepted Answer

Direction of $\ell$: $(1,2,\lambda)$; normal $(1,2,3)$.
$$\cos\theta=\frac{5+3\lambda}{\sqrt{14}\sqrt{5+\lambda^2}}=\frac{3}{\sqrt{14}} \implies (5+3\lambda)^2=9(5+\lambda^2).$$
Solving gives $\lambda=\dfrac23$.

Parametrize $\ell$ as $(t,\ 1+2t,\ 3+\lambda t)$; on the plane:
$$t+2(1+2t)+3\left(3+\frac23t\right)=4 \implies 7t+11=4 \implies t=-1.$$
$$(\alpha,\beta,\gamma)=\left(-1,\ -1,\ \frac73\right).$$
$$\alpha+2\beta+6\gamma=-1-2+14=11.$$

Solutions · 1