If the angle between the line 2(x+1)=y=z+4 and the plane 2x-y+ z+4=0 is 6, then the value …

Question

If the angle between the line $2(x+1)=y=z+4$ and the plane $2x-y+\sqrt\lambda z+4=0$ is $\dfrac\pi6$, then the value of $\lambda$ is:

Accepted Answer

The line is $\dfrac{x+1}{1}=\dfrac y2=\dfrac{z+4}2$ (direction $(1,2,2)$). With plane normal $(2,-1,\sqrt\lambda)$:
$$\sin\frac\pi6=\frac12=\frac{|2-2+2\sqrt\lambda|}{3\sqrt{5+\lambda}}=\frac{2\sqrt\lambda}{3\sqrt{5+\lambda}}\implies 16\lambda=9(5+\lambda)\implies7\lambda=45\implies\lambda=\frac{45}7.$$

Solutions · 1