For a,b Z and |a-b|10, let the angle between the plane P: ax+y-z=b and the line x-1=a-y=z+…

Question

For $a,b\in\mathbb Z$ and $|a-b|\leq10$, let the angle between the plane $P:\ ax+y-z=b$ and the line $x-1=a-y=z+1$ be $\cos^{-1}\left(\dfrac13\right)$. If the distance of the point $(6,-6,4)$ from the plane $P$ is $3\sqrt6$, then $a^4+b^2$ is equal to:

Accepted Answer

Normal $(a,1,-1)$, line direction $(1,-1,1)$. Since the angle between plane and line satisfies $\sin\theta=2\sqrt2/3$ (from $\cos\theta=1/3$):
$$\sin\theta=\frac{|a-2|}{\sqrt3\sqrt{a^2+2}}=\frac{2\sqrt2}{3} \implies 3(a-2)^2=8(a^2+2) \implies 5a^2+12a+4=0.$$
$$a=\frac{-12\pm8}{10}=-\frac25\ \text{or}\ -2 \implies a=-2\ (\text{integer}).$$
Distance condition:
$$\frac{|{-2(6)-6-4-b}|}{\sqrt6}=3\sqrt6 \implies |-22-b|=18 \implies b=-40\text{ or }-4.$$
Since $|a-b|\leq10$: $b=-4$ (as $b=-40$ gives $|a-b|=38$).
$a^4+b^2=16+16=32.$

Solutions · 1