The foot of the perpendicular from the origin O to a plane P, which meets the coordinate a…

Question

The foot of the perpendicular from the origin $O$ to a plane $P$, which meets the coordinate axes at $A,B,C$, is $(2,a,4)$, $a\in\mathbb N$. If the volume of tetrahedron $OABC$ is $144$ cubic units, which point is NOT on $P$?

Accepted Answer

Plane equation (normal through foot $(2,a,4)$): $2x+ay+4z=20+a^2$, giving $A=\left(\dfrac{20+a^2}2,0,0ight)$, $B=\left(0,\dfrac{20+a^2}a,0ight)$, $C=\left(0,0,\dfrac{20+a^2}4ight)$.

Volume $=\dfrac16\cdot\dfrac{20+a^2}2\cdot\dfrac{20+a^2}a\cdot\dfrac{20+a^2}4 = 144 \implies (20+a^2)^3 = 144	imes48	imes a$

Solving gives $a=2$, so the plane is $2x+2y+4z=24$, i.e. $x+y+2z=12$.

Checking each option: $(3,0,4)$ gives $3+0+8=11
e12$ --- not on the plane.

Solutions · 1