Let (,2,1) be a point on the plane which passes through the point (4,-2,2).

Question

Let $(\lambda,2,1)$ be a point on the plane which passes through the point $(4,-2,2)$. If the plane is perpendicular to the line joining the points $(-2,-21,29)$ and $(-1,-16,23)$, then $\left(\dfrac{\lambda}{11}\right)^2-\dfrac{4\lambda}{11}-4$ is equal to:

Answer: 8

Accepted Answer

$\vec{AB}=(4-\lambda,-4,1)$ (with $A=(\lambda,2,1)$, $B=(4,-2,2)$); direction of the given line $\vec{PQ}=(1,5,-6)$.

Perpendicularity of $\vec{AB}$ to the plane's containing directions requires $\vec{AB}\cdot\vec{PQ}=0$ (since the plane's normal is along $PQ$, and $AB$ lies in the plane):
$$(4-\lambda)(1)+(-4)(5)+(1)(-6)=0 \implies -\lambda-22=0 \implies \lambda=-22.$$
$$\left(\frac{-22}{11}\right)^2-\frac{4(-22)}{11}-4=4+8-4=8.$$

Solutions · 1