Let the foot of perpendicular from the point A(4,3,1) on the plane P:x-y+2z+3=0 be N.

Question

Let the foot of perpendicular from the point $A(4,3,1)$ on the plane $P:x-y+2z+3=0$ be N. If B$(5,\alpha ,\beta ),\alpha ,\beta \in Z$ is a point on plane P such that the area of the triangle ABN is $3\sqrt{2}$, then ${\alpha }^{2}+{\beta }^{2}+\alpha \beta $ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

We have, equation of plane $P$ is $x-y+2z+3=0$

Perpendicular distance from $A$ to a plane $P$ i.e. N

$=\frac{|4-3+2(1)+3|}{\sqrt{{1}^{2}+(-1{)}^{2}+{2}^{2}}}$

$B(5,\alpha ,\beta )$ lies on the plane $P$, so

$5-\alpha +2\beta +3=0\Rightarrow \alpha =2\beta +8$ .........(i)

Direction ratio of $AN$

Solutions · 1