Three Dimensional Geometry Question 33
Question 33 - 01 February - Shift 2
Let the plane $P$ pass through the intersection of the planes $2 x+3 y-z=2$ and $x+2 y+3 z=6$, and be perpendicular to the plane $2 x+y-z+1=0$. If $d$ is the distance of $P$ from the point $(-7,1,1)$, then $d^{2}$ is equal to :
(1) $\frac{250}{83}$
(2) $\frac{15}{53}$
(3) $\frac{25}{83}$
(4) $\frac{250}{82}$
Show Answer
Answer: (1)
Solution:
Formula: Family of Planes, A Plane and A Point
$P \equiv P_1+\lambda P_2=0$
$(2+\lambda) x+(3+2 \lambda) y+(3 \lambda-1) z-2-6 \lambda=0$
Plane $P$ is perpendicular to $P_3 $
$\therefore \overrightarrow{{}n} \cdot \overrightarrow{{}n}_3=0$
$2(\lambda+2)+(2 \lambda+3)-(3 \lambda-1)=0$
$\lambda=-8$
$P \equiv-6 x-13 y-25 z+46=0$
$6 x+13 y+25 z-46=0$
Distance from $(-7,1,1)$
$d=|\frac{-42+13+25-46}{\sqrt{36+169+625}}|=\frac{50}{\sqrt{830}}$
$d^{2}=\frac{50 \times 50}{830}=\frac{250}{83}$
