Three Dimensional Geometry Question 5
Question 5 - 24 January - Shift 2
Let the plane containing the line of intersection of the planes
P1: $x+(\lambda+4) y+z=1$ and
P2: $2 x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2 \lambda, \lambda,-\lambda)$ from the plane $P 2$ is
(1) $5 \sqrt{6}$
(2) $4 \sqrt{6}$
(3) $2 \sqrt{6}$
(4) $3 \sqrt{6}$
Show Answer
Answer: (4)
Solution:
Formula: Family of Planes, A Plane and A Point
Equation of plane passing through point of intersection of P1 and P2
$P=P1+kP 2$
$(x+(\lambda+4) y+z-1)+k(2 x+y+z-2)=0$
Passing through $(0,1,0)$ and $(1,0,1)$
$(\lambda+4-1)+k(1-2)=0$
$(\lambda+3)-k=0$
Also passing $(1,0,1)$
$(1+1-1)+k(2+1-2)=0$
$1+k=0$
$k=-1$
put in (1)
$\lambda+3+1=0$
$\lambda=-4$
Then point $(2 \lambda, \lambda,-\lambda)$
$ (-8,-4,4) $
$d=|\frac{-16-4+4-2}{\sqrt{6}}|$
$d=\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6}$
