Three Dimensional Geometry Question 15
Question 15 - 29 January - Shift 2
The plane $2 x-y+z=4$ intersects the line segment joining the points $A(a,-2,4)$ and $B(2, b,-3)$ at the point $C$ in the ratio $2: 1$ and the distance of the point $C$ from the origin is $\sqrt{5}$. If $ab<0$ and $P$ is the point $(a-b, b, 2 b-a)$ then $CP^{2}$ is equal to :
(1) $\frac{17}{3}$
(2) $\frac{16}{3}$
(3) $\frac{73}{3}$
(4) $\frac{97}{3}$
Show Answer
Answer: (1)
Solution:
Formula: Section Formula, Distance of $P$ from coordinate axes, Equation Of A Plane
$A(a,-2,4), B(2, b,-3)$
$AC: CB=2: 1$
$\Rightarrow C \equiv(\frac{a+4}{3}, \frac{2 b-2}{3}, \frac{-2}{3})$
$C$ lies on $2 x-y+2=4$
$\Rightarrow \frac{2 a+8}{3}-\frac{2 b-2}{3}-\frac{2}{3}=4$
$\Rightarrow a-b=2 \ldots$
Also $OC=\sqrt{5}$
$\Rightarrow(\frac{a+4}{3})^{2}+(\frac{2 b-2}{3})^{2}+\frac{4}{9}=5$
Solving, (1) and (2)
$(b+6)^{2}+(2 b-2)^{2}=41$
$\Rightarrow 5 b^{2}+4 b-1=0$
$\Rightarrow b=-1$ or $\frac{1}{5}$
$\Rightarrow a=1$ or $\frac{11}{5}$
But $ab<0 \Rightarrow(a, b)=(1,-1)$
$C \equiv(\frac{5}{3}, \frac{-4}{3}, \frac{-2}{3}), P \equiv(2,-1,-3)$
$CP^{2}=\frac{1}{9}+\frac{1}{9}+\frac{49}{9}=\frac{51}{9}=\frac{17}{3}$
