Question

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 :

• A. $\frac{17}{3}$
• B. $\frac{16}{3}$
• C. $\frac{73}{3}$
• D. $\frac{97}{3}$

Answer 1

Answer: (A)

$A ( a ,-2,4), B (2, b ,-3)$
$ AC : CB =2: 1$
$ \Rightarrow C \equiv\left(\frac{ a +4}{3}, \frac{2 b -2}{3}, \frac{-2}{3}\right)$
$C$ 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 \text { (1) }$
Also $OC =\sqrt{5}$
$\Rightarrow\left(\frac{ a +4}{3}\right)^2+\left(\frac{2 b-2}{3}\right)^2+\frac{4}{9}=5$....(2)
Solving, (1) and (2)
$ (b+6)^2+(2 b-2)^2=41 $
$\Rightarrow 5 b^2+4 b-1=0 $
$\Rightarrow b=-1 \text { or } \frac{1}{5}$
$\Rightarrow a=1 \text { or } \frac{11}{5}$
But $ab < 0 \Rightarrow( a , b )=(1,-1)$
$ C \equiv\left(\frac{5}{3}, \frac{-4}{3}, \frac{-2}{3}\right), P \equiv(2,-1,-3)$
$CP ^2=\frac{1}{9}+\frac{1}{9}+\frac{49}{9}=\frac{51}{9}=\frac{17}{3}$