Question

The plane $2x-y+z=4$ 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 $(a-b,b,2b-a)$ then $C{P}^2$ is equal to :

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

Answer 1

Answer: (A)

$A(a,-2,4),B(2,b,-3)$
$AC:CB=2:1$
$\Rightarrow C\equiv \left(\frac{a+4}{3},\frac{2b-2}{3},\frac{-2}{3}\right)$
$C$ lies on $2x-y+2=4$
$\Rightarrow \frac{2a+8}{3}-\frac{2b-2}{3}-\frac{2}{3}=4$
$\Rightarrow a-b=2\dots \text{ (1) }$
Also $OC=\sqrt{5}$
$\Rightarrow {\left(\frac{a+4}{3}\right)}^2+{\left(\frac{2b-2}{3}\right)}^2+\frac{4}{9}=5$....(2)
Solving, (1) and (2)
$(b+6{)}^2+(2b-2{)}^2=41$
$\Rightarrow 5b^2+4b-1=0$
$\Rightarrow b=-1\text{ or }\frac{1}{5}$
$\Rightarrow a=1\text{ or }\frac{11}{5}$
But $ab<0\Rightarrow (a,b)=(1,-1)$
$C\equiv \left(\frac{5}{3},\frac{-4}{3},\frac{-2}{3}\right),P\equiv (2,-1,-3)$
$C{P}^2=\frac{1}{9}+\frac{1}{9}+\frac{49}{9}=\frac{51}{9}=\frac{17}{3}$