Question

Let P be a plane $lx+my+nz=0$ containing the line, $\frac{\left[1-x\right]}{1}=\frac{\left[y+4\right]}{2}=\frac{\left[z+2\right]}{3}$. If plane $P$ divides the line segment $AB$ joining points $A\left(-3,-6,1\right)$ and $B\left(2,4,-3\right)$ in ratio $k:1$ then the value of $k$ is equal to :

• A. $1.5$
• B. $2$
• C. $4$
• D. $3$

Answer 1

The correct option is B

$2$

Explanation for the correct option:

$L:\frac{\left[1-x\right]}{1}=\frac{\left[y+4\right]}{2}=\frac{\left[z+2\right]}{3}$

$P:lx+my+nz=0$

Line lies on plane

$\Rightarrow -l+2m+3n=0$ …….(i)

Point on line $(1,-4,-2)$ lies on a plane.

$\Rightarrow l-4m-2n=0$ ………(ii)

from (i) and (ii)

$$\begin{gathered} -2m+n=0 \\ \Rightarrow 2m=n \end{gathered}$$

and from (i)

$$\begin{gathered} l=3n+2m \\ \Rightarrow l=4n \end{gathered}$$

$$\begin{gathered} \therefore l:m:n=4n:\frac{n}{2}:n \\ \Rightarrow l:m:n=8n:n:2n \\ \Rightarrow l:m:n=8:1:2 \end{gathered}$$

Now equation of plane is $8x+y+2z=0$.

Let point $R$ divide $AB$ in ratio $k:1$.

$\Rightarrow R\left(\frac{-3+2k}{k+1},\frac{-6+4k}{k+1},\frac{1-3k}{k+1}\right)$ lies on the plane.

$$\begin{gathered} \Rightarrow 8\left(\frac{-3+2k}{k+1}\right)+\left(\frac{-6+4k}{k+1}\right)+2\left(\frac{1-3k}{k+1}\right)=0 \\ \Rightarrow -24+16k-6+4k+2-6k=0 \\ \Rightarrow -28+14k=0 \\ \Rightarrow k=2 \end{gathered}$$

Hence, the correct answer is option (B).