Prove the following
Question:

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

  1. (1) $1.5$

  2. (2) 3

  3. (3) 2

  4. (4) 4


Correct Option: , 3

Solution:

Point $\mathrm{C}$ is

$\left(\frac{2 k-3}{k+1}, \frac{4 k-6}{k+1}, \frac{-3 k+1}{k+1}\right)$

$\frac{x-1}{-1}=\frac{y+4}{2}=\frac{z+2}{3}$

Plane $l_{\mathrm{X}}+\mathrm{my}+\mathrm{nz}=0$

$l(-1)+m(2)+n(3)=0$

$-l+2 m+3 n=0$

It also satisfy point $(1,-4,-2) l-4 m-2 n=0$

Solving (1) and (2) $2 m+3 n=4 m+2 n$

$\mathrm{n}=2 \mathrm{~m}$

$l-4 m-4 m=0$

$l=8 \mathrm{~m}$

$\frac{l}{8}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{2}$

$l: \mathrm{m}: \mathrm{n}=8: 1: 2$

Plane is $8 x+y+2 z=0$ It will satisfy point $\mathrm{C}$

$8\left(\frac{2 \mathrm{k}-3}{\mathrm{k}+1}\right)+\left(\frac{4 \mathrm{k}-6}{\mathrm{k}+1}\right)+2\left(\frac{-3 \mathrm{k}+1}{\mathrm{k}+1}\right)=0$

$16 \mathrm{k}-24+4 \mathrm{k}-6-6 \mathrm{k}+2=0$

$14 \mathrm{k}=28 \quad \therefore \quad \mathrm{k}=2$

 

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