Let P be a plane lx+my+nz=0 containing the line 1-x1=y+42=z+23.

Question

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

Accepted Answer

Line lies on plane

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

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

$l - 4m - 2n = 0$ .... (2)

from (1) \& (2)

$$ - 2m + n = 0 \Rightarrow 2m = n$$

$$l = 3n + 2m \Rightarrow l = 4n$$

$$l:m:n::4n:{n \over 2}:n$$

$l:m:n::8n:n:2n$

$l:m:n::8:1:2$

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

R divide AB is ratio k : 1

$$R:\left( {{{ - 3 + 2k} \over {k + 1}},{{ - 6 + 4k} \over {k + 1}},{{1 - 3k} \over {k + 1}}} \right)$$ lies on plane

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

$- 24 + 16k - 6 + 4k + 2 - 6k = 0$

$- 28 + 14k = 0$

$k = 2$

Solutions · 1