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Let P be a plane lx + my + nz = 0 containing the line ....

Let P be a plane lx + my + nz = 0 containing the line, $\frac{{1 - x}}{1} = \frac{{y + 4}}{2} = \frac{{z + 2}}{3}$. If 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:

(A) 4
(B) 2
(C) 1.5
(D) 3

Solution

The plane lx + my + nz = 0 contains the line $\frac{{x - 1}}{{ - 1}} = \frac{{y + 4}}{2} = \frac{{z + 2}}{3}$ having direction ratios $\equiv $ -1, 2, 3. Thus,

-l + 2m + 3n = 0

Since the line passes through (1, -4, -2) the plane would also contain this point.

So, l - 4m -2n = 0

Thus, l = 8m & n = 2m

The equation of plane can be written as, 8mx + my + 2mz = 0 or 8x + y + 2z = 0.

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

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

$ \Rightarrow 8( - 3 + 2k) + ( - 6 + 4k) + 2(1 - 3k) = 0$

$ \Rightarrow k = 2$

Answer: (B)