### If for a > 0, the feet of perpendiculars from the points ....

If for a > 0, the feet of perpendiculars from the points A (a, -2a, 3) and B (0, 4, 5) on the plane lx + my + nz = 0 are points C (0, -a, -1) and D respectively, then the length of line segment CD is equal to:

(A) $\sqrt {41}$
(B) $\sqrt {31}$
(C) $\sqrt {66}$
(D) $\sqrt {55}$

Solution

Point C (0, -a, -1) lies on the plane lx + my + nz = 0.

So, 0 - am - n = 0

Or, n = -am

Normal to the plane lx + my + nz = 0 is parallel to AC.

So, $\frac{{a - 0}}{l} = \frac{{ - 2a + a}}{m} = \frac{{3 + 1}}{n}$

$\Rightarrow \frac{a}{l} = - \frac{a}{m} = \frac{4}{n} = \frac{4}{{ - am}}$

$\Rightarrow a = 2$ (a = -2 is rejected since a > 0)

Direction ratios of AC $\equiv a, - a,4 \equiv 2, - 2,4 \equiv$ Direction ratios of BD

Point D $\equiv (2r + 0, - 2r + 4,4r + 5)$

Since point D lies on the plane lx + my + nz = 0, we have

l (2r) + m (-2r + 4) + n (4r + 5) = 0

But, as obtained earlier l = -m & n = -2m

m (-2r) + m (-2r + 4) - 2m (4r + 5) = 0

So, -2r -2r + 4 - 8r - 10 = 0

Thus, $r=- \frac {1}{2}$

Point D $\equiv (2r + 0, - 2r + 4,4r + 5) \equiv (-1,5,3)$

Point C $\equiv (0, -a, -1) \equiv (0, -2, -1)$

Distance CD $= \sqrt {{1^2} + {{( - 7)}^2} + {{( - 4)}^2}} = \sqrt {66}$ unit