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

Let P be a plane lx + my + nz = 0 containing the line, 1x1=y+42=z+23. 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 x11=y+42=z+23 having direction ratios -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(3+2kk+1,6+4kk+1,13kk+1) lies on the plane.

8(3+2kk+1)+6+4kk+1+2(13kk+1)=0

8(3+2k)+(6+4k)+2(13k)=0

k=2

Answer: (B)

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