Let P be a plane lx + my + nz = 0 containing the line, 1–x1=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 x–1–1=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+1,1–3kk+1) lies on the plane.
∴8(–3+2kk+1)+–6+4kk+1+2(1–3kk+1)=0
⇒8(–3+2k)+(–6+4k)+2(1–3k)=0
⇒k=2
Answer: (B)