Question

If lines [(x – 1)/2] = [(y – 3)/4] = z and [(x – 4)/3] = [(1 – y)/2] = [(z – 1)/1] are co-planer, then the equation of plane containing these two lines is ____

(A) 6x + y + 16z = 9

(B) 6x + y – 16z = 9

(C) 6x – y – 16z = 9

(D) 6x – y + 16z = 9

Answer 1

The correct option (B) 6x + y – 16z = 9  

Explanation:

lines: [(x – 1)/2] = [(y – 3)/4] = (z/1) ⇒ vector a = (1, 3, 0) & vector  ℓ = (2, 4, 1)

also [(x – 4)/3] = [(y – 1)/(– 2)] = [(z – 1)/1] ⇒ vector  b = (4, 1, 1) &

vector  m = (3, – 2, 1)

vector  a – b = (– 3, 2, – 1) and

vector l x vector m

Now vector  (a – b) ∙ vector  (ℓ × m) = (– 3, 2, – 1) ∙ (6, 1, – 16) = – 18 + 2 + 16 = 0

Lines are coplanar.

equation of plane is vector (r – a) ∙ vector  (ℓ × m) = 0 

∴ (x – 1)(6) – (y – 3)(– 1) + z(– 16) = 0

∴ 6x – 9 + y – 16z = 0

∴ 6x + y – 16z = 9.