Q 66. If the angle between the planes r.(mi- 2j + 3k) 1 + 3k )+4=0 and r.( 3i-mj +2k) -6 = 0 is n / 2 then m = ? A .- 3/2 B.- 5/2 C. - 6 / 5 D. - 7/5

The angle between two planes is given by the formula:

cos θ = |A1.A2 + B1.B2 + C1.C2| / √[(A1² + B1² + C1²) * (A2² + B2² + C2²)]

where A1, B1, C1 are the coefficients of i, j, k in the first plane and A2, B2, C2 are the coefficients of i, j, k in the second plane.

Given the planes r.(mi- 2j + 3k) +4=0 and r.( 3i-mj +2k) -6 = 0, we can identify the coefficients as:

For the first plane: A1 = m, B1 = -2, C1 = 3 For the second plane: A2 = 3, B2 = -m, C2 = 2

Substituting these into the formula, we get:

cos θ = |m3 + (-2)(-m) + 3*2| / √[(m² + (-2)² + 3²) * (3² + (-m)² + 2²)]

Given that the angle between the planes is π/2, cos θ = 0. Therefore, the numerator of the above expression must be zero:

|m3 + (-2)(-m) + 3*2| = 0 |3m + 2m + 6| = 0 |5m + 6| = 0

Solving this equation for m, we get m = -6/5. Therefore, the correct answer is:

C. - 6 / 5