The distance of line 3y - 2z - 1 = 0 = 3x - z + 4 from the point (2, -1, 6) is :

Solution

Maths sir · 4 steps

Turn the two planes into one line with a point and a direction, then drop a perpendicular from P.

1. Point and direction. From 3x - z + 4 = 0 we get x = (z-4)/3, and from 3y - 2z - 1 = 0 we get y = (2z+1)/3. Putting z = 1 gives the point A(-1, 1, 1). Each unit rise in z moves x by 1/3 and y by 2/3, so the direction is (1, 2, 3).

   Line: (x, y, z) = (-1 + s, 1 + 2s, 1 + 3s)

2. Find the foot. For P(2, -1, 6), AP⃗ = (3, -2, 5). The foot is at s = (AP⃗ · (1,2,3))/(1²+2²+3²) = (3 - 4 + 15)/14 = 1, which is the point M(0, 3, 4). It satisfies both planes: 9 - 8 - 1 = 0 and 0 - 4 + 4 = 0.
3. Measure PM. MP⃗ = (2, -4, 2), so PM = √(4 + 16 + 4) = √24. As a check, MP⃗ · (1,2,3) = 2 - 8 + 6 = 0, so PM really is perpendicular to the line.
4. So. The distance is √24 = 2√6, which is option (c).

The idea to keep. A line given as two planes is easiest to handle after you write it as a point plus a direction. Then project onto the direction to find the foot of the perpendicular, and measure from there.

Answer from the NTA final answer key. Solution by PrepWiser.