Let a straight line L pass through the point P(2, -1, 3) and be perpendicular to the lines (x - 1)/2 = (y + 1)/1 = (z - 3)/-2 and (x - 3)/1 = (y - 2)/3 = (z + 2)/4 . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is:

Solution

Maths sir · 4 steps

A line perpendicular to two lines runs along their cross product, so find that vector first.

1. Direction of L. The given lines have directions (2, 1, -2) and (1, 3, 4). A vector perpendicular to both is their cross product.

   |i, j, k; 2, 1, -2; 1, 3, 4| = 10 i - 10 j + 5 k

2. Write the line. Divide by 5 to get the direction (2, -2, 1). Any point on L is (2 + 2t, -1 - 2t, 3 + t).
3. Meet the yz-plane. On the yz-plane x = 0, so 2 + 2t = 0 and t = -1. This gives Q = (0, 1, 2).
4. So. PQ = √((2-0)² + (-1-1)² + (3-2)²) = √(4 + 4 + 1) = 3, which is option (d).

The idea to keep. Perpendicular to two lines means the direction is the cross product of their direction vectors. To meet a coordinate plane, set that coordinate to zero in the parametric point.

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