If the lines (x - 1)/1 = (y - 2)/2 = (z + 3)/1 and (x - a)/2 = (y + 2)/3 = (z - 3)/1 intersect at the point P, then the distance of the point P from the plane z = a is :

Solution

Maths sir · 4 steps

Write both lines with a parameter. The y and z equations do not involve a, so solve those two first.

1. General points. A point on the first line is (1+t, 2+2t, -3+t). A point on the second line is (a+2s, -2+3s, 3+s). At P these two points are the same.
2. Match y and z. From y: 2+2t = -2+3s, so 2t-3s=-4. From z: -3+t = 3+s, so t = s+6. Substituting gives 2s+12-3s=-4, so s=16 and t=22.
3. Find P and a. With t=22, P is (23, 46, 19). Now match x: 23 = a+2(16), so a=-9. Check with the second line: (-9+32, -2+48, 3+16) = (23, 46, 19).

   Distance of P from z=-9 = |19-(-9)| = 28

4. So. The plane z=a is z=-9, and the distance from P is the gap in z-coordinates, which is 28. The answer is (a).

The idea to keep. For intersecting lines with an unknown, first solve with the two coordinates that do not contain the unknown. The third coordinate then gives you the unknown. The distance of a point from z=c is just |z₀-c|.

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