Let the line x/1=(6-y)/2=(z+8)/5 intersect the lines (x-5)/4=(y-7)/3=(z+2)/1 and (x+3)/6=(3-y)/3=(z-6)/1 at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2 x-2 y+z=14 is :

Solution

Maths sir · 5 steps

Write the common line with one parameter, meet each line to fix A and B, then use the plane distance formula.

1. Parametrise the line. Set x/1=(6-y)/2=(z+8)/5=t. Then any point on it is (t, 6-2t, -8+5t).
2. Find A. A point on the first line is (5+4s, 7+3s, -2+s). Matching x gives t=5+4s, and then matching y gives 6-10-8s=7+3s, so s=-1 and t=1. Both lines give z=-3 there, so A=(1, 4, -3).
3. Find B. A point on the second line is (-3+6u, 3-3u, 6+u). Matching x gives t=-3+6u, and then matching y gives 12-12u=3-3u, so u=1 and t=3. Both lines give z=7 there, so B=(3, 0, 7).
4. Mid-point. The mid-point of AB is M=((1+3)/2, (4+0)/2, (-3+7)/2)=(2, 2, 2).

   d=(|2(2)-2(2)+2-14|)/√(2²+(-2)²+1²)=12/3=4

5. So. The mid-point of AB is at a distance of 4 from the plane, which is option (c).

The idea to keep. When a line meets other lines, give it one parameter and give each other line its own. Match two coordinates to solve, and use the third coordinate to check that the lines really meet.

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