Let P and Q be the points on the line (x+3)/8=(y-4)/2=(z+1)/2 which are at a distance of 6 units from the point R(1,2,3). If the centroid of the triangle PQR is (α, β, γ), then α²+β²+γ² is :

Solution

Maths sir · 5 steps

Write a general point on the line with one parameter t, then let the distance condition find t.

1. General point. Set each fraction equal to t. Any point on the line is (-3+8t, 4+2t, -1+2t).
2. Distance from R. The differences from R(1,2,3) are 8t-4, 2t+2 and 2t-4. Square them, add, and set the total equal to 6² = 36.

   (8t-4)²+(2t+2)²+(2t-4)² = 72t²-72t+36 = 36 ⇒ 72t(t-1)=0

3. Find P and Q. So t=0 or t=1. These give P(-3,4,-1) and Q(5,6,1).
4. Centroid. α = (-3+5+1)/3 = 1, β = (4+6+2)/3 = 4, γ = (-1+1+3)/3 = 1.
5. So. α²+β²+γ² = 1+16+1 = 18, which is option (a).

The idea to keep. For any point on a line in 3D that satisfies a condition, write the point in terms of one parameter. The condition becomes an equation in that parameter alone.

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