Let the point, on the line passing through the points P(1,-2,3) and Q(5,-4,7), farther from the origin and at a distance of 9 units from the point P, be (α, β, γ). Then α²+β²+γ² is equal to :

Solution

Maths sir · 2 steps

Turn PQ⃗ into a unit vector, step 9 units both ways from P, and keep the point farther from O.

1. Direction of the line. PQ⃗ = (5-1, -4+2, 7-3) = (4, -2, 4), and its length is √(16+4+16) = 6. So the unit vector along the line is (2/3, -1/3, 2/3).

   (α, β, γ) = (1, -2, 3) ± 9(2/3, -1/3, 2/3) = (1, -2, 3) ± (6, -3, 6)

   Sign	Point	α²+β²+γ²
   +	(7, -5, 9)	49 + 25 + 81 = 155
   -	(-5, 1, -3)	25 + 1 + 9 = 35

2. So. The sum of squares is the square of the distance from the origin, so the farther point is (7, -5, 9). That gives α²+β²+γ² = 155, option (b).

The idea to keep. A point at distance d from P along a line is P ± d u, and there are always two of them. Normalise the direction first, then use the extra condition to pick one.

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