Let a line with direction ratios a,-4 a,-7 be perpendicular to the lines with direction ratios 3,-1,2 b and b, a,-2. If the point of intersection of the line (x+1)/(a²+b²)=(y-2)/(a²-b²)=z/1 and the plane x-y+z=0 is (α, β, γ), then α+β+γ is equal to _________.

Solution

Maths sir · 4 steps

Two perpendicular conditions give two equations in a and b, and the line only needs a² and b².

1. First perpendicularity. Perpendicular lines have zero dot product of direction ratios. With (3,-1,2b) this gives 3a + 4a - 14b = 0, so 7a = 14b and a = 2b.
2. Second perpendicularity. With (b, a, -2) the dot product gives ab - 4a² + 14 = 0. Put a = 2b: 2b² - 16b² + 14 = 0, so b² = 1 and a² = 4.
3. Write the line. Now a² + b² = 5 and a² - b² = 3. Set each ratio equal to t: a general point is (5t - 1, 3t + 2, t). The signs of a and b do not matter, since only their squares appear.

   (5t - 1) - (3t + 2) + t = 0 ⇒ 3t = 3 ⇒ t = 1

4. So. The point is (α, β, γ) = (4, 5, 1), and α + β + γ = 10.

The idea to keep. Perpendicular lines mean the dot product of their direction ratios is zero. To meet a plane, write the line's general point in terms of t and put it into the plane equation.

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