Let the line (x-3)/7=(y-2)/-1=(z-3)/-4 intersect the plane containing the lines (x-4)/1=(y+1)/-2=z/1 and 4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a ∈ ℝ at the point P(α, β, γ). Then the value of α+β+γ equals _____________.

Solution

Maths sir · 5 steps

Write the plane through the second line as P₁ + λ P₂ = 0, then make the first line sit inside it.

1. Family of planes. Every plane through the line 4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3 has the form (4ax - y + 5z - 7a) + λ(2x - 5y - z - 3) = 0. Its normal is (4a + 2λ, -1 - 5λ, 5 - λ).
2. Contain the other line. The line (x-4)/1=(y+1)/-2=z/1 lies in the plane only if its point (4, -1, 0) satisfies the plane and its direction (1, -2, 1) is perpendicular to the normal. The point gives 9a + 1 + 10λ = 0. The direction gives 4a + 11λ + 7 = 0.
3. Solve and simplify. From the first condition, a = (-1 - 10λ)/9. Substituting gives 59λ + 59 = 0, so λ = -1 and a = 1. The plane becomes 2x + 4y + 6z - 4 = 0, that is x + 2y + 3z = 2.

   (3 + 7t) + 2(2 - t) + 3(3 - 4t) = 2 ⇒ 16 - 7t = 2 ⇒ t = 2

4. Meet the line. A general point of (x-3)/7=(y-2)/-1=(z-3)/-4 is (3 + 7t, 2 - t, 3 - 4t). Putting it into the plane gives t = 2, so P = (17, 0, -5).
5. So. α + β + γ = 17 + 0 - 5 = 12.

The idea to keep. To make a plane contain a line, use two conditions: a point of the line satisfies the plane, and the line's direction is perpendicular to the plane's normal.

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