A plane P is parallel to two lines whose direction ratios are -2,1,-3 and -1,2,-2 and it contains the point (2,2,-2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts α, β, γ. If V is the volume of the tetrahedron OABC, where O is the origin, and p=α+β+γ, then the ordered pair (V, p) is equal to :

Solution

Maths sir · 5 steps

The plane is parallel to both lines, so its normal is their cross product. Find that first.

1. Find the normal. A normal to the plane is perpendicular to both direction vectors. So take the cross product of (-2, 1, -3) and (-1, 2, -2).

   n⃗ = |î, ĵ, k̂; -2, 1, -3; -1, 2, -2| = (-2+6) î - (4-3) ĵ + (-4+1) k̂ = 4î - ĵ - 3k̂

2. Write the plane. Through (2, 2, -2): 4(x-2) - (y-2) - 3(z+2) = 0, which gives 4x - y - 3z = 12. Check: 8 - 2 + 6 = 12.
3. Read the intercepts. Divide by 12 to get x/3 + y/-12 + z/-4 = 1. So α = 3, β = -12, γ = -4, and p = 3 - 12 - 4 = -13.
4. Tetrahedron volume. OA, OB, OC are mutually perpendicular edges, so V = 1/6|αβγ| = 1/6 × 3 × 12 × 4 = 24.
5. So. (V, p) = (24, -13), which is option (b).

The idea to keep. Plane parallel to two lines means normal = cross product of their directions. For a plane cutting the axes, V = 1/6|αβγ|, but keep the signs of the intercepts when you add them.

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