If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x-3 y+ 2 z-1=0=4 x-y+z is A x+B y+C z=1, then 140(C-B+A) is equal to ___________.

Solution

Maths sir · 4 steps

The line is the meeting of two planes, so cross their normals. That direction is your plane's normal.

1. Direction of the line. The line lies in both planes, so it is perpendicular to both normals n⃗₁ = (1, -3, 2) and n⃗₂ = (4, -1, 1). Its direction is their cross product.

   n⃗₁ × n⃗₂ = |î, ĵ, k̂; 1, -3, 2; 4, -1, 1| = (-3+2) î - (1-8) ĵ + (-1+12) k̂ = -î + 7ĵ + 11k̂

2. Write the plane. The plane is perpendicular to the line, so (-1, 7, 11) is its normal. Through (1, 1, 2) the plane is -x + 7y + 11z = -1 + 7 + 22 = 28.
3. Match the form. Divide by 28 to make the right side 1. This gives A = -1/28, B = 7/28, C = 11/28.
4. So. C - B + A = (11 - 7 - 1)/28 = 3/28, and 140 × 3/28 = 15.

The common slip: 170. Dropping the minus sign on the ĵ term gives the normal (-1, -7, 11), the plane -x - 7y + 11z = 14, and 170 instead of 15.

The idea to keep. A line given as two plane equations has direction n⃗₁ × n⃗₂. Watch the minus sign on the middle ĵ term of the determinant.

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