Let the equation of the plane P containing the line x+10=(8-y)/2=z be ax+by+3z=2(a+b) and the distance of the plane P from the point (1, 27, 7) be c. Then a²+b²+c² is equal to __________.

Solution

Maths sir · 4 steps

Containing a line means two things: the normal is perpendicular to it, and its point lies on the plane.

1. Read the line. Put x+10 = (8-y)/2 = z = t, so x = t-10, y = 8-2t, z = t. The line passes through (-10, 8, 0) with direction (1, -2, 1), where the minus sign comes from 8-y.
2. Two conditions. The normal (a, b, 3) is perpendicular to (1, -2, 1), so a - 2b + 3 = 0. The point (-10, 8, 0) lies on the plane, so -10a + 8b = 2a + 2b, which gives b = 2a.
3. Find the plane. Substituting gives a - 4a + 3 = 0, so a = 1 and b = 2. The plane is x + 2y + 3z = 6.

   c = (|1 + 2(27) + 3(7) - 6|)/√(1² + 2² + 3²) = 70/√14 = 5√14

4. So. c² = 25 × 14 = 350, and a² + b² + c² = 1 + 4 + 350 = 355.

The idea to keep. For a plane containing a line, write two conditions: normal dot direction is zero, and the line's point satisfies the plane. Rewrite the line in standard form first to get the signs right.

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