If the equation of a plane P, passing through the intersection of the planes, x + 4y - z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b in R, then the distance of the point (3, 2, -1) from the plane P is...........

Equation of plane P is (x + 4y - z + 7) + lambda (3x + y + 5z - 8) = 0 => x(1 + 3lambda ) + y(4 + lambda ) + z( - 1 + 5lambda ) + (7 - 8lambda ) \ = 0 1 + 3lambda / a = 4 + lambda / b = 5lambda - 1 / 6 = 7 - 8lambda / - 15 Therefore 15 - 75 lambda = 42 - 48 lambda => -27 = 27 lambda => lambda = -1 Therefore Plane is (x + 4y - z + 7) - 1 (3x + y + 5z - 8) \ = 0 => 2x - 3y + 6z - 15 = 0 Distance of (3, 2, -1) from the plane P = | 6 - 6 - 6 - 15 | / 7 = 21 / 7 = 3

If the equation of a plane P passing through the inters... | null (Mathematics)

Equation of plane P is

(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0

\Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) \\ = 0

{{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}

\therefore 15 - 75 \lambda

= 42 - 48 \lambda

\Rightarrow -27 = 27 \lambda

\Rightarrow \lambda = -1

\therefore \text{ Plane is } (x + 4y - z + 7) - 1 (3x + y + 5z - 8) \\ = 0

\Rightarrow 2x - 3y + 6z - 15 = 0

Distance of (3, 2, -1) from the plane P

= {{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3

Explanation

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