The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y

Engineering

Mathematics

Plane and Its Different Forms

Question

The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is

14x – 2y + 15z = 27

14x + 2y – 15z = 1

14x + 2y + 15z = 31

– 14x + 2y + 15z = 3

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$∵$ Plane is perpendicular to both planes.

$∴$ normal of required plane, $\vec{n} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 3 & - 6 & - 2 \end{matrix} | = - 14 \hat{i} - 2 \hat{j} - 15 \hat{k}$

$∴$ Plane will be 14 (x – 1) + 2 (y – 1) + 15 (z – 1) = 0 ⇒ 14x + 2y + 15z = 31. Ans.