The equation of the plane passing through the point ( 1 , 2 , 1 ) and perpendicular to the planes x + 2 y + 2 z − 7 = 0 and 3 x + 3 y + 2 z − 5 = 0 is

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Let the normal vector to the required plane be $n$ .

Since the required plane is perpendicular to the given planes, its normal vector is perpendicular to the normal vectors of the given planes, $n_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $n_{2} = 3 \hat{i} + 3 \hat{j} + 2 \hat{k}$ .

$n = n_{1} \times n_{2} = \hat{i} 13 \hat{j} 23 \hat{k} 22$

$n = \hat{i} (4 - 6) - \hat{j} (2 - 6) + \hat{k} (3 - 6) = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k}$

The equation of the plane passing through $(1, 2, 1)$ with normal vector $n$ is:

$- 2 (x - 1) + 4 (y - 2) - 3 (z - 1) = 0$

$- 2 x + 2 + 4 y - 8 - 3 z + 3 = 0$

$- 2 x + 4 y - 3 z - 3 = 0$

$2 x - 4 y + 3 z + 3 = 0$

Answer: $2 x - 4 y + 3 z + 3 = 0$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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