A plane passes through the point (1, -2, 1) a | vedclass

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(A) Let the normal vector of the required plane be $\vec{n} = (a, b, c)$.Since the plane is perpendicular to the planes with normal vectors $\vec{n_1}

Vedclass · Accepted Answer

$2\sqrt{2}$

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(A) Let the normal vector of the required plane be $\vec{n} = (a, b, c)$.Since the plane is perpendicular to the planes with normal vectors $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$,the vector $\vec{n}$ is parallel to $\vec{n_1} 	imes \vec{n_2}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, 0)$.The equation of the plane passing through $(1, -2, 1)$ is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.The distance of the plane $x + y + 1 = 0$ from the point $(1, 2, 2)$ is given by $d = \frac{|1(1) + 1(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

$A$ plane passes through the point $(1, -2, 1)$ and is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. Find the distance of the plane from the point $(1, 2, 2)$.

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