A plane which is perpendicular to two planes | vedclass

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(B) $1$. Let the equation of the plane passing through $(1, 2, 1)$ be $a(x - 1) + b(y - 2) + c(z - 1) = 0$.
$2$. The plane is perpendicular to the pl

Vedclass · Accepted Answer

$\sqrt{2}$ units

Vedclass · Answer

(B) $1$. Let the equation of the plane passing through $(1, 2, 1)$ be $a(x - 1) + b(y - 2) + c(z - 1) = 0$.
$2$. The plane is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The normal vectors are $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$.
$3$. The normal vector $\vec{n}$ of the required plane is $\vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
$4$. We can take the normal vector as $(1, 1, 0)$. The equation of the plane is $1(x - 1) + 1(y - 2) + 0(z - 1) = 0$,which simplifies to $x + y - 3 = 0$.
$5$. The distance $d$ from the point $(2, 3, 4)$ to the plane $x + y - 3 = 0$ is $d = \frac{|2 + 3 - 3|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ units.

$A$ plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$ passes through $(1, 2, 1)$. The distance of the plane from the point $(2, 3, 4)$ is

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