A plane E is perpendicular to the two planes | vedclass

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(C) The normal vectors of the given planes are $\vec{n}_1 = (2, -2, 1)$ and $\vec{n}_2 = (1, -1, 2)$.Since plane $E$ is perpendicular to both,its norm

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) The normal vectors of the given planes are $\vec{n}_1 = (2, -2, 1)$ and $\vec{n}_2 = (1, -1, 2)$.Since plane $E$ is perpendicular to both,its normal vector $\vec{n}_3$ is $\vec{n}_1 	imes \vec{n}_2$.$\vec{n}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4+1) - \hat{j}(4-1) + \hat{k}(-2+2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, 0)$.The equation of plane $E$ passing through $P(1, -1, 1)$ is $1(x-1) + 1(y+1) + 0(z-1) = 0$,which simplifies to $x + y = 0$.The distance of $Q(a, a, 2)$ from $x + y = 0$ is $\frac{|a + a|}{\sqrt{1^2 + 1^2}} = \frac{|2a|}{\sqrt{2}} = |a|\sqrt{2}$.Given $|a|\sqrt{2} = 3\sqrt{2}$,so $|a| = 3$,which means $a = \pm 3$.If $a = 3$,$Q = (3, 3, 2)$. Then $PQ^2 = (3-1)^2 + (3+1)^2 + (2-1)^2 = 2^2 + 4^2 + 1^2 = 4 + 16 + 1 = 21$.If $a = -3$,$Q = (-3, -3, 2)$. Then $PQ^2 = (-3-1)^2 + (-3+1)^2 + (2-1)^2 = (-4)^2 + (-2)^2 + 1^2 = 16 + 4 + 1 = 21$.Thus,$(PQ)^2 = 21$.

$A$ plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$,then $(PQ)^2$ is equal to

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