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Question

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{3}=\frac{y}{2}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular

Vedclass · Accepted Answer

$6x - 67y + 29z = 0$

Vedclass · Answer

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{3}=\frac{y}{2}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular to $\vec{v}_1 = (3, 2, 4)$.Let $P_2$ be the plane containing $L_2: \frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $L_3: \frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$.The normal vector $\vec{n}_2$ of $P_2$ is $\vec{v}_2 	imes \vec{v}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 3 & 2 \ 2 & -4 & 3 \end{vmatrix} = \hat{i}(9+8) - \hat{j}(12-4) + \hat{k}(-16-6) = 17\hat{i} - 8\hat{j} - 22\hat{k}$.Since $P_1 \perp P_2$,$\vec{n}_1$ is perpendicular to $\vec{n}_2 = (17, -8, -22)$.Thus,$\vec{n}_1 = \vec{v}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 4 \ 17 & -8 & -22 \end{vmatrix} = \hat{i}(-44+32) - \hat{j}(-66-68) + \hat{k}(-24-34) = -12\hat{i} + 134\hat{j} - 58\hat{k}$.Dividing by $-2$,we get the normal vector $(6, -67, 29)$.The equation of the plane is $6x - 67y + 29z = 0$.

The equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is:

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