Find the equation of the plane perpendicular | vedclass

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(NONE) Let the lines be $L_1, L_2, L_3$ with direction vectors $\vec{v_1} = (2, 3, 4)$,$\vec{v_2} = (3, 4, 2)$,and $\vec{v_3} = (4, 2, 3)$ respectivel

Vedclass · Accepted Answer

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

Vedclass · Answer

(NONE) Let the lines be $L_1, L_2, L_3$ with direction vectors $\vec{v_1} = (2, 3, 4)$,$\vec{v_2} = (3, 4, 2)$,and $\vec{v_3} = (4, 2, 3)$ respectively.
All these lines pass through the origin $(0, 0, 0)$.
The plane containing $L_1$ and $L_2$ has a normal vector $\vec{n_1} = \vec{v_1} 	imes \vec{v_2} = (-10, 8, -1)$.
The plane containing $L_1$ and $L_3$ has a normal vector $\vec{n_2} = \vec{v_1} 	imes \vec{v_3} = (1, 10, -8)$.
The plane perpendicular to both these planes will have a normal vector $\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -10 & 8 & -1 \ 1 & 10 & -8 \end{vmatrix} = \hat{i}(-64+10) - \hat{j}(80+1) + \hat{k}(-100-8) = (-54, -81, -108)$.
Dividing by $-27$,we get the normal vector $(2, 3, 4)$.
Since the plane passes through the origin $(0, 0, 0)$,the equation is $2x + 3y + 4z = 0$.

Find the equation of the plane perpendicular to the plane containing the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$.

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