Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the direction vectors of the given lines be $\overrightarrow{d_1}=3\hat{i}+2\hat{j}+4\hat{k}$, $\overrightarrow{d_2}=4\hat{i}+3\hat{j}+2\hat{k}$, and $\overrightarrow{d_3}=2\hat{i}-4\hat{j}+3\hat{k}$.

The normal vector $\overrightarrow{n_1}$ to the plane containing the lines with direction vectors $\overrightarrow{d_2}$ and $\overrightarrow{d_3}$ is given by their cross product:

$\overrightarrow{n_1}=\overrightarrow{d_2}\times \overrightarrow{d_3}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 3& 2\\ 2& -4& 3\end{array}\right|$

$\overrightarrow{n_1}=\hat{i}(9-(-8))-\hat{j}(12-4)+\hat{k}(-16-6)=17\hat{i}-8\hat{j}-22\hat{k}$

The required plane contains the first line (which passes through the origin and has direction $\overrightarrow{d_1}$) and is perpendicular to the plane with normal $\overrightarrow{n_1}$. Therefore, the normal vector $\vec{n}$ of the required plane is perpendicular to both $\overrightarrow{d_1}$ and $\overrightarrow{n_1}$:

$\vec{n}=\overrightarrow{d_1}\times \overrightarrow{n_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 4\\ 17& -8& -22\end{array}\right|$

$\vec{n}=\hat{i}(-44-(-32))-\hat{j}(-66-68)+\hat{k}(-24-34)$

$\vec{n}=-12\hat{i}+134\hat{j}-58\hat{k}$

Dividing by $-2$ to simplify the direction ratios, we get the normal vector as $6\hat{i}-67\hat{j}+29\hat{k}$.

Since the required plane contains the first line, it must pass through the origin $(0,0,0)$. The equation of the plane is:

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

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

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