Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ be $P_1$.

The direction vectors of these lines are ${\vec{d}}_2=3\hat{i}+4\hat{j}+2\hat{k}$ and ${\vec{d}}_3=4\hat{i}+2\hat{j}+3\hat{k}$.

The normal vector to the plane $P_1$ is given by their cross product:
${\vec{n}}_1={\vec{d}}_2\times {\vec{d}}_3=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|=\hat{i}(12-4)-\hat{j}(9-8)+\hat{k}(6-16)=8\hat{i}-\hat{j}-10\hat{k}$

Let the required plane be $P$. It contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$, which passes through the origin $(0,0,0)$ and has the direction vector ${\vec{d}}_1=2\hat{i}+3\hat{j}+4\hat{k}$.

Since plane $P$ is perpendicular to plane $P_1$, its normal vector $\vec{n}$ must be perpendicular to both ${\vec{d}}_1$ and ${\vec{n}}_1$.

$\vec{n}={\vec{d}}_1\times {\vec{n}}_1=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 8& -1& -10\end{array}\right|=\hat{i}(-30+4)-\hat{j}(-20-32)+\hat{k}(-2-24)=-26\hat{i}+52\hat{j}-26\hat{k}$

Dividing by $-26$, the direction ratios of the normal to the required plane are $(1,-2,1)$.

Since the plane $P$ passes through the origin $(0,0,0)$, its equation is:
$1(x-0)-2(y-0)+1(z-0)=0$

$x-2y+z=0$

Answer: $x-2y+z=0$