Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The DR's of normal to the plane containing $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. $\Rightarrow {\overrightarrow{\mathbf{n}}}_1=\left|\begin{array}{ccc}\hat{\mathbf{i}}& \hat{\mathbf{j}}& \hat{\mathbf{k}}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|=(8\hat{\mathbf{i}}-\hat{\mathbf{j}}-10\hat{\mathbf{k}})$ Also, equation of plane containing $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and DR's of normal to be ${\overrightarrow{\mathbf{n}}}_2=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}+c\hat{\mathbf{k}}$
$$\begin{array}{ll}\Rightarrow & ax+by+cz=0\\ \text{ where, }{\overrightarrow{\mathbf{n}}}_1\cdot {\overrightarrow{\mathbf{n}}}_2=0& \\ \Rightarrow & 8a-b-10c=0\\ \text{ and }& {\overrightarrow{\mathbf{n}}}_2\perp (2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}})\\ \Rightarrow & 2a+3b+4c=0\end{array}$$ From Eqs. (ii) and (iii), we get $$\begin{array}{rl}& \frac{a}{-1-10}=\frac{b}{8}=\frac{c}{-1}\\ \Rightarrow \frac{a}{-4+30}& =\frac{b}{-20-32}=\frac{c}{24+2}\\ \Rightarrow \frac{a}{26}& =\frac{b}{-52}=\frac{c}{26}\\ \Rightarrow \frac{a}{1}& =\frac{b}{-2}=\frac{c}{1}\end{array}$$ From Eqs. (i) and (iv), required equation of plane, is $x-2y+z=0$.