Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

We have lines $y-2=0$ and $4x-3z+5=0$ $$\begin{array}{rl}& \therefore 4x=3z-5=3\left[z-\left(\frac{5}{3}\right)\right]\\ & \therefore \frac{4x}{12}=\frac{3\left[z-\left(\frac{5}{3}\right)\right]}{12}\Rightarrow \frac{x}{3}=\frac{z-\left(\frac{5}{3}\right)}{4},y=2\end{array}$$ Thus line passes through the point $\left(0,2,\frac{5}{3}\right)$ i.e. a point having position vector $2\hat{\mathrm{j}}+\frac{5}{3}\hat{\mathrm{k}}$ Also direction ratios of a line are $3,0,4$ Hence required vector equation is $$\overline{\mathrm{r}}=\left(2\hat{\mathrm{j}}+\frac{5}{3}\hat{\mathrm{k}}\right)+\lambda (3\hat{\mathrm{i}}+4\hat{\mathrm{k}})$$