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

Any point on the line $\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}=t_1$ (say) is $\left(t_1,t_1-a,t_1\right)$ and any point on the line $\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}=t_2($ say $)$ is $\left(2t_2-a,t_2,t_2\right)$. Now direction cosine of the lines intersecting the above lines is proportional to $\left(2t_2-a-t_1,t_2-t_1+a,t_2-t_1\right)$. Hence $2{\mathrm{t}}_2-\mathrm{a}-{\mathrm{t}}_1=2\mathrm{k},{\mathrm{t}}_2-{\mathrm{t}}_1+\mathrm{a}=\mathrm{k}$ and ${\mathrm{t}}_2-{\mathrm{t}}_1=2\mathrm{k}$ On solving these, we get $t_1=3a,t_2=a$. Hence points are $(3a,2a,3a)$ and $(a,a,a)$