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

$\vec{r}=3\vec{i}+8\vec{j}+3\vec{k}+\lambda (3\vec{i}-\vec{j}+\vec{k})$.............(i) $\overrightarrow{\mathrm{r}}=-3\overrightarrow{\mathrm{i}}-7\overrightarrow{\mathrm{j}}+6\overrightarrow{\mathrm{k}}+\mu (-3\overrightarrow{\mathrm{i}}+2\overrightarrow{\mathrm{j}}+4\overrightarrow{\mathrm{k}})$..............(ii)

Let L and M be points on the line (i) and (ii) respectively

So that LM is perpendicular to both the lines

Let position vector of $L$ be $3\vec{i}+8\vec{j}+3\vec{k}+{\lambda }_0(3\vec{i}-\vec{j}+\vec{k})$

and the position vector of $M$ be $-3\vec{i}-7\vec{j}+6\vec{k}+{\mu }_0(-3\vec{i}+2\vec{j}+4\vec{k})$ then $\overrightarrow{\mathrm{LM}}=-6\overrightarrow{\mathrm{i}}-15\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}-{\lambda }_0(3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})+{\mu }_0(-3\overrightarrow{\mathrm{i}}+2\overrightarrow{\mathrm{j}}+4\overrightarrow{\mathrm{k}})$ since $\overrightarrow{\mathrm{LM}}$ is perpendicular to both the lines (i) and (ii)

$\therefore \overrightarrow{\mathrm{LM}}\cdot (3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})=0$ and $\overrightarrow{\mathrm{LM}}\cdot (-3\overrightarrow{\mathrm{i}}+2\overrightarrow{\mathrm{j}}+4\overrightarrow{\mathrm{k}})=0$

Thus $-18+15+3-{\lambda }_0(9+1+1)+{\mu }_0(-9-2+4)=0$

i.e. $-11{\lambda }_0-7{\mu }_0=0$ ............(iii)

and $18-30+12-{\lambda }_0(-9-2+4)+{\mu }_0(9+4+16)=0$

i.e. $7{\lambda }_0+29{\mu }_0=0$ ............(iv)

from (iii) and (iv) we get

${\lambda }_0={\mu }_0=0$ $\therefore \overrightarrow{\mathrm{LM}}=-6\overrightarrow{\mathrm{i}}-15\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}$ $\therefore |\overrightarrow{\mathrm{LM}}|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}$

Position vector of $L$ is $3\vec{i}+8\vec{j}+3\vec{k}$

$\therefore$ equation of the line of shortest distance (LM) is $\overrightarrow{\mathrm{r}}=3\overrightarrow{\mathrm{i}}+8\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}+\lambda (-6\overrightarrow{\mathrm{i}}-15\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}})$

$\overrightarrow{\mathrm{r}}=3\overrightarrow{\mathrm{i}}+8\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}+\lambda (-6\overrightarrow{\mathrm{i}}-15\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}})$