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

Equation of first line, $\begin{array}{l}\frac{x-6}{1}=\frac{y-2}{-2}=\frac{z-2}{2}=k\text{ (say) }\\ \therefore x=k+6,y=-2k+2,z=2k+2\end{array}$ Hence, general point on the first line, $\mathrm{P}\equiv (\mathrm{k}+6,-2\mathrm{k}+2,2\mathrm{k}+2)$ Equation of second line, $\begin{array}{l}\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}}{-2}=\frac{\mathrm{Z}+1}{-2}=l\text{ (say) }\\ \therefore \mathrm{x}=3l-4,\mathrm{y}=-2l,\mathrm{z}=-2l-1\end{array}$ Hence, general point on the second line, $Q\equiv (3l-4,-2l,-2l-1)$ Direction ratios of $\mathrm{PQ}$ are $3l-4-\mathrm{k}-6,-2l+2\mathrm{k}-2,-2l-1-2\mathrm{k}-2$ i.e. $3l-\mathrm{k}-10,-2l+2\mathrm{k}-2,-2l-2\mathrm{k}-3$ Now $|\mathrm{PQ}|$ will be the shortest distance between the two lines if PQ is perpendicular to both the lines. Hence, $\begin{array}{l}1(3l-\mathrm{k}-10)+(-2)\\ (-2l+2\mathrm{k}-2)+2(-2l-2\mathrm{k}-3)=0\end{array}$ and $3(3l-\mathrm{k}-10)+(-2)(-2l$ $+2\mathrm{k}-2)+(-2)(-2\mathrm{l}-2\mathrm{k}-3)=0$ i.e. $3l-9\mathrm{k}=12$ or $l-3\mathrm{k}=4$ ...(i) and $17l-3\mathrm{k}=20$ ...(ii) Subtracting equation (i) from (ii), we get $16\mathrm{l}=16$ $\therefore l=1$ Putting this value of 1 in equation (i), we get $\begin{array}{rl}-3\mathrm{k}& =3,\therefore \mathrm{k}=-1\\ \therefore \mathrm{P}& \equiv (-1+6,-2(-1)+2,2(-1)+2)\\ & \equiv (5,4,0)\end{array}$ Similarly, $\mathrm{Q}=(-1,-2,-3)$ Hence, shortest distance, PQ, $\begin{array}{l}=\sqrt{(-1-5{)}^2+(-2-4{)}^2+(-3-0{)}^2}\\ =\sqrt{(-6{)}^2+(-6{)}^2+(-3{)}^2}=\sqrt{36+36+9}\\ =9\text{ units }\end{array}$