Question

Let $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{l}$ and $\overrightarrow{r}=\overrightarrow{b}+\mu \overrightarrow{m}$ be two lines in space where $\overrightarrow{a}=5\hat{i}+\hat{j}+2\hat{k}$. $\overrightarrow{b}=-\hat{i}+7\hat{j}+8\hat{k}$, $\overrightarrow{l}=4\hat{i}+\hat{j}-\widehat{k]and\overrightarrow{m}=2\hat{i}-5\hat{j}-7\hat{(}k}$ then the p.v. of a point which lies on both of these lines, is

• A. $\hat{i}+2\hat{j}+\hat{k}$
• B. $2\hat{i}+\hat{j}+\hat{k}$
• C. $\hat{i}+\hat{j}+2\hat{k}$
• D. non existent as the lines are skew

Answer 1

The correct option is D non existent as the lines are skew

$1^{st}$ line $\overline{r}=\overline{a}+\lambda \overline{l}$

$\frac{x-5}{4}=\frac{y-1}{1}=\frac{z-2}{-1}=RR$ is constant

$2^{nd}$ line $\overline{r}=\overline{b}+\mu \overline{m}$

$\frac{x+1}{2}=\frac{y-7}{-5}=\frac{z-8}{-7}=\mu \mu$ is constant

$\therefore x=4R+5y=R+1z=R+2$

$x=2\mu -1y=-5\mu +7z=-7\mu +8$

$\therefore 4R+5=2\mu -\mathrm{1....}\left(1\right)$

$R+1=-5\mu +\mathrm{7....}\left(2\right)$

$-R+2=-7\mu +\mathrm{8....}\left(3\right)$

From $\left(1\right)$ & $\left(2\right)$

$\frac{4R-2\mu =-6+R+5\mu =6}{5R+3\mu =0}\therefore R=-\frac{3\mu }{5}$

$\therefore$ Substituting $R=-\frac{3\mu }{5}$ in $\left(3\right)$

$LHS=-R+2=-(-\frac{3\mu }{5})+2=\frac{3\mu }{5}+2$

$RHS=-7\mu +8\therefore LHS\ne RHS$

$\therefore$ Lines do not intresed

$\therefore$ They are skew lines.