Question

The distance of point of intersection of lines $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}$ and $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ from $(1,-4,7)$ is

• A. $\sqrt{15}$
• B. $\sqrt{27}$
• C. $\sqrt{26}$
• D. $\sqrt{14}$

Answer 1

The correct option is C $\sqrt{26}$

$\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}={\lambda }_1$
and $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}={\lambda }_2$
For intersection point,
$4+{\lambda }_1=1+2{\lambda }_2$
$\Rightarrow {\lambda }_1-2{\lambda }_2=-3\cdots \left(1\right)$
$-3-4{\lambda }_1=-1-3{\lambda }_2$
$\Rightarrow 4{\lambda }_1-3{\lambda }_2=-2\cdots \left(2\right)$
$-1+7{\lambda }_1=-10+8{\lambda }_2$
$\Rightarrow 7{\lambda }_1-8{\lambda }_2=-9\cdots \left(3\right)$

From equations $\left(1\right)$ and $\left(2\right),$ we get
${\lambda }_1=1,{\lambda }_2=2$
which satisfies equation $\left(3\right).$
$\Rightarrow$ Point of intersection is $(4+1,-3-4,-1+7)$
i.e., $(5,-7,6)$
Distance of $(5,-7,6)$ from $(1,-4,7)$ is $\sqrt{16+9+1}=\sqrt{26}$