Question

The length of the perpendicular (in units) from the point $\left(1,2,4\right)$ on the straight line $\frac{x-2}{1}=\frac{y-7}{2}=\frac{z-3}{-1}$ lies in the interval

• A. $\left(1,\frac{3}{2}\right)$
• B. $\left(2,3\right)$
• C. $(0,2]$
• D. $[4,5)$

Answer 1

Answer: (C)

Let, point $A$ is $\left(1,2,4\right)$ and any general point on the line is $B\left(\lambda +2,2\lambda +7,-\lambda +3\right)$ ,
$\overrightarrow{\mathrm{AB}}=\left(\lambda +1\right)\hat{i}+\left(2\lambda +5\right)\hat{j}+\left(-\lambda -1\right)\hat{k}$
and a vector parallel to the line is $\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$
$\overrightarrow{\mathrm{AB}}\cdot \overrightarrow{b}=0\Rightarrow \lambda +1+4\lambda +10+\lambda +1=0$
$\Rightarrow \lambda =-2$
The length of the perpendicular $=\left|\overrightarrow{\mathrm{AB}}\right|$
$=\sqrt{{\left(-2+1\right)}^2+{\left(-4+5\right)}^2+{\left(2-1\right)}^2}=\sqrt{3}$ units