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