Question

The length of perpendicular drawn from the point (5,4, -1) to the line $\overrightarrow{r}=\hat{i}+\lambda (2\hat{i}+9\hat{j}+5\hat{k})$ is

• A. $\sqrt{\frac{2001}{112}}$
• B. $\sqrt{\frac{2309}{112}}$
• C. $\sqrt{\frac{2109}{110}}$
• D. none of these

Answer 1

The correct option is C

$\sqrt{\frac{2109}{110}}$

Let P be the foot of the perpendicular from the point A(5,4,-1) to the line (1) whose equation is
$\overrightarrow{r}=\hat{i}+\lambda (2\hat{i}+9\hat{j}+5\hat{k})......\left(1\right)$
Coordinates of P are given by $(1+2\lambda ,9\lambda ,5\lambda )$ for some value of $\lambda$.
The direction ratios of AP are
$1+2\lambda -5,9\lambda -4,5\lambda -(-1)i.e.2\lambda -4,9\lambda -4,5\lambda +1$
The direction ratios of line (1) are (2,9,5).
$\because AP\perp \left(1\right),2(2\lambda -4)+9(9\lambda -4)+5(5\lambda +1)=0\Rightarrow 4\lambda -8+81\lambda -36+25\lambda +5=0\Rightarrow 110\lambda -39=0\Rightarrow \lambda =\frac{39}{110}Now,A{P}^2=(1+2\lambda -5{)}^2+(9\lambda -4{)}^2+(5\lambda -(-1){)}^2=(2\lambda -4{)}^2+(9\lambda -4{)}^2+(5\lambda +1{)}^2=4{\lambda }^2-16\lambda +16+81{\lambda }^2-72\lambda +16+25{\lambda }^2+10\lambda +1=110{\lambda }^2-78\lambda +33\left({\frac{39}{110}}^2\right)-78\left(\frac{39}{110}\right)+33=\frac{{39}^2-78\times 39+33\times 110}{110}=\frac{2109}{110}\Rightarrow AP=\sqrt{\frac{2109}{110}}$