Question

Find the co-ordinates of the foot of perpendicular drawn from the point A (1, 8, 4) to the line joining the points B (0, -1, 3) and C (2, -3, -1)___

• A. $(\frac{-5}{3},\frac{2}{3},\frac{19}{3})$
  
• B. $(2,-3,-1)$
• C. $(1,-2,1)$
• D. $(\frac{1}{2},-\frac{3}{2},2)$

Answer 1

The correct option is A

$(\frac{-5}{3},\frac{2}{3},\frac{19}{3})$

Let L be the foot of perpendicular drawn from the points A (1, 8, 4) to the line passing through B and C as shown in the fig. The equation of line BC by using formula
$\overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a}),$ the equation of the line BC is
$\overrightarrow{r}=(-\hat{j}+3\hat{k})+\lambda (2\hat{i}-2\hat{j}-4\hat{k})\Rightarrow x\hat{i}+y\hat{i}+z\hat{k}=2\lambda \hat{i}-(2\lambda +1)\hat{j}+(3-4\lambda )\hat{k}$
Comparing both sides, we get
$x=2\lambda ,y=-(2\lambda +1),z=3-4\lambda$
Thus, the co-ordinate of L are $(2\lambda ,-(2\lambda +1),(3-4\lambda \left)\right).\left(1\right)$
So that the direction ratios of the line AL are$(1-2\lambda ),8+(2\lambda +1),4-(3-4\lambda )i.e.1-2\lambda ,2\lambda +9,1+4\lambda$
Since AL is perpendicular to BC, we have,
$(1-2\lambda )(2-0)+(2\lambda +9)(-3+1)+(4\lambda +1)(-1-3)=0$


$\Rightarrow \lambda =\frac{-5}{6}$
The required point is obtained by substituting the value of $\lambda ,$ in (1) which is $(\frac{-5}{3},\frac{2}{3},\frac{19}{3})$