Question

The coordinates of the foot of perpendicular drawn from the point $A\left( {1,8,4} \right)$ to the line joining the points $B\left( {0, - 1,3} \right)$ and $C\left( {2, - 3, - 1} \right)$
A.$\left( {\dfrac{{ - 5}}{3},\dfrac{2}{3},\dfrac{{19}}{3}} \right)$
B.$\left( {\dfrac{5}{3},\dfrac{1}{3},\dfrac{{16}}{3}} \right)$
C.$\left( {\dfrac{2}{3},\dfrac{{19}}{3},\dfrac{{16}}{3}} \right)$
D.$\left( {1,2,3} \right)$

Answer 1

Hint: First, find the equation of the lines joining the points, $B\left( {0, - 1,3} \right)$ and $C\left( {2, - 3, - 1} \right)$. Then write the direction ratios of the line $BC$. Now, find the coordinates of any general point on line $BC$ using a variable say, $\lambda $. Write the direction ratios of the perpendicular drawn from $A\left( {1,8,4} \right)$ to line $BC$. Take the dot product of direction ratios equal to 0 to find the value of $\lambda $ for the foot of perpendicular. Substitute the value of $\lambda $ in the general point of $BC$ to find the coordinates of the foot of perpendicular.

Complete step-by-step answer:
Let $D$ be the foot of perpendicular from the point $A\left( {1,8,4} \right)$ to the line joining the points $B\left( {0, - 1,3} \right)$ and $C\left( {2, - 3, - 1} \right)$
First of all, draw the diagram corresponding to the given question.

We can find the equation of line from two points. If $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ are two points on the line, then the equation of line is $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$.
Then equation of line $BC$ is,
$
  \dfrac{{x - 0}}{{2 - 0}} = \dfrac{{y - \left( { - 1} \right)}}{{ - 3 - \left( { - 1} \right)}} = \dfrac{{z - 3}}{{\left( { - 1} \right) - 3}} \\
   \Rightarrow \dfrac{x}{2} = \dfrac{{y + 1}}{{ - 2}} = \dfrac{{z - 3}}{{ - 4}} \\
 $
The direction ratio of $BC$ is $\left\langle {2, - 2, - 4} \right\rangle $
Equate the equation of $BC$ to some variable, say $\lambda $
So, let $\dfrac{x}{2} = \dfrac{{y + 1}}{{ - 2}} = \dfrac{{z - 3}}{{ - 4}} = \lambda $
Then, general point on line $BC$ is given by,
 $
  \dfrac{x}{2} = \lambda ,\dfrac{{y + 1}}{{ - 2}} = \lambda ,\dfrac{{z - 3}}{{ - 4}} = \lambda \\
  x = 2\lambda ,y = - 2\lambda - 1,z = - 4\lambda + 3 \\
  D = \left( {2\lambda , - 2\lambda - 1, - 4\lambda + 3} \right) \\
$
Find the direction ratios of $AD$.
$\left\langle {2\lambda - 1, - 2\lambda - 1 - 8, - 4\lambda + 3 - 4} \right\rangle = \left\langle {2\lambda - 1, - 2\lambda - 9, - 4\lambda - 1} \right\rangle $
As we know, $BC \bot AD$ , the dot product of direction ratios will be equal to 0.
$
  \left\langle {2, - 2, - 4} \right\rangle .\left\langle {2\lambda - 1, - 2\lambda - 9, - 4\lambda - 1} \right\rangle = 0 \\
   \Rightarrow 2\left( {2\lambda - 1} \right) - 2\left( { - 2\lambda - 9} \right) - 4\left( { - 4\lambda - 1} \right) = 0 \\
   \Rightarrow 4\lambda - 2 + 4\lambda + 18 + 16\lambda + 4 = 0 \\
   \Rightarrow 24\lambda + 20 = 0 \\
   \Rightarrow \lambda = - \dfrac{{20}}{{24}} \\
   \Rightarrow \lambda = - \dfrac{5}{6} \\
$
On substituting the value of $\lambda = - \dfrac{5}{6}$ in the general point of $D\left( {2\lambda , - 2\lambda - 1, - 4\lambda + 3} \right)$ , we get,
$
  \left( {2\left( { - \dfrac{5}{6}} \right), - 2\left( { - \dfrac{5}{6}} \right) - 1, - 4\left( { - \dfrac{5}{6}} \right) + 3} \right) \\
   = \left( { - \dfrac{5}{3},\dfrac{5}{3} - 1,\dfrac{{10}}{3} + 3} \right) \\
   = \left( { - \dfrac{5}{3},\dfrac{2}{3},\dfrac{{19}}{3}} \right) \\
$
Thus, the coordinates of foot of perpendicular are $\left( {\dfrac{{ - 5}}{3},\dfrac{2}{3},\dfrac{{19}}{3}} \right)$.
Hence, option A is the correct option.

Note: If $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ are two points on the line, then the equation of line is $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$. The dot product of direction ratios of two perpendicular lines is always zero. Also, the direction ratios of two parallel lines are the same.