Question

Find the coordinates 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).      

Answer 1

The Cartesian equation of the line joining points B(0, −1, 3) and C(2, −3, −1) is \[\frac{x - 0}{2 - 0} = \frac{y - \left( - 1 \right)}{- 3 - \left( - 1 \right)} = \frac{z - 3}{- 1 - 3}\]

\[\text { Or  }  \frac  {   x}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{- 4}\] Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line  \[\frac{x}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{- 4}\] 

The coordinates of general point on the line  \[\frac{x}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{- 4}\]  are given by  \[\frac{x}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{- 4} = \lambda\]

\[Or \text { x } = 2\lambda, y = - 2\lambda - 1, z = - 4\lambda + 3\]

Let the coordinates of L be  \[\left( 2\lambda, - 2\lambda - 1, - 4\lambda + 3 \right)\] 

Therefore, the direction ratios of AL are proportional to  \[2\lambda - 1, - 2\lambda - 1 - 8, - 4\lambda + 3 - 4\] or  \[2\lambda - 1, - 2\lambda - 9, - 4\lambda - 1\] 

Direction ratios of the given line are proportional to 2, −2, −4.  But, AL is perpendicular to the given line.

\[\therefore 2 \times \left( 2\lambda - 1 \right) + \left( - 2 \right) \times \left( - 2\lambda - 9 \right) + \left( - 4 \right) \times \left( - 4\lambda - 1 \right) = 0\]

\[ \Rightarrow 4\lambda - 2 + 4\lambda + 18 + 16\lambda + 4 = 0\]

\[ \Rightarrow 24\lambda + 20 = 0\]

\[ \Rightarrow \lambda = - \frac{5}{6}\]

Putting , 

\[\lambda = - \frac{5}{6}\] in 

\[\left( 2\lambda, - 2\lambda - 1, - 4\lambda + 3 \right)\] 

we get , 

\[\left( 2 \times \left( - \frac{5}{6} \right), - 2 \times \left( - \frac{5}{6} \right) - 1, - 4 \times \left( - \frac{5}{6} \right) + 3 \right) = \left( - \frac{5}{3}, \frac{2}{3}, \frac{19}{3} \right)\]

Thus, the required coordinates of the foot of the perpendicular are \[\left( - \frac{5}{3}, \frac{2}{3}, \frac{19}{3} \right)\].