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Question
Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) .\] Also, find the coordinates of the foot of the perpendicular from P.
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Solution
Let L be the foot of the perpendicular drawn from the point P ( -1, 3, 2) to the line \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) .\]

Let the position vector L be \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 2\lambda \hat{i} + \left( 2 + \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \] ...........(1)
Now,
\[\overrightarrow{PL} = \text{ Position vector of L - Position vector of P } \]
\[ \Rightarrow \overrightarrow{PL} = \left\{ 2\lambda \hat{i} + \left( 2 + \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \right\} - \left( - \hat{i} + 3 \hat{j} + 2 \hat{k} \right)\]
\[ \Rightarrow \overrightarrow{PL} = \left( 2\lambda + 1 \right) \hat{i} + \left( \lambda - 1 \right) \hat{j} + \left( 3\lambda + 1 \right) \hat{k} . . . (2)\]
Since \[\overrightarrow{PL}\] is perpendicular to the given line, which is parallel to \[\overrightarrow{b} = 2 \hat{i} + \hat{j} + 3 \hat{k} \]
we have ,
\[\overrightarrow{PL} . \overrightarrow{b} = 0\]
\[ \Rightarrow \left\{ \left( 2\lambda + 1 \right) \hat{i} + \left( \lambda - 1 \right) \hat{j} + \left( 3\lambda + 1 \right) \hat{k} \right\} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 0 \]
\[ \Rightarrow 2\left( 2\lambda + 1 \right) + 1\left( \lambda - 1 \right) + 3\left( 3\lambda + 1 \right) = 0\]
\[ \Rightarrow \lambda = - \frac{2}{7}\]
Substituting \[ \Rightarrow \lambda = - \frac{2}{7}\] in (1),
we get the position vector of L as \[- \frac{4}{7} \hat{i} + \frac{12}{7} \hat{j} + \frac{15}{7} \hat{k} \] So, the coordinates of the foot of the perpendicular from P to the given line is L \[\left( - \frac{4}{7}, \frac{12}{7}, \frac{15}{7} \right)\]
Substituting \[\lambda = - \frac{2}{7}\] in (2), we get \[\overrightarrow{PL} = \frac{3}{7} \hat{i} - \frac{9}{7} \hat{j} + \frac{1}{7} \hat{k} \]
Equation of the perpendicular drawn from P to the given line is
\[\overrightarrow{r} = \text{ Position vector of P } + \lambda\left( \vec{PL} \right)\]
\[ = \left( - \hat{i} + 3 \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i} - 9 \hat{j} + \hat{k} \right)\]
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