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Find the Equation of the Perpendicular Drawn from Point P (−1, 3, 2) to Line → R = ( 2 ^ J + 3 ^ K ) + λ ( 2 ^ I + ^ J + 3 ^ K ) . Also, Find the Coordinates of Foot of the Perpendicular from P. - Mathematics

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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.

Sum
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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 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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Chapter 28: Straight Line in Space - Exercise 28.4 [Page 30]

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RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.4 | Q 9 | Page 30

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