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Find the Equations of the Line Passing Through the Point (2, 1, 3) and Perpendicular to the Lines X − 1 1 = Y − 2 2 = Z − 3 3 a N D X − 3 = Y 2 = Z 5 - Mathematics

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Question

Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines  \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{  and  } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]

Sum
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Solution

Let:

\[\overrightarrow{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}  \]

\[ \overrightarrow{b_2} = - 3 \hat{i} + 2 \hat{j} + 5 \hat{k} \]

Since the required line is perpendicular to the lines parallel to the vectors  \[\overrightarrow{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}  \text { and }  \overrightarrow{b_2} = - 3 \hat{i}  + 2 \hat{j}  + 5 \hat{k}  \]  it is parallel to the vector  \[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}\]  Now, 

\[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2} \]

\[ = \begin{vmatrix}\hat{i} & \hat{j}  & \hat{k} \\ 1 & 2 & 3 \\ - 3 & 2 & 5\end{vmatrix}\]

\[ = 4 \hat{i} - 14 \hat{j} + 8 \hat{k} \]

\[ = 2\left( 2 \hat{i} - 7 \hat{j} + 4 \hat{k}  \right)\]

Thus, the direction ratios of the required line are proportional to 2, -7, 4. The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2,-7, 4 is \[\frac{x - 2}{2} = \frac{y - 1}{- 7} = \frac{z - 3}{4}\]

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Chapter 28: Straight Line in Space - Exercise 28.2 [Page 16]

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RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.2 | Q 15 | Page 16

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