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Find the Equation of the Line Passing Through the Point (1, −1, 1) and Perpendicular to the Lines Joining the Points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1). - Mathematics

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Question

Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).

Sum
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Solution

The direction ratios of the line joining the points (4, 3, 2), (1, -1, 0) and (1, 2, -1), (2, 1, 1) are -3,-4,-2 and 1, -1,2 ,respectively. 

Let:

\[\overrightarrow{b_1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k} \]

\[ \overrightarrow{b_2} = \hat{i} - \hat{j} + 2 \hat{k}\] 
Since the required line is perpendicular to the lines parallel to the vectors 

\[\overrightarrow{b_1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k}  \text{ and } \overrightarrow{b_2} = \hat{i} - \hat{j} + 2 \hat{k} \] it is parallel to the vector

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

Now,

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

\[ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 3 & - 4 & - 2 \\ 1 & - 1 & 2\end{vmatrix}\]

\[ = - 10 \hat{i} + 4 \hat{j} + 7 \hat{k}\]

So, the direction ratios of the required line are proportional to

-10, 4, 7.

The equation of the required line passing through the point (1,-1, 1) and having direction ratios proportional to -10, 4, 7 is 

\[\frac{x - 1}{- 10} = \frac{y + 1}{4} = \frac{z - 1}{7}\]

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Chapter 28: Straight Line in Space - Exercise 28.2 [Page 17]

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RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.2 | Q 17 | Page 17

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