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