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Question
Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{ and } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]
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Solution
\[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7}\]
\[ \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]
Let:
\[\overrightarrow{b_1} = 8 \hat{i} - 16 \hat{j} + 7 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k}\]
Since the required line is perpendicular to the lines parallel to the vectors
\[\overrightarrow{b_1} = 8 \hat{i} - 16 \hat{j} + 7 \hat{k} \text{ and } \overrightarrow{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \] it is parallel to the vector
\[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}\]
Now,
\[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2} \]
\[ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 8 & - 16 & 7 \\ 3 & 8 & - 5\end{vmatrix}\]
\[ = 24 \hat{i} + 61 \hat{j} + 112 \hat{k} \]
The direction ratios of the required line are proportional to 24, 61, 112. The equation of the required line passing through the point (1, 2,-4) and having direction ratios proportional to 24, 61, 112 is
\[\frac{x - 1}{24} = \frac{y - 2}{61} = \frac{z + 4}{112}\]
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