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Determine the Equations of the Line Passing Through the Point (1, 2, −4) and Perpendicular to the Two Lines X − 8 8 = Y + 9 − 16 = Z − 10 7 a N D X − 15 3 = Y − 29 8 = Z − 5 − 5

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प्रश्न

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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उत्तर

\[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7}\]

   \[ \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]

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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अध्याय 27: Straight Line in Space - Exercise 28.2 [पृष्ठ १७]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Exercise 28.2 | Q 18 | पृष्ठ १७

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