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Find the Angle Between the Following Pairs of Lines: − X + 2 − 2 = Y − 1 7 = Z + 3 − 3 a N D X + 2 − 1 = 2 Y − 8 4 = Z − 5 4

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प्रश्न

Find the angle between the following pair of line:

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]

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उत्तर

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  }\frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]

The equations of the given lines can be re-written as

\[\frac{x - 2}{2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{y - 4}{2} = \frac{z - 5}{4}\]

Let

\[\overrightarrow{b_1}\] and \[\overrightarrow{b_2}\] be vectors parallel to the given lines.

Now,

\[\overrightarrow{b_1} = 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \]

\[ \overrightarrow{b_2} = - 1 \hat{i} + 2 \hat{j} + 4 \hat{k}\]

If θ   is the angle between the given lines, then

\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]

\[ = \frac{\left( 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \right) . \left( - 1 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)}{\sqrt{2^2 + 7^2 + \left( - 3 \right)^2} \sqrt{\left( - 1 \right)^2 + 2^2 + 4^2}}\]

\[ = \frac{- 2 + 14 - 12}{\sqrt{62} \sqrt{21}}\]

\[ = 0\]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

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अध्याय 27: Straight Line in Space - Exercise 28.2 [पृष्ठ १६]

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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 9.6 | पृष्ठ १६

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