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प्रश्न
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]
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उत्तर
\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]
Let
\[\overrightarrow{b_1} \text{ and } \overrightarrow{b_2}\] be vector parallel to the given line.
Now,
\[\overrightarrow{b_1} = \hat{i} + 2 \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \]
If θ is the angle between the given line, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) . \left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{3^2 + 2^2 + 6^2}}\]
\[ = \frac{3 + 4 + 12}{3 \times 7}\]
\[ = \frac{19}{21}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{19}{21} \right)\]
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