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Find the Angle Between the Pairs of Lines with Direction Ratios Proportional to 1, 2, −2 and −2, 2, 1 .

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प्रश्न

Find the angle between the pairs of lines with direction ratios proportional to  1, 2, −2 and −2, 2, 1 .

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उत्तर

1, 2, −2 and −2, 2, 1

\[\text{ Let }  \overrightarrow{m_1} \text{ and }  \overrightarrow{m_2} \text{ be vectors parallel to the two given lines } . \]

\[ \text{ Then, the angle between the two given lines is same as the angle between } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2 .} \]

\[\text{ Now }, \]

\[ \overrightarrow{m_1} = \text{ Vector parallel to the line having direction ratios proportional to 1, 2, - 2} \]

\[ \overrightarrow{m_2} = \text{ Vector parallel to the line having direction ratios proportional to - 2, 2, 1} \]

\[ \therefore \overrightarrow{m_1} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]

\[ \overrightarrow{m_2} = - 2 \hat{i} + 2 \hat{j} + \hat{k} \]

\[\text{ Let }  \theta \text{ be the angle between the lines } . \]

\[Now, \]

\[\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}\]

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

\[ = \frac{- 2 + 4 - 2}{3 \times 3}\]

\[ = 0\]

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

 

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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 10.3 | पृष्ठ १६

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