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Find the Angle Between the Pairs of Lines with Direction Ratios Proportional To A, B, C And B − C, C − A, A − B.

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प्रश्न

Find the angle between the pairs of lines with direction ratios proportional to   abc and b − cc − aa − b.

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उत्तर

abc and b − cc − aa − b

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

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

\[\text{ Now } , \]

\[ \overrightarrow{m_1} = \text{ Vector parallel to the line having direction ratios proportional to a, b, c} \]

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

\[ \therefore \overrightarrow{m_1} = a \hat{i} + b \hat{j} + c \hat{k}  \text{ and } \overrightarrow{m_2} = \left( b - c \right)  \hat{ i }+ \left( c - a \right) \hat{j} + \left( a - b \right) \hat{k}  \]

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

\[Now, \]

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

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

\[ = \frac{ab - ac + bc - ba + ca - cb}{\sqrt{a^2 + b^2 + c^2} \sqrt{\left( b - c \right)^2 + \left( c - a \right)^2 + \left( a - b \right)^2}}\]

\[ = 0\]

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

 

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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 10.4 | पृष्ठ १६

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