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Question
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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Solution
a, b, c and b − c, c − a, a − 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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