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Question
Write the value of λ for which the lines \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\] are perpendicular to each other.
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Solution
We have ,
\[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \]
\[\frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\]
The given lines are parallel to vectors
\[\vec{b_1} = - 3 \hat{i} + 2\lambda \hat{j} + 2 \hat{k} \text{ and } \vec{b_2} = 3\lambda \hat{i} + \hat{j} - 5 \hat{k} \]
For \[\vec{b_1} \perp \vec{b_2}\]
we must have ,
\[\vec{b_1} . \vec{b_2} = 0\]
\[ \Rightarrow \left( - 3 \hat{i} + 2\lambda \hat{j} + 2 \hat{k} \right) . \left( 3\lambda \hat{i} + \hat{j} - 5 \hat{k} \right) = 0\]
\[ \Rightarrow - 7\lambda - 10 = 0\]
\[ \Rightarrow \lambda = - \frac{10}{7}\]
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