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Write the Value of λ for Which the Lines X − 3 − 3 = Y + 2 2 λ = Z + 4 2 a N D X + 1 3 λ = Y − 2 1 = Z + 6 − 5 Are Perpendicular to Each Other. - Mathematics

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प्रश्न

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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उत्तर

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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अध्याय 28: Straight Line in Space - Very Short Answers [पृष्ठ ४१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Very Short Answers | Q 12 | पृष्ठ ४१

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