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Condition for a Conic to Be a Pair of Straight Lines

The shortest ...

Question

The shortest distance between the skew lines $\to r =_{1} + t_{1} and \to r =_{2} + n - \to a_{2}, where_{1} = 9 j + 2 k,_{1} = 3 i - j + k,_{2} = - 6 i - 5 j + 10 k,_{2} = - 3 i + 2 j + 4 k$ is $3 \sqrt{a}$ . Then the value of $a$ is:

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Solution

Given $_{1} = 9 j + 2 k,_{1} = 3 i - j + k,_{2} = - 6 i - 5 j + 10 k,_{2} = - 3 i + 2 j + 4 k$ , we have: $_{2} -_{1} = - 6 i - 14 j + 8 k$
and
$_{1} \times {\to a}_{2} = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 3 & - 1 & 1 - 3 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ = - 6 i - 15 j + 3 k,$
So,
$d = \frac{(_{2} -_{1}) . (_{1} \times {\to a}_{2})}{|_{1} \times {\to a}_{2} |} \Rightarrow d = \frac{(- 6) (- 6) + (- 14) (- 15) + (8) (3)}{\sqrt{36 + 225 + 9}} \Rightarrow d = \frac{270}{\sqrt{270}} = \sqrt{270} = 3 \sqrt{30} \Rightarrow a = 30$

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[NCERT EXEMPLAR]

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Condition for a Conic to Be a Pair of Straight Lines

Standard XII Mathematics

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