The shortest distance in units between the lines whose equations are r-2-7k- -2-3 k- and r-2-3k-s7-6- 3k- is-

We know, shortest distance =|(a c)·(b×d)||b×d| From the given lines, nbsp; r=î+2ĵ+7k̂ +λ(î+2ĵ+3 k̂ ) and nbsp;r= î 2ĵ+3k̂+s(7î+6ĵ+ 3k̂ ) we have a=î+2ĵ+7k̂ ...

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Standard XII

Mathematics

Shortest Distance between Two Lines

The shortest ...

Question

The shortest distance (in units) between the lines whose equations are $\to r =^i + 2^j + 7^k + λ (^i + 2^j + 3^k)$ and $\to r = -^i - 2^j + 3^k + s (7^i + 6^j + 3^k)$ is:

\frac{30}{\sqrt{133}}

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\frac{8}{\sqrt{133}}

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\frac{40}{\sqrt{113}}

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\frac{30}{\sqrt{113}}

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Solution

The correct option is B $\frac{8}{\sqrt{133}}$
We know, shortest distance $= \frac{| (\to a - \to c) \cdot (\to b \times \to d) |}{| \to b \times \to d |}$
From the given lines,
$\to r =^i + 2^j + 7^k + λ (^i + 2^j + 3^k)$ and $\to r = -^i - 2^j + 3^k + s (7^i + 6^j + 3^k)$
we have $\to a =^i + 2^j + 7^k; \to c = -^i - 2^j + 3^k; \to b =^i + 2^j + 3^k; \to d = 7^i + 6^j + 3^k$
On solving, we have
$(\to a - \to c) = 2^i + 4^j + 4^k$
and $\to b \times \to d = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 2 & 3 7 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 12^i + 18^j - 8^k$
$\Rightarrow | \to b \times \to d | = \sqrt{532}$
Hence, shortest distance $= \frac{16}{2 \sqrt{133}} = \frac{8}{\sqrt{133}}$

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The shortest distance (in units) between the lines whose equations are →r=^i+2^j+7^k+λ(^i+2^j+3^k) and →r=−^i−2^j+3^k+s(7^i+6^j+3^k) is:

The shortest distance (in units) between the lines whose equations are $\to r =^i + 2^j + 7^k + λ (^i + 2^j + 3^k)$ and $\to r = -^i - 2^j + 3^k + s (7^i + 6^j + 3^k)$ is: