The shortest distance between the lines r-3i-5j-7k- i-2j-k- and r-i-j-k- 7i-6j-k- is

D=| ( a⃗_⃗2⃗ a⃗_⃗1⃗ ). (b⃗_⃗1⃗×b⃗_⃗2⃗ )|/|b⃗_⃗1⃗×b⃗_⃗2⃗| d=46/5√(5) ...

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Distance Formula

The shortest ...

Question

The shortest distance between the lines $\to r = 3 \to i + 5 \to j + 7 \to k + λ (\to i + 2 \to j + \to k) a n d \to r = - \to i - \to j - \to k + μ (7 \to i - 6 \to j + \to k) i s$

$\frac{16}{5 \sqrt{5}}$

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$\frac{26}{5 \sqrt{5}}$

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$\frac{36}{5 \sqrt{5}}$

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$\frac{46}{5 \sqrt{5}}$

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\to r = \to i + 2 \to j + 3 \to k + λ (\to i - \to j + \to k) a n d \to r = 2 \to i - \to j - \to k + μ (\to - i + \to j - \to k) i s

\to r = 2 \to i - 2 \to j + 3 \to k + λ (\to i - \to j + 4 \to k)

\to r \cdot (\to i + 5 \to j + \to k) = 5

The lines $\to r = \to i - \to j + λ (2 \to i + \to k)$ and $\to r = (2 \to i - \to j) + μ (\to i + \to j - \to k)$ intersect for

\to r = 3 \to i + 2 \to j - 5 \to k

\to a = 2 \to i - \to j + \to k

\to b = \to i + 3 \to j - 2 \to k

\to c = - 2 \to i + \to j - 3 \to k

\to r = λ \to a + μ \to b + ν \to c

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k,

\to C = 4 \to i - 3 \to j + 7 \to k

\to R

\to R \times \to B = \to C \times \to B

\to R . \to A = 0

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