Find the shortest distance between the pair of parallel lines r- - i- - j- - - 2-i- -j- - k- and r- - 2-i -j- - k- - - 4- -j- - k- is

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Family of Planes Passing through the Intersection of Two Planes

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Question

Find the shortest distance between the pair of parallel lines
$\to r = \to i + \to j + λ (\to 2 i - \to j + \to k) a n d \to r = \to 2 i + \to j - \to k + μ (\to 4 + \to j - \to k) i s$

\frac{\sqrt{1} 0}{6}

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\frac{\sqrt{1} 3}{6}

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\frac{\sqrt{1} 9}{6}

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\frac{\sqrt{1} 1}{6}

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Solution

The correct option is D $\frac{\sqrt{1} 1}{6}$
the ques is wrong if the lines are parallel so assuming the following ques and solution
given lines
$\to r =^i +^j + λ (2^i -^j +^k)$
$\to r = 2^i +^j -^k + μ (2^i -^j +^k)$
both lines are parallel
position vector of line
$\to a =^i +^j$
$\to c = 2^i +^j -^k$
normal vector
$\to b = 2^i -^j +^k$

$S D = ∣ ∣ ∣ ∣ \frac{(\to c - \to a) \times \to b}{∣ ∣ \to b ∣ ∣} ∣ ∣ ∣ ∣$

$S D = ∣ ∣ ∣ ∣ \frac{(2^i +^j -^k - (^i +^j) \times (2^i -^j +^k)}{\sqrt{2^{2} + 1^{2} + 1^{2}}} ∣ ∣ ∣ ∣$

$S D = ∣ ∣ ∣ ∣ \frac{(^i -^k) \times (2^i -^j +^k)}{\sqrt{4 + 1 + 1}} ∣ ∣ ∣ ∣$

$S D = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 0 & - 1 2 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣}{\sqrt{6}}$
$S D = \frac{∣ ∣ -^i - 3^j -^k ∣ ∣}{\sqrt{6}}$

$S D = \frac{\sqrt{1 + 9 + 1}}{\sqrt{6}}$
$S D = \sqrt{\frac{11}{6}}$

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Find the shortest distance between the pair of parallel lines →r=→i+→j+λ(→2i−→j+→k)and→r=→2i+→j−→k+μ(→4+→j−→k)is

Find the shortest distance between the pair of parallel lines
$\to r = \to i + \to j + λ (\to 2 i - \to j + \to k) a n d \to r = \to 2 i + \to j - \to k + μ (\to 4 + \to j - \to k) i s$