Find the shortest distance between the pair of parallel lines r- - i- - 2j- - 3k- - - i- -j- - k- and r- - 2i- - j- - k- - - -i- -j- - k- is

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Shortest Distance between Two Skew Lines

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Question

Find the shortest distance between the pair of parallel lines
$\to r = \to i + 2 \to j + 3 \to k + λ (\to i - \to j + \to k) a n d \to r = 2 \to i - \to j - \to k + μ (\to - i + \to j - \to k) i s$

\frac{77}{\sqrt{3}}

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\frac{11}{\sqrt{14}}

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\frac{78}{\sqrt{3}}

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\frac{68}{\sqrt{3}}

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Solution

The correct option is B $\frac{11}{\sqrt{14}}$
Shortest distance between lines with vector equations
$\to r =_{1} + λ_{1}$ and $\to r =_{2} + μ_{2}$ is
$∣ ∣ ∣ ∣ ∣ \frac{(_{1} \times_{2}) . (_{2} -_{1})}{∣ ∣_{1} \times_{2} ∣ ∣} ∣ ∣ ∣ ∣ ∣$

Given: $\to r =^i + 2^j + 3^k + λ (^i -^j +^k)$ and $\to r = 2^i -^j -^k + μ (-^i +^j -^k)$

$_{1} =^i + 2^j + 3^k,_{2} = 2^i -^j -^k$ and $_{1} = 2^i -^j -^k,_{2} = -^i +^j -^k$

$_{1} \times_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 1 & - 1 - 1 & 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$_{1} \times_{2} = (1 + 1)^i - (- 2 - 1)^j + (2 - 1)^k$

$_{1} \times_{2} = 2^i + 3^j +^k$

$∣ ∣_{1} \times_{2} ∣ ∣ = \sqrt{2^{2} + 3^{2} + 1^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}$

$_{2} -_{1} = 2^i -^j -^k -^i - 2^j - 3^k =^i - 3^j - 4^k$

Now $(_{1} \times_{2}) . (_{2} -_{1}) = (2^i + 3^j +^k) (^i - 3^j - 4^k) = 2 - 9 - 4 = - 11$

Now,Shortest distance between lines $= ∣ ∣ ∣ ∣ ∣ \frac{(_{1} \times_{2}) . (_{2} -_{1})}{∣ ∣_{1} \times_{2} ∣ ∣} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ \frac{- 11}{\sqrt{14}} ∣ ∣ ∣ = \frac{11}{\sqrt{14}}$ units by substituting the above values

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Similar questions

Q. 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

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

Q. The point of intersection of the lines

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

and

\to r = (2 \to i + 3 \to j + 5 \to k) + s (\to i + 2 \to j + 3 \to k)

is:

Q. If

\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

and

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

such that

\to r = λ \to a + μ \to b + ν \to c

then

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

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Find the shortest distance between the pair of parallel lines →r=→i+2→j+3→k+λ(→i−→j+→k)and→r=2→i−→j−→k+μ(→−i+→j−→k)is

Find the shortest distance between the pair of parallel lines
$\to r = \to i + 2 \to j + 3 \to k + λ (\to i - \to j + \to k) a n d \to r = 2 \to i - \to j - \to k + μ (\to - i + \to j - \to k) i s$