Find the shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equation are given by
$\to r =^i +^j + λ (2^i -^j +^k)$
$\to r = 2^i +^j -^k + μ (3^i - 5^j + 2^k)$

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Solution

As we know that the vector equation of a line is of form r = a + mb where a is the position vector through which line is passing, b is a vector parallel to line and m is a constant.

If two equations of line

r= a₁ + mb₁

r= a₂ + nb_{2 are given,}

then to find the distance between these lines first we have to find,

b₁ X b₂
a₂ – a₁

| b₁ X b₂|

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Similar questions

l_{1}

l_{2}

\to r =^i +^j + μ (2^i -^j +^k)

\to r = 2^i +^i -^k + μ (3^i - 5^j + 2^k)

\to r =^i +^j + λ (2^i -^j +^k)

\to r = 2^i +^j -^k + μ (3^i - 5^j + 2^k)

\to r =^i +^j + l (2^i -^j +^k)

2^i +^j -^k + m (3^i - 5^j + 2^k)

\to r =^i + 2^j +^k + λ (^i -^j +^k)

\to r = 2^i -^j -^k + μ (2^i +^j + 2^k)

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

\to r = 2^i -^j -^k + μ (2^i +^j + 2^k)

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