If r → = i ^ + j ^ + t ( 2 i ^ - j ^ + k ^ ) and r → = 2 i ^ - j ^ - k ^ + t ( 3 i ^ - 5 j ^ + 2…

If r→=i^+j^+t(2i^-j^+k^) and r→=2i^-j^-k^+t(3i^-5j^+2k^) are the vector equations of two lines L1 and L2 then the shortest distance between them is
  1. $\frac{9}{\sqrt{59}}$
  2. 1059
  3. $\frac{11}{\sqrt{59}}$
  4. 0

Solution

Here, $a_1 = i + j$ and $b_1 = 2i - j + k$ and $a_2 = 2i + j - k$ and $b_2 = 3i - 5j + 2k$. $\Rightarrow a_2 - a_1 = i - k$ and $b_1 \times b_2 = \begin{vmatrix} i & j & k \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = 3i - j - 7k$. The shortest distance between lines $L_1$ and $L_2 = \left| \frac{(a_2 - a_1) \cdot (b_1 \times b_2)}{|b_1 \times b_2|} \right|$. $= \left| \frac{(i - k) \cdot (3i - j - 7k)}{3i - j - 7k} \right|$. $= \frac{10}{\sqrt{59}}$.

Asked in: MHT CET Full Test 1

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