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

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Solution

Given, $\to r = 2^i +^j +^k + λ (^i - 2^j +^k)$ and
$\to r =^i - 3^j -^k + λ (5^i + 2^j -^k)$
Comparing above equation with $\to r = {\to a}_{1} + λ {\to b}_{1}$ and $\to r = {\to a}_{2} + λ {\to b}_{2}$ , we get
${\to a}_{1} = 2^i +^j +^k, {\to b}_{1} =^i - 2^j +^k$
${\to a}_{2} =^i - 3^j -^k, {\to b}_{2} = 5^i + 2^j -^k$

Therefore, ${\to a}_{2} - {\to a}_{1} = -^i - 4^j - 2^k$

${\to b}_{1} \times {\to b}_{2} = (^i - 2^j +^k) \times (5^i + 2^j -^k)$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 2 & 1 5 & 2 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 6^j + 12^k$
$| {\to b}_{1} \times {\to b}_{2} | = \sqrt{36 + 144} = \sqrt{180} = 6 \sqrt{5}$

Shortest distance is given by,
$d = ∣ ∣ ∣ ∣ \frac{({\to b}_{1} \times {\to b}_{2}) . ({\to a}_{2} - {\to a}_{1})}{| {\to b}_{1} \times {\to b}_{2} |} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \frac{(6^j + 12^k) . (-^i - 4^j - 2^k)}{6 \sqrt{5}} ∣ ∣ ∣ ∣$
$= \frac{| - 24 - 24 |}{6 \sqrt{5}}$
$= \frac{48}{6 \sqrt{5}}$
$= \frac{8}{\sqrt{5}}$

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Find the shortest distance between the lines l1 and l2 whose vector equations are →r=2^i+^j+^k+λ(^i−2^j+^k) and →r=^i−3^j−^k+λ(5^i+2^j−^k).

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