Find the shortest distance between the lines whose vector equations are $\to r = (^i + 2^j + 3^k) + λ (^i - 3^j + 2^k)$ and $\to r = (4^i + 5^j + 6^k) + μ (2^i + 3^j +^k)$ .

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Solution

The given lines are $\to r = (^i + 2^j + 3^k) + λ (^i - 3^j + 2^k)$ and
$\to r = 4^i + 5^j + 6^k + μ (^2 i + 3^j +^k)$
It is known that the shortest distance between the lines, $\to r =_{1} + λ_{1}$ and $\to r =_{2} + μ_{2}$ is given by,
$d = ∣ ∣ ∣ ∣ ∣ \frac{(_{1} \times_{2}) (_{1} -_{2})}{∣ ∣_{1} \times_{2} ∣ ∣} ∣ ∣ ∣ ∣ ∣ . . . . . (1)$
Comparing the given equations with $\to r =_{1} + λ_{1}$ and $\to r =_{1} + μ_{1}$ , we obtain
$_{1} =^i + 2^j + 3^k$
$_{1} =^i - 3^j + 2^k$
$_{2} = 4^i + 5^j + 6^k$
$_{2} =^2 i + 3^j +^k$
$_{2} -_{1} = (4^i + 5^j + 6^k) - ((^i + 2^j + 3^k) = 3^i + 3^j + 3^k$
$_{1} \times_{2} = \sqrt{{(- 9)}^{2} + {(3)}^{2} + {(9)}^{2}} = \sqrt{18 + 9 + 81} = \sqrt{171} = 3 \sqrt{19}$
$(_{1} \times_{2}) . (_{1} -_{2}) = (- 9^i + 3^j + 9^k) . (3^i + 3^j + 3^k)$
$= - 9 \times 3 + 3 \times 3 + 9 \times 3 = 9$
Substituting all the values in equation (1), we obtain
$d = ∣ ∣ ∣ \frac{9}{3 \sqrt{19}} ∣ ∣ ∣ = \frac{3}{\sqrt{19}}$
Therefore, the shortest distance between the two given lines is $\frac{3}{\sqrt{19}}$ units.

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Similar questions

Find the shortest distance between the lines whose vector equations are

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

\to r = (4^i + 5^j + 6^k) + t (2^i + 3^j +^k)

\to r = (4^i -^j) + λ (^i + 2^j - 3^k), λ ϵ R

\to r = (-^i -^j + 2^k) + μ (2^i + 4^j - 5^k), μ ϵ R

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 + 2^j +^k + λ (^i -^j +^k)

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

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