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Shortest Distance between Two Skew Lines

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Question

Find the shortest distance between the skew lines $\to r = (6 ˆ i + 2 ˆ j + 2 ˆ k) + λ (ˆ i - 2 ˆ j + 2 ˆ k)$ and $\to r = (- 4 ˆ i - ˆ k) + μ (3 ˆ i - 2 ˆ j - 2 ˆ k)$

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Solution

$\begin{matrix} W e h a v e t h e g i v e n e q u a t i o n \to r = (6 ˆ i + 2 ˆ j + 2 ˆ k) + λ (ˆ i - 2 ˆ j + 2 ˆ k) c o m p a r i n g w i t h \to r = \to a_{1} + λ \to b_{1} \to a_{1} = (6 ˆ i + 2 ˆ j + 2 ˆ k) & \to b_{1} = (ˆ i - 2 ˆ j + 2 ˆ k) s i m i l a r l y \to r = (- 4 ˆ i - ˆ k) + μ (3 ˆ i - 2 ˆ j - 2 ˆ k) c o m p a r i n g w i t h \to r = \to a_{2} + μ \to b_{2} \to a_{2} = (- 4 ˆ i - ˆ k) & \to b_{2} = (3 ˆ i - 2 ˆ j - 2 ˆ k) N o w, (\to a_{2} - \to a_{2}) = (- 4 ˆ i - ˆ k) - (6 ˆ i + 2 ˆ j + 2 ˆ k) = - 10 ˆ i - 2 ˆ j - 3 ˆ k (\to b_{1} \times \to b_{2}) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 1 & - 2 & 2 3 & - 2 & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ˆ i [(- 2 \times - 2) - (- 2 \times 2)] - ˆ j [(1 \times - 2) - (3 \times 2)] + ˆ k [(1 \times - 2) - (3 \times - 2)] = ˆ i [4 + 4] + ˆ j [- 2 - 6] + ˆ k [- 2 + 6] = 8 ˆ i + 8 ˆ j + 4 ˆ k N o w M a g n i t u d e o f \to b_{1} \times \to b_{2} = \sqrt{8^{2} + 8^{2} + 4^{2}} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 A l s o, (\to b_{1} \times \to b_{2}) . (\to a_{2} - \to a_{2}) = (- 10 ˆ i - 2 ˆ j - 3 ˆ k) (8 ˆ i + 8 ˆ j + 4 ˆ k) = (8 \times - 10) + (8 \times - 2) + (4 \times - 3) = - 80 - 16 - 12 = - 108 S h o r t e s t d i s tan c e = ∣ ∣ ∣ ∣ \frac{(\to b_{1} \times \to b_{2}) . (\to a_{2} - \to a_{2})}{(\to b_{1} \times \to b_{2})} ∣ ∣ ∣ ∣ = ∣ ∣ \frac{- 108}{12} ∣ ∣ = | - 9 | = 9 H e n c e, t h e s h o r t e s t d i s tan c e b w t w e e n t h e g i v e n l i n e i s 9. \end{matrix}$

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Q. Find the shortest distance between the skew lines

r = (6 i + 2 j + 2 k) + t (i - 2 j + 2 k)

r = (- 4 i - k) + s (3 i - 2 j - 2 k)

Q. Find the shortest distance between the skew lines

r = (6 i + 2 j + 2 k) + t (i - 2 j + 2 k)

F = (- 4 i - k) + s (3 i - 2 j - 2 k)

where s,t are scalars.

Q. Find the shortest distance between the lines
(i)

\vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k})

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})

\vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}) and \vec{r} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k})

Q. Find the angle between the lines

\to r = 3 i + 2 j - 4 k + λ (i + 2 j + 2 k)

\to r = (5 j - 2 k) + μ (3 i + 2 j + 6 k)

Q. Find the shortest distance between the lines

¯ ¯ ¯ r = 4 ¯ i - ¯ j + λ (¯ i + 2 ¯ j - 5 ¯ ¯ ¯ k)

¯ ¯ ¯ r = ¯ i - ¯ j + 2 ¯ ¯ ¯ k + μ (¯ i + 2 ¯ j - 5 ¯ ¯ ¯ k)

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