12. Find the distance between the parallel linesx = -4 – tx = 6 + s4:{y = -3+ 2t, teR and l2:{ y = -5 – 2s, s ER.z = 3 – 2tz = 5+ 2sDo not use a formula from the textbook. Draw a diagram and use projections to findthe distance.

Given, l1=x=-4-ty=-3+2tz=3-2t, t∈ℝ &l2=x=6+sy=-5-2sz=5+2s, s∈ℝ represented as, l1=-4, -3,…

Answered: 12. Find the distance between the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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12. Find the distance between the parallel lines
x = -4 – t
x = 6 + s
4:{y = -3+ 2t, teR and l2:{ y = -5 – 2s, s ER.
z = 3 – 2t
z = 5+ 2s
Do not use a formula from the textbook. Draw a diagram and use projections to find
the distance.

Expert Solution

Step 1

Given,

$l_{1} = \{\begin{cases} x = - 4 - t \\ y = - 3 + 2 t \\ z = 3 - 2 t \end{cases}, t \in ℝ & l_{2} = \{\begin{cases} x = 6 + s \\ y = - 5 - 2 s \\ z = 5 + 2 s \end{cases}, s \in ℝ$

represented as,

$l_{1} = (- 4, - 3, 3) + t (- 1, 2, - 2) l_{2} = (6, - 5, 5) + s (1, - 2, 2)$

The projection of line PQ will give minimum distance between lines i.e., $\hat{n}$

$\begin{array}{rcl} \vec{n} & = & |\begin{matrix} i & j & k \\ - 1 & 2 & 2 \\ 1 & - 2 & 2 \end{matrix}| \\ = & i (8) - j (- 4) + k (0) \\ = & 8 i + 4 j + 0 k \\ | \vec{n} | & = & \sqrt{{(8)}^{2} + {(4)}^{2} + 0} \\ = & \sqrt{64 + 16} \\ = & \sqrt{80} \\ = & 4 \sqrt{5} \end{array}$

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