Equations of Lines — Question 9

Section 13-2 · Calculus pages 2417–2418

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Question 9

Show that L1:r→=⟨2,−1,3⟩+t⟨4,2,−2⟩andL2:r→=⟨8,2,0⟩+s⟨−2,−1,1⟩L_1:\vec r=\left\langle 2,-1,3\right\rangle+t\left\langle 4,2,-2\right\rangle\quad\text{and}\quad L_2:\vec r=\left\langle 8,2,0\right\rangle+s\left\langle-2,-1,1\right\rangle describe the same geometric line. Find the parameter conversion between tt and ss.

Question 9 – Solution

Strategy Check parallel directions and show that one base point lies on the other line.

Diagram shown in the original SVG worksheet.

Same point set The second direction is −12⟨4,2,−2⟩-\tfrac 12\left\langle 4,2,-2\right\rangle. Also ⟨8,2,0⟩=⟨2,−1,3⟩+32⟨4,2,−2⟩\left\langle 8,2,0\right\rangle=\left\langle 2,-1,3\right\rangle+\tfrac 32\left\langle 4,2,-2\right\rangle, so the second base point lies on L1L_1. Thus the lines coincide.

Parameter conversion Equating the representations gives r→=r→0+td→=r→0+32d→−s2d→,\vec r=\vec r_0+t\vec d=\vec r_0+\frac 32\vec d-\frac{s}{2}\vec d, so t=32−s2\boxed{t=\frac 32-\frac{s}{2}}, equivalently s=3−2ts=3-2t.

Verification The conversion makes all three coordinates identical.