Equations of Lines — Question 7

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Question 7

Reflect the point P=(4,0,1)P=(4,0,1) across the line L:r→=⟨1,1,1⟩+t⟨1,−1,0⟩L:\vec r=\left\langle 1,1,1\right\rangle+t\left\langle 1,-1,0\right\rangle. That is, find P′P' such that LL perpendicularly bisects PP′¯\overline{PP'}.

Original worksheet page 1: question and worked solution for 6-2-007
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Question 7 – Solution

Strategy First find the perpendicular foot HH on the line; then use H=(P+P′)/2H=(P+P')/2.

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Projection point Write H=(1+t,1−t,1)H=(1+t,1-t,1). The condition (P−H)⋅⟨1,−1,0⟩=0(P-H)\cdot\left\langle 1,-1,0\right\rangle=0 gives ⟨3−t,t−1,0⟩⋅⟨1,−1,0⟩=4−2t=0,\left\langle 3-t,t-1,0\right\rangle\cdot\left\langle 1,-1,0\right\rangle=4-2t=0, so t=2t=2 and H=(3,−1,1)H=(3,-1,1).

Reflection Therefore P′=2H−P=(2,−2,1)P'=2H-P=\boxed{(2,-2,1)}.

Verification The midpoint of PP and P′P' is HH, and P−H=⟨1,1,0⟩P-H=\left\langle 1,1,0\right\rangle is perpendicular to the direction of LL.

Original worksheet page 2: question and worked solution for 6-2-007

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