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Physics / Chapter 3: Vector / Section 3-5

Adding Vectors by Components — Question 2

Section 3-5 · Physics pages 394–395

Question 2 Subtraction joins two endpoints Positions relative to a common origin are⃗rP = (−2ˆı + 5ˆȷ) m and⃗rQ = (4ˆı −3ˆȷ) m. P Q⃗ rQ −⃗rP O The relative displacement points from the initial endpoint P to the final endpoint Q. (a) Find the displacement from P to Q, including magnitude and direction. (b) Find the reverse displacement. Why is subtracting the position magnitudes not a distance between P and Q? (c) If every position is remeasured from an origin translated by (10, −6) m, verify that the P-to-Q displacement is unchanged.
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Question 2 - Solution Subtraction joins two endpoints P Q⃗ rQ −⃗rP O (a) Subtract the starting position.⃗ dPQ = (4, −3) −(−2, 5) = (6, −8) m , d = 10 m, θ = −53.130◦⃗ dPQ =⃗rQ −⃗rP = (6ˆı −8ˆȷ) m. Its magnitude is 10 m and its direction is 53.13◦south of east. The vertical subtraction is −3 −5 = −8, not −3 + 5. (b) Reverse the endpoint order.⃗ dQP = −⃗dPQ, |⃗rQ| −|⃗rP | = 5 − √ 29 m̸ = 10 m⃗ dQP = (−6ˆı + 8ˆȷ) m, with the same length. In contrast, |⃗rQ| −|⃗rP | = 5 − √ 29 ≈−0.385 m depends on distances from the chosen origin and is not the endpoint separation. (c) Translate both position records consistently.⃗ r′ P = (−12, 11),⃗ r′ Q = (−6, 3),⃗ r′ Q −⃗r′ P = (6, −8) m The new positions are⃗rP ′ = (−12, 11) m and⃗rQ ′ = (−6, 3) m. Their difference is again (6, −8) m. The common origin shift cancels; changing only one position would mix coordinate systems.

Question 2

Subtraction joins two endpoints

Positions relative to a common origin are r→P=(−2ı̂+5ȷ̂)\vec r_P=(-2\hat\imath+5\hat\jmath) m and r→Q=(4ı̂−3ȷ̂)\vec r_Q=(4\hat\imath-3\hat\jmath) m.

Diagram shown in the original SVG worksheet.

  1. Find the displacement from P to Q, including magnitude and direction.

  2. Find the reverse displacement. Why is subtracting the position magnitudes not a distance between P and Q?

  3. If every position is remeasured from an origin translated by (10,−6)(10,-6) m, verify that the P-to-Q displacement is unchanged.

Question 2 - Solution

Subtraction joins two endpoints

Diagram shown in the original SVG worksheet.

(a) Subtract the starting position.

d→PQ=(4,−3)−(−2,5)=(6,−8)m,d=10m,θ=−53.130∘\boxed{\vec d_{PQ}=(4,-3)-(-2,5)=(6,-8)\,\mathrm m},\quad d=10\,\mathrm m,\quad\theta=-53.130^\circ d→PQ=r→Q−r→P=(6ı̂−8ȷ̂)\vec d_{P Q}=\vec r_Q-\vec r_P=(6\hat\imath-8\hat\jmath) m. Its magnitude is 10 m and its direction is 53.13∘53.13^\circ south of east. The vertical subtraction is −3−5=−8-3-5=-8, not −3+5-3+5.

(b) Reverse the endpoint order.

d→QP=−d→PQ,|r→Q|−|r→P|=5−29m≠10m\vec d_{QP}=-\vec d_{PQ},\quad|\vec r_Q|-|\vec r_P|=5-\sqrt{29}\,\mathrm m\ne10\,\mathrm m d→QP=(−6ı̂+8ȷ̂)\vec d_{Q P}=(-6\hat\imath+8\hat\jmath) m, with the same length. In contrast, |r→Q|−|r→P|=5−29≈−0.385|\vec r_Q|-|\vec r_P|=5-\sqrt{29}\approx-0.385 m depends on distances from the chosen origin and is not the endpoint separation.

(c) Translate both position records consistently.

r→P′=(−12,11),r→Q′=(−6,3),r→Q′−r→P′=(6,−8)m\vec r_P'=(-12,11),\quad\vec r_Q'=(-6,3),\quad\vec r_Q'-\vec r_P'=(6,-8)\quad\mathrm m The new positions are r→P′=(−12,11)\vec r_P\,{}^{\prime}=(-12,11) m and r→Q′=(−6,3)\vec r_Q\,{}^{\prime}=(-6,3) m. Their difference is again (6,−8)(6,-8) m. The common origin shift cancels; changing only one position would mix coordinate systems.

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