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Unit Vectors — Question 9

Section 3-4 · Physics pages 388–389

Question 9 Sum first, then normalize Two dimensionless vectors are⃗a = 3ˆı and⃗b = 4ˆȷ. Compare normalizing their sum with adding their separately normalized directions. (3, 4) (1, 1) direction x The unequal original weights tilt the resultant farther toward north. (a) Find the unit direction of⃗a +⃗b. (b) Find ˆa+ˆb, its magnitude, and its unit direction. Why does this direction differ from part (a)? (c) Give a condition under which the two procedures give the same final direction, and a case in which neither summed direction exists.
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Question 9 - Solution Sum first, then normalize (3, 4) (1, 1) direction x (a) Retain the relative sizes before normalization. [⃗ a +⃗b = (3, 4)/5 , θ = tan−1(4/3) = 53.130◦⃗ a +⃗b = 3ˆı + 4ˆȷ has magnitude 5. Its unit direction is 3ˆı/5 + 4ˆȷ/5, at 53.13◦from east. (b) Normalization discards the original weights. ˆa + ˆb = (1, 1), |ˆa + ˆb| = √ 2, [ ˆa + ˆb = (1, 1)/ √ 2 ˆa + ˆb = ˆı + ˆȷ has magnitude √ 2, not one. Its unit direction is (ˆı + ˆȷ)/ √ 2, at 45◦. Separately normalizing the inputs replaces their 3:4 weighting by 1:1 and therefore changes the summed direction. (c) State the exceptions precisely. |⃗a| = |⃗b| = q > 0 ⇒⃗a +⃗b = q(ˆa + ˆb);⃗ b = −⃗a ⇒both sums vanish Equal nonzero input magnitudes make the two sums proportional by a positive factor, provided their sum is nonzero. If the two inputs have equal magnitude and opposite directions, both sums vanish, so no unit direction exists. Unit vectors are not closed under addition.

Question 9

Sum first, then normalize

Two dimensionless vectors are a→=3ı̂\vec a=3\hat\imath and b→=4ȷ̂\vec b=4\hat\jmath. Compare normalizing their sum with adding their separately normalized directions.

Diagram shown in the original SVG worksheet.

  1. Find the unit direction of a→+b→\vec a+\vec b.

  2. Find â+b̂\hat a+\hat b, its magnitude, and its unit direction. Why does this direction differ from part (a)?

  3. Give a condition under which the two procedures give the same final direction, and a case in which neither summed direction exists.

Question 9 - Solution

Sum first, then normalize

Diagram shown in the original SVG worksheet.

(a) Retain the relative sizes before normalization.

a→+b→̂=(3,4)/5,θ=tan⁡−1(4/3)=53.130∘\boxed{\widehat{\vec a+\vec b}=(3,4)/5},\quad\theta=\tan^{-1}(4/3)=53.130^\circ a→+b→=3ı̂+4ȷ̂\vec a+\vec b=3\hat\imath+4\hat\jmath has magnitude 5. Its unit direction is 3ı̂/5+4ȷ̂/53\hat\imath/5+4\hat\jmath/5, at 53.13∘53.13^\circ from east.

(b) Normalization discards the original weights.

â+b̂=(1,1),|â+b̂|=2,â+b̂̂=(1,1)/2\hat a+\hat b=(1,1),\quad|\hat a+\hat b|=\sqrt2,\quad\widehat{\hat a+\hat b}=(1,1)/\sqrt2 â+b̂=ı̂+ȷ̂\hat a+\hat b=\hat\imath+\hat\jmath has magnitude 2\sqrt2, not one. Its unit direction is (ı̂+ȷ̂)/2(\hat\imath+\hat\jmath)/\sqrt2, at 45∘45^\circ. Separately normalizing the inputs replaces their 3:4 weighting by 1:1 and therefore changes the summed direction.

(c) State the exceptions precisely.

|a→|=|b→|=q>0⇒a→+b→=q(â+b̂);b→=−a→⇒both sums vanish|\vec a|=|\vec b|=q>0\Rightarrow\vec a+\vec b=q(\hat a+\hat b);\quad\vec b=-\vec a\Rightarrow\text{both sums vanish} Equal nonzero input magnitudes make the two sums proportional by a positive factor, provided their sum is nonzero. If the two inputs have equal magnitude and opposite directions, both sums vanish, so no unit direction exists. Unit vectors are not closed under addition.

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