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What is the angle θbetween vectors A→and B→in each part of Exercise12?

Short Answer

Expert verified

a. θ=1250

b.θ=680

Step by step solution

01

Given information (part a)

A→=3i^+4j^andB→=2i^-6j^

02

Explanation (part a)

The angle between A→and B→, we have

θ=cos-1A→.B→A→B→

we know that

role="math" localid="1649877276481" A→·B→=AxBx+AyBy=(3i^)(2i^)+(4j^)(-6j^)=6i^2-24j^2=6-24A→·B→=-18A→=9+16=25=5B→=4+36=40=210

Substitute the values of A→.B→,A→andB→in the above equation, we get

θ=cos-1-185×210=cos-1-9510=cos-1(-0.569)=1250

03

Given information (part b)

A→=3i^-2j^

B→=6i^+4j^

04

Explanation (part b)

The angle betweenA→ andB→ , we have

θ=cos-1A→.B→A→B→

we know that

A→·B→=AxBx+AyBy=(3i^)(6i^)+(-2j^)(4j^)=18i^2-8j^2=18-8A→·B→=10A→=9+4=13B→=36+16=52=213

Substitute the values of A→.B→,A→andB→in the above equation, we get

θ=cos-11013×213=cos-11026=cos-1(0.384)=680

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