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Chapter 9: Q. 12 (page 228)

Evaluate the dot product A鈫�路B鈫�if

a). role="math" localid="1647759176820" A鈫�=3i^+4j^and B鈫�=2i^-6j^

b).A鈫�=3i^-2j^andB鈫�=6i^+4j^

Short Answer

Expert verified

a). A鈫�路B鈫�=-18

b).A鈫�路B鈫�=10

Step by step solution

01

Step 1. Given information

We have,

A鈫�=3i^+4j^andB鈫�=2i^-6j^

also A鈫�=3i^-2j^andB鈫�=6i^+4j^

02

Step 2. Part a). dot product A→·B→ if A→=3i^+4j^ and B→=2i^-6j^

The dot product for these two vectors is in the form

A鈫�路B鈫�=AxBx+AyBy

Now, we put the x-components and y-components of the two vectors

A鈫�路B鈫�=AxBx+AyBy=(3i^)(2i^)+(4j^)(-6j^)=6i^2-24j^2

The magnitude of i^2=1and j^2=1. So, the dot product will be

A鈫�路B鈫�=6i^2-24j^2=6-24=-18

03

Step 3. Part b). Dot product A→·B→ if A→=3i^-2j^ and B→=6i^+4j^

We use the same step as in part (a) to get the dot products by putting the x-components and y-components of the two vectors

A鈫�路B鈫�=AxBx+AyBy=(3i^)(6i^)+(-2j^)(4j^)=18i^2-8j^2=18-8=10

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Most popular questions from this chapter

A gardener pushes a 12kg lawnmower whose handle is tilted up 37掳 above horizontal. The lawnmower鈥檚 coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2m/s? Assume his push is parallel to the handle.

Evaluate the dot product A鈬赌.B鈬赌,if

a. A鈬赌=3i^+4j^and B鈬赌=2i^-6j^

b.A鈬赌=3i^-2j^andB鈬赌=6i^+4j^

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A 150 g particle at x=0 is moving at 2.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.250N)sin(x/2.00m). What is the particle鈥檚 speed when it reaches x=3.14 m?

The driver of a car traveling at 60 mph slams on the brakes, and the car skids to a halt. What happened to the kinetic energy the car had just before stopping?
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