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For the vectors in Fig. 3-32, with a=4, b=3, and c=5, what are (a) the magnitude and (b) the direction of ab, (c) the magnitude and (d) the direction ofac, and (e) the magnitude and (f) the direction ofbc? (The z-axis is not shown)

Short Answer

Expert verified

(a) The magnitude of abis 12.00 units .

(b) The direction of the cross product abis along the +z-direction.

(c) The magnitude ofac is 12.00 units .

(d) The direction of the cross product acis along the -z-direction.

(e) The magnitude of bcis 12.00 units .

(f) The direction of the cross productbc is along the +z-direction.

Step by step solution

01

Step 1:Understanding the magnitude and direction of the cross product

If the magnitude of the vectors and the angle between them is known, it is possible to find the magnitude of the cross product. The formula for the cross product is,

ab=absin (i)

The direction of the cross product can be found using the right-hand rule.

The values of the vectors are given as, a=4,b=3,c=5. The angle between aand bis 90.From the configuration of the figure,the sum of all the vectors is zero. Therefore,

a+b+c=0 (ii)

02

Step 2:(a) Calculating the magnitude of the cross product |a→×b→|

Calculations forab

The angle between thea and bis 90. Calculate the magnitude is of the cross product using the equation (i).

ab=43sin90=12.00

Thus, the magnitude of cross product abis 12.00 units .

03

(b) Finding the direction of cross product |a→×b→|

Using the Right-HandRule, it can be seen that the cross product abpoints in +z -direction.

04

(c) Calculating the magnitude of cross product |a→×c→|

First, calculate the cfrom vectora and busing,

c=-a-b (iii)

To calculate the cross product, substitute the value of vector cfrom equation (iii)

ac=a-a-b=-ab

(Since aa=0)

The magnitude of abis 12.00 . Therefore,

ac=12.00

Therefore, the magnitude ofac is 12.00 units .

05

(d) Finding the direction of cross product |a→×c→|

Using the Right Hand Rule, it can be seen that acpoints in -z-direction.

06

(e) Calculating the magnitude of cross product |b→×c→|

Use the value cof from equation (iii) to calculate the magnitude of cross product bc.

bc=b-a-b=-ba=ab=12

Therefore, the magnitude of cross product ofbc is 12 units .

07

(f) Finding the direction of cross product |b→×c→|

Using the Right Hand Rule, it can be seen that the cross product, bcpoints in +z-direction.

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