Question

Find vector A dot vector B ,if |A|=2,|B|=5,|A×B|=8​

Answer 1

Answer:

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Explanation:

|A|=2,|B|=5

The magnitude of a vector product(or cross product) of vector A and vector B would be

|A||B|sinø

Where ø is the angle between the vectors A and B

Therefore, here

|A||B|sinø=8

.:2∗5sinø=8

.:10sinø=8

.:sinø=8/10

.:sinø=4/5

Dot product of vector A and vector B i.e A.B would be

|A||B|cosø

From previous calculations , we got sinø=4/5

sin2ø+cos2ø=1(it is a trignometrical identity)

.:(4/5)2+cos2ø=1

.:cos2ø=1−16/25

.:cos2ø=9/25

.:cosø=3/5(plusorminus)

Therefore the dot product A.B would be

|A||B|cosø

=(2∗5)(3/5)(plusorminus)=10(3/5)(plusorminus)=2(3)(plusorminus)=6(plusorminus)

Therefore the answer is A.B= 6(plusorminus)