Question

Find the angle between vector a and vector b,if
a.b=axb
From which book this question is

Answer 1

Answer:

Magnitude of vector a = 3

Magnitude of vector b = 4

magnitude of vector 4a = 12

magnitude of vector 3b = 12

angle between vector a & vector b = 120

angle between vector 4a & vector 3b = 120

Magnitude\ of\ 4a+3b=|4a+3b|=[(4a)^{2}+(3b)^{2}+2(4a)(3b)cos120]^{\frac{1}{2}}

\rightarrow |4a+3b|=[(12)^{2}+(12)^{2}-2(12)(12)cos60]^{\frac{1}{2}}

\rightarrow |4a+3b|=[(12)^{2}+(12)^{2}-2(12)^{2}\frac{1}{2}]^{\frac{1}{2}}

\rightarrow |4a+3b|=[(12)^{2}]^{\frac{1}{2}}

\rightarrow |4a+3b|=12

Ans: 12

Explanation:

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Answer 2

This question taken from 11th physics or otherwise 12th physics books