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Question

Find the angle which the vector A=3^i2^j+4^k makes with zaxis.

A
cos1(229)
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B
cos1(429)
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C
cos1(49)
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D
0
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Solution

The correct option is B cos1(429)
Given, A=3^i2^j+4^k and
other vector will be B=^k, as it is zaxis
We know that,
cosθ=A.B|A||B|
cosθ=(3^i2^j+4^k).(^k)32+(2)2+42×12=429
θ=cos1(429)

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