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cos theta =4/21 then what is the value of theta
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Answer:
Step-by-step explanation:
a=2i + 2j - k
b=6i - 3j +2k
|a| =root(2^2 + 2^2 + 1^2)
=root(9)=3
|b|=root(6^2 + 3^2 + 2^2)
=root(49) = 7
a . b = |a| |b| cos @ where @ is angle btw vectors
a . b = (2i + 2j - k) . (6i - 3j +2k)
=12 - 6 -2
=4
cos@ = a . b / (|a| |b|)
cos @ = 4/(7*3)=4/21
@= cos^-1(4/21)
Hope it's correct and helps you
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your solution is in image
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