The vectors a=3i-2j+2k and b = -j-2k are the adjacent sides of a parallelogram.find the angle between its diagonals
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a=3i-2j+2k,b=-j-2k
a+b &a-b are the diagonals of the parallelogram.
a+b=3i-3j. ; a-b=3i-j+4k
Angle between the diagonals is
Cos theta=(a+b)*(a-b) / |a+b|*|a-b|
Cos theta= (3i-3j) *(3i-j+4k) /√18*√26
Cos theta=9+3 / 3√2*√2√13
Cos theta=12/ 2 *3*√13
Cos theta=2/√13
theta=cos^-1(2/√13)
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