prove that thw angle subatended between vector 3i+j+2k and 2i+2j+4k,is
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We're asked to find the angle between two vectors, given their unit vector notations.
To do this, we can use the equation
→A⋅→B=ABcosθ
rearranging to solve for angle, θ:
cosθ=→A⋅→BAB
θ=arccos⎛⎝→A⋅→BAB⎞⎠
where
→A⋅→B is the dot product of the two vectors, which is
→A⋅→B=AxBx+AyBy+AzBz
=(2)(1)+(3)(2)+(1)(−4)=4
A and B are the magnitudes of vectors →Aand →B, which are
A=√22+32+12=√14
B=√12+22+(−4)2=√21
Therefore, we have
θ=arccos(4√14√21)=arccos(47√6)=76.5o
To do this, we can use the equation
→A⋅→B=ABcosθ
rearranging to solve for angle, θ:
cosθ=→A⋅→BAB
θ=arccos⎛⎝→A⋅→BAB⎞⎠
where
→A⋅→B is the dot product of the two vectors, which is
→A⋅→B=AxBx+AyBy+AzBz
=(2)(1)+(3)(2)+(1)(−4)=4
A and B are the magnitudes of vectors →Aand →B, which are
A=√22+32+12=√14
B=√12+22+(−4)2=√21
Therefore, we have
θ=arccos(4√14√21)=arccos(47√6)=76.5o
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