The angle between a=i+j and b=i+k 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
=(1)(1)+(1)(1)+(1)(0)=2
A and B are the magnitudes of vectors →Aand →B, which are
A=√12+12+12=√3
B=√12+12+02=√2
Therefore, we have
θ=arccos(2√3√2)=arccos(2√6)=35.3o
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
=(1)(1)+(1)(1)+(1)(0)=2
A and B are the magnitudes of vectors →Aand →B, which are
A=√12+12+12=√3
B=√12+12+02=√2
Therefore, we have
θ=arccos(2√3√2)=arccos(2√6)=35.3o
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