Find the angle between unit vector I and a=√3i+j
1 answer:
Recall that for two vectors x and y,
x • y = ||x|| ||y|| cos(θ)
where θ is the angle between them.
We have
i • (√3 i + j) = √3
||i|| = 1
||√3 i + j|| = √((√3)² + 1²) = √4 = 2
Solve for θ :
√3 = 2 cos(θ)
cos(θ) = √3/2
θ = arccos(√3/2) = π/6
(Note that we assume θ is the <u>acute</u> angle between the vectors.)
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