$θ = \cos^{- 1} (\hat{A} \cdot \hat{B})$ $theta = cos^-1(hatA * hatB)$

vCalc Reviewed

θ = \cos^{- 1} (\hat{A} \cdot \hat{B})

$theta = cos^-1(hatA * hatB)$

(x_{a}) X component of vector A

$(x_a)"X component of vector A"$

(y_{a}) Y component of vector A

$(y_a)"Y component of vector A"$

(x_{b}) X component of the vector B

$(x_b)"X component of the vector B"$

(y_{b}) Y component of the vector B

$(y_b)"Y component of the vector B"$

Tags

The angle between vectors (2D) formula, $θ = \cos^{- 1} (\hat{A} \cdot \hat{B})$ $theta = cos^-1(hatA * hatB)$ , /attachments/3871f723-e6f5-11e5-9770-bc764e2038f2/AdotB.png Sum of vectors A and B computes the angle between two vectors. To do so, this function:

computes the magnitude of each vector
uses the magnitudes to convert both vectors to unit vectors,
computes the dot product of the two unit vectors, and finally
takes the arc-cosine of the dot product to provide the angle between the two input vectors.

Note: the input vectors do NOT need to be unit vectors.

θ=cos−1(ˆA⋅ˆB)θ=cos-1(A^⋅B^)theta = cos^-1(hatA * hatB)

$θ = \cos^{- 1} (\hat{A} \cdot \hat{B})$ $theta = cos^-1(hatA * hatB)$