Question

Vector A has components Ax = −2.2, Ay = 4, Az = 2.2, while vector B has components Bx = 3.4, By = −6.1, Bz = 6.5. What is the angle θAB between these vectors? (Answer between 0◦ and 180◦.) Answer in units of ◦.

Answer 1

you might use: \(\vec A \bullet \vec B = |\vec A||\vec B| \cos \phi\) where ø is the angle between the vectors.

so for \(\vec A = < −2.2, 4, 2.2>\) and \(\vec B = < 3.4, −6.1, 6.5>\), you calculate the dot product.......

Answer 2

\(\large \vec A \bullet \vec B = A_x B_x + A_y B_y + A_z B_z\)

\(\large |\vec A| = \sqrt{A_x^2 + A_y^2 + A_z^2}\)
\(\large |\vec B| = \sqrt{B_x^2 + B_y^2 + B_z^2}\)

Answer 3

so:
\[ \cos \phi = {|\vec A \bullet \vec B | \over |\vec A||\vec B|}\]

Answer 4

Oh wow thank you so much!