Question

Find the angle between u =

a. 65.9o

Б. 98.50

53-8; and v= 6i+j. Round to the nearest tenth of a degree.

90.4o

d. 33.30

Please select the best answer from the choices provided

0

A

ооо

Answer 1

Answer is not listed.

Explanation:

To begin with the dot product between the vectors

`a = 53i - 8j\\b = 6i+j`

`a \bullet b = a_x (b_y) + a_y( b_y) = 53 (6) + (-8) ( 1) = 318 - 8 = 310`

Then you compute the magnitude of the vectors:

`|a| = √(a_x^2 + a_y^2) = √(532 + (-8)^2 )= √(2809 + 64) = √(2873) = 13√(17)\\|b| = √(b_x^2 + b_y^2) = √(62 + 12) = √(36 + 1) = √(37)`

Then, remember that there is a theorem which states that

`{\displaystyle \cos(\text{Angle between vectors}) = \cos(\alpha ) = \frac{a\bullet{b}}a`

Now for our problem

`\cos(\alpha) = 0.9508084195267653\\\alpha = \arccos(0.9508084195267653) = 18.0`