Question

Find the angle between u = (8.- 3) and v = (-3,- 8) Round to the nearest tenth of a degree.

Answer 1

Answer:

B

Step-by-step explanation:

edge answer

Answer 2

Answer:

90°

Step-by-step explanation:

First you must calculate the module or the magnitude of both vectors

The module of u is:

$|u|=\sqrt{(8)^2 + (-3)^2} \\\\|u|=\sqrt{64 + 9}\\\\|u|=8.544$

The module of v is:

$|v|=\sqrt{(-3)^2 + (-8)^2} \\\\|u|=\sqrt{9 + 64}\\\\|u|=8.544$

Now we calculate the scalar product between both vectors

$u*v = 8*(-3) + (-3)*(-8)\\\\u*v = -24+ 24=0$

Finally we know that the scalar product of two vectors is equal to:

$u*v = |u||v|*cos(\theta)$

Where $\theta$ is the angle between the vectors u and v. Now we solve the equation for $\theta$

$0 = 8.544*8.544*cos(\theta)\\\\0 = cos(\theta)\\\\\theta= arcos(0)\\\\\theta=90\°$

the answer is 90°

Whenever the scalar product of two vectors is equals to zero it means that the angle between them is 90 °