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Mrac [35]
4 years ago
12

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

Mathematics
2 answers:
trasher [3.6K]4 years ago
6 0

Answer:

B

Step-by-step explanation:

edge answer

Nimfa-mama [501]4 years ago
5 0

Answer:

<h2>90°</h2>

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 °

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