Question

find the angle between ( u=langle-2,-1
angle ) and ( v=langle 1,-2
angle ) to the nearest tenth of a degree.
the angle between ( u ) and ( v ) is ( square^{circ} ).
(round to the nearest tenth.)

Explanation:

Step1: Calculate dot product of u and v

$\mathbf{u} \cdot \mathbf{v} = (-2)(1) + (-1)(-2) = -2 + 2 = 0$

Step2: Calculate magnitudes of u and v

$||\mathbf{u}|| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236$
$||\mathbf{v}|| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.236$

Step3: Find cosine of the angle

$\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} = \frac{0}{\sqrt{5} \cdot \sqrt{5}} = 0$

Step4: Compute the angle

$\theta = \arccos(0) = 90.0^\circ$

Answer:

90.0