Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <-5, -4>, v = <-4, -3>

Answer 1

Answer on Question #44313 – Math - Analytic Geometry

Find the angle between the given vectors to the nearest tenth of a degree.

$$u = \angle -5, -4 >, v = \angle -4, -3 >$$

Solution:

$$u = \angle -5, -4 >$$ $$v = \angle -4, -3 >$$

Magnitudes of the vectors:

$$|u| = \sqrt{(-5)^2 + (-4)^2} = \sqrt{41}$$ $$|v| = \sqrt{(-4)^2 + (-3)^2} = \sqrt{25} = 5$$

The scalar product of vectors ($\alpha$ – angle between vectors)

$$\begin{array}{l} u \cdot v = |u| \cdot |v| \cdot \cos \alpha \\ \cos \alpha = \dfrac{u \cdot v}{|u| \cdot |v|} \\ \alpha = \arccos \left( \dfrac{u \cdot v}{|u| \cdot |v|} \right) = \arccos \left( \dfrac{u \cdot v}{|u| \cdot |v|} \right) \\ = \arccos \left( \dfrac{(-5) \cdot (-4) + (-4) \cdot (-3)}{\sqrt{41} \cdot 5} \right) = \arccos \left( \dfrac{32\sqrt{41}}{41 \cdot 5} \right) = 1.8{}^\circ \end{array}$$

Answer: angle between the given vectors is $1.8{}^\circ$.

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