Question

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

u = <6, -1>, v = <7, -4>


what can i do in this

Answer 1

Answer on quaestion 42413 - Math - Analytic Geometry

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

$u = <6, -1>, v = <7, -4>$

Let $\alpha$ - the angle beetween u and v. Then by wellknown formula

$$\cos(\alpha) = \frac{u \cdot v}{|u||v|}$$

But

$$|u| = \sqrt{u_1^2 + u_2^2} = \sqrt{6^2 + (-1)^2} = \sqrt{37}$$ $$|v| = \sqrt{v_1^2 + v_2^2} = \sqrt{7^2 + (-4)^2} = \sqrt{65}$$ $$u \cdot v = u_1 * v_1 + u_2 * v_2 = 6 * 7 + (-1) * (-4) = 46$$

So

$$\cos(\alpha) = \frac{46}{\sqrt{37 * 65}}$$ $$\alpha = \arccos\left(\frac{46}{\sqrt{37 * 65}}\right)$$

Then $\alpha$ approximately equal to 20.3 degree

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