gainejavima

2021-11-13

Use the dot product to find an angle between $u=5i+2j$ and $v=8i-3j$ to the nearest tenth of a degree.

Feas1981

Beginner2021-11-14Added 16 answers

We will use the dot product to find the angle between u and v, given

$u=5i+2j,\text{}v=8i-3j$

Remember that the dot product of two nonzero vectors u and v is defined to be

$u\cdot v=\left|\left|u\right|\right|\left|\left|v\right|\right|\mathrm{cos}\theta$

where$\theta$ is the angle between u and v.

Using the dot product formula, solving for$\mathrm{cos}\theta$ gives

$\mathrm{cos}\theta =\frac{u\cdot v}{\left|\left|u\right|\right|\cdot \left|\left|v\right|\right|}$

$=\frac{\left(5\right)\left(8\right)+\left(2\right)(-3)}{\sqrt{{5}^{2}+{2}^{2}}\sqrt{{8}^{2}+{(-3)}^{2}}}$

$=\frac{34}{\sqrt{29}\cdot \sqrt{73}}$

$=\frac{34}{\sqrt{2117}}$

Solving for the value of$\theta$ , we get

$\theta ={\mathrm{cos}}^{-\left(\frac{34}{\sqrt{2117}}\right)}$

$={42.4}^{\circ}$

Remember that the dot product of two nonzero vectors u and v is defined to be

where

Using the dot product formula, solving for

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