Question

Given vectors u = ⟨–3, 2⟩ and v = ⟨2, 1⟩, what is the measure of the angle between the vectors?

Answer 1

The measure of the angle between the vectors

$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }$.

What is the measure of the angle between the vectors?

Given:

$\mathrm{u}=\langle -3,2\rangle$ and $v=\langle 2,1\rangle$

Computing the angle between the vectors, we get

$\quad \cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$

To estimate the lengths of the vectors, we get

Computing the Euclidean Length of a vector,

$\left|\left(x_{1}, \ldots, x_{n}\right)\right|=\sqrt{\sum_{i=1}^{n}\left|x_{i}\right|^{2}}$

Let, $\mathrm{u} &=\langle -3,2\rangle$ and $\mathrm{v} &=\langle 2,1\rangle$

If $\mathrm{u} &=\langle -3,2\rangle$

$|u| &=\sqrt{-3^{2}+(2)^{2}}$

$&=\sqrt{5}i$ and

$\mathrm{v} &=\langle 2,1\rangle$

$|v| &=\sqrt{2^{2}+(1)^{2}}$

$&=\sqrt{5}$

Finally, the angle is given by:

Computing the angle between the vectors, we get

$\cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$

$&\cos (\Phi)=-\sqrt{13 } i/ \sqrt{5 }$

simplifying the above equation, we get

$&\Phi=\arccos (\cos (\Phi))$

$=\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }$

Therefore, the measure of the angle between the vectors

$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }$.

To learn more about vectors refer to:

brainly.com/question/19708440

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