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Find the angle between the vectors, approximate your answer to the nearest tenth: v-)->), w = (1,8)67.69135.045.0112.4°

Find the angle between the vectors, approximate your answer to the nearest tenth: v-example-1

1 Answer

6 votes


135

Explanation

Step 1

let


\begin{gathered} v=\langle-3,-2\rangle \\ w=\langle1,5\rangle \end{gathered}

we can use the product scalar formula


\begin{gathered} \vec{v}\cdot\vec{w}=\lvert\vec{v}\rvert\lvert\vec{w}\rvert\cos \emptyset \\ \end{gathered}

where theta, is the angle between the vectors, so

a) find the measure of the vector v


\begin{gathered} \lvert\vec{v}\rvert=\sqrt[]{(-3)^2+(-2)^2} \\ \lvert\vec{v}\rvert=\sqrt[]{9+4} \\ \lvert\vec{v}\rvert=√(13) \end{gathered}

b) now the other vector (w)


\begin{gathered} \lvert\vec{w}\rvert=\sqrt[]{(1)^2+(5)^2} \\ \lvert\vec{w}\rvert=\sqrt[]{1+25} \\ \lvert\vec{w}\rvert=\sqrt[]{26} \end{gathered}

Step 2

find the scalar product.

The angle between two vectors can be found using vector multiplication (scalar product)


\begin{gathered} \text{scalar product} \\ a=\langle x_1,y_1\rangle \\ b=\langle x_2,y_2\rangle \\ a\cdot b=(x_1\cdot x_2+y_1\cdot y_2) \end{gathered}
\vec{v}\cdot\vec{w}=(-3\cdot1-2\cdot5)=-3-10=-13

Step 3

Finally, replace in the formula


\begin{gathered} \vec{v}\cdot\vec{w}=\lvert\vec{v}\rvert\lvert\vec{w}\rvert\cos \emptyset \\ -13=\sqrt[]{13}\sqrt[]{26}\cos \text{ }\emptyset \\ -13=18.38\text{ cos}\emptyset \\ \text{divide both sides by 18.38} \\ (-13)/(18.38)=\frac{18.38\text{ cos}\emptyset}{18.38} \\ -0.07054=\text{ cos }\emptyset \\ \emptyset\text{= }\cos ^(-1)(-0.070) \\ \emptyset=135.01 \\ \text{rounded} \\ \emptyset=135 \end{gathered}

I hope this helps you

User Alexander Nikolov
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