Respuesta :
Dot product between vectors
We have two different kinds of products between vectors.
The dot product is given by two expressions:
[tex]\begin{gathered} (1)\hat{u}\cdot\hat{v}=u_xv_x+u_yv_y \\ (2)\hat{u}\cdot\hat{v}=\hat{|u|}|\hat{v|}\cos \theta \end{gathered}[/tex]First expression
We have that
[tex]\begin{gathered} \hat{u}=\langle u_x,u_y\rangle=\langle6,-5\rangle \\ \hat{v}=\langle v_x,v_y\rangle=\langle11,8\rangle \end{gathered}[/tex]Then,
[tex]\begin{gathered} u_x=6\text{ and }u_y=-5 \\ v_x=11\text{ and }v_y=8 \end{gathered}[/tex]Then, for the first expression we have that:
[tex]\begin{gathered} \hat{u}\cdot\hat{v}=u_xv_x+u_yv_y \\ \downarrow \\ \hat{u}\cdot\hat{v}=6\cdot11+(-5)\cdot8 \\ =66-40=26 \end{gathered}[/tex]Then
[tex]\hat{u}\cdot\hat{v}=26[/tex]Second expression
We have that it can be expressed in another way. It will help us to find the answer:
[tex]\hat{u}\cdot\hat{v}=\hat{|u|}|\hat{v|}\cos \theta[/tex]We have that
|u| and |v| are the magnitudes of each vector.
They are given by:
[tex]\begin{gathered} \hat{|u|}=\sqrt{u^2_x_{}+u^2_y} \\ |\hat{v|}=\sqrt{v^2_x+v^2_y} \end{gathered}[/tex]Then, we have that
[tex]\begin{gathered} \hat{|u|}=\sqrt{u^2_x_{}+u^2_y} \\ \downarrow\text{ since }u_x=6\text{ and }u_y=-5 \\ \hat{|u|}=\sqrt[]{6^2+(-5)^2}=\sqrt[]{36+25}=\sqrt[]{61} \end{gathered}[/tex]and
[tex]\begin{gathered} \hat{|v|}=\sqrt[]{v^2_x+v^2_y} \\ \downarrow\text{ since }v_x=11\text{ andv}_y=8 \\ \hat{|v|}=\sqrt[]{11^2+8^2}=\sqrt[]{121+64}=\sqrt[]{185} \end{gathered}[/tex]Then, using the second expression:
[tex]\begin{gathered} \hat{u}\cdot\hat{v}=\hat{|u|}|\hat{v|}\cos \theta \\ \downarrow \\ \hat{u}\cdot\hat{v}=\sqrt[]{61}\cdot\sqrt[]{185}\cos \theta=\sqrt[]{11,285}\cos \theta \end{gathered}[/tex]Equation for the angle
Then, we have that both expressions are equal:
[tex]\begin{gathered} \hat{u}\cdot\hat{v}=26 \\ \hat{u}\cdot\hat{v}=\sqrt[]{11,285}\cos \theta \end{gathered}[/tex]Then,
[tex]26=\sqrt[]{11,285}\cos \theta[/tex]Using this equation we can find θ.
We solve this equation for θ by "leaving it alone" on the right isde of the equation:
[tex]\begin{gathered} 26=\sqrt[]{11,285}\cos \theta \\ \downarrow\text{ taking }\sqrt[]{11,285}\text{ to the left side} \\ \frac{26}{\sqrt[]{11,285}}=\cos \theta \end{gathered}[/tex]Using the calculator we have that:
[tex]\begin{gathered} \sqrt{11,285}\cong106.23 \\ \downarrow \\ \frac{26}{\sqrt[]{11,285}}\cong\frac{26}{106.23}\cong0.244 \\ \downarrow \\ 0.244\cong\cos \theta \end{gathered}[/tex]To solve the equation for θ, we take the cos to the left side of the equation using its inverse function arccos:
[tex]\begin{gathered} 0.244\cong\cos \theta \\ \downarrow \\ \arccos (0.244)\cong\theta \end{gathered}[/tex]Using a calculator for arccos(0.244):
[tex]\arccos (0.244)\cong75.83[/tex]Then
θ ≅ 75.83º
