Question

# Find the angle between each of the following pairs of vectors \vec{A}=A_x\hat{i}+A_y\hat{j} and \vec{B}=B_x\hat{i}+B_y\hat{j} A_{x_1}=-2.20,\ A_{y_1}=6.60\ B_{x_1}=2.00,\ B_{y_1}=-2.30

Vectors
Find the angle between each of the following pairs of vectors $$\vec{A}=A_x\hat{i}+A_y\hat{j}$$ and $$\vec{B}=B_x\hat{i}+B_y\hat{j}$$
$$A_{x_1}=-2.20,\ A_{y_1}=6.60\ B_{x_1}=2.00,\ B_{y_1}=-2.30$$

2021-05-24
The expression for the vector form of $$\vec{A}$$ is,
$$\vec{A}=A_x\hat{i}+A_y\hat{j}$$
The expression for the vector form of $$\vec{B}$$ is,
$$B=B_x\hat{i}+B_y\hat{j}$$
The expression for the magnitude of the vector $$\vec{A}$$ is,
$$|\vec{A}|=\sqrt{A_x^2+A_y^2}$$
The expression for the magnitude of the vector $$\vec{B}$$ is,
$$|\vec{B}|=\sqrt{B_x^2+B_y^2}$$
The vector form of the $$\vec{A_2}$$ is,
$$\vec{A_1}=\vec{A_{x,1}\hat{i}+A_{y,1}\hat{j}}$$
$$=(-2.20)\hat{i}+(6.60)\hat{j}$$
$$=-(2.20)\hat{i}+(6.60)\hat{j}$$
The vector form of $$\vec{B_2}$$ is,
$$\vec{B_2}=B_{x,2}\hat{i}+B_{y,2}\hat{j}$$
$$=(11.8)\hat{i}+(6.80)\hat{j}$$
The expression for the dot product of the vectors $$\vec{A_2}$$ and $$\vec{B_2}$$ is,
$$\vec{A_2}\cdot\vec{B_2}=|\vec{A_1}||\vec{B_1}|\cos\theta_{A_1B_1}$$
$$\theta_{A_2B_2}=\cos^{-1}(\frac{\vec{A}\cdot\vec{B}}{|\vec{A_2}||\vec{B_2}})$$
$$=\cos^{-1}\left(\frac{(-(2.20)\hat{i}+(6.60)\hat{j})\cdot((2.00)\hat{i}-(2.30)\hat{j})}{(\sqrt{(-2.20)^2+(6.60)^2})(\sqrt{(2.00)^2+(-2.30)^2})}\right)$$
$$=22.57^\circ$$