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
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asked 2021-05-23
Find the angle between each of the following pairs of vectors \(\vec{A}=A_x\hat{i}+A_y\hat{j}\) and \(\vec{B}=B_x\hat{i}+B_y\hat{j}\)
\(A_{x_1}=-2.20,\ A_{y_1}=6.60\ B_{x_1}=2.00,\ B_{y_1}=-2.30\)

Answers (1)

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,
\(\vec{A_2}\cdot\vec{B_2}=|\vec{A_1}||\vec{B_1}|\cos\theta_{A_1B_1}\)
\(\theta_{A_2B_2}=\cos^{-1}(\frac{\vec{A}\cdot\vec{B}}{|\vec{A_2}||\vec{B_2}})\)
\(=\cos^{-1}\left(\frac{(-(2.20)\hat{i}+(6.60)\hat{j})\cdot((2.00)\hat{i}-(2.30)\hat{j})}{(\sqrt{(-2.20)^2+(6.60)^2})(\sqrt{(2.00)^2+(-2.30)^2})}\right)\)
\(=22.57^\circ\)
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