Question

Using the definition of the scalar product, find the angles between

(a) $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{3}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\hat{\mathbf{j}}$and $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{4}\hat{\mathbf{i}}\mathbf{-}\mathbf{4}\hat{\mathbf{j}}$,

(b) $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{2}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\hat{\mathbf{j}}$and $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{3}\hat{\mathbf{i}}\mathbf{-}\mathbf{4}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\hat{\mathbf{k}}$, and

(c) $\overrightarrow{\mathbf{A}}\mathbf{=}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\hat{\mathbf{k}}$and $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{3}\hat{\mathbf{j}}\mathbf{+}\mathbf{4}\hat{\mathbf{k}}$.

Short answer

(a) The angle between vectors $\overrightarrow{A}$ and $\overrightarrow{\mathrm{B}}$ is $11°$.

(b)The angle between vectors $\overrightarrow{A}$ and $\overrightarrow{\mathrm{B}}$ is $155.94°$.

(c) The angle between vectors $\overrightarrow{A}$ and $\overrightarrow{\mathrm{B}}$ is $82.34°$.

Step-by-step solution

Given data

The provided vectors are

(a)$\overrightarrow{A}=3\hat{i}-2\hat{j}$and $\overrightarrow{B}=4\hat{i}-4\hat{j}$

(b) $\overrightarrow{A}=-2\hat{i}+4\hat{j}$ and $\overrightarrow{B}=3\hat{i}-4\hat{j}+2\hat{k}$

(c) $\overrightarrow{A}=\hat{i}-2\hat{j}+2\hat{k}$and $\overrightarrow{B}=3\hat{j}+4\hat{k}$

Angle between two vectors

The angle between two vectors $\overrightarrow{A}$ and $\overrightarrow{\mathrm{B}}$ is

$\mathit{\theta }\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}}{\left|\overrightarrow{\mathit{A}}\right|\left|\overrightarrow{\mathit{B}}\right|}\mathbf{\right)}$ .....(I)

Determining the angle between vectors in the first case

The dot product of the two vectors is

$\begin{array}{c}\overrightarrow{A}·\overrightarrow{B}=\left(3\hat{i}-2\hat{j}\right)·\left(4\hat{i}-4\hat{j}\right)\\ =3\times 4-2\times \left(-4\right)\\ =20\end{array}$

The magnitude of vector $\overrightarrow{A}$is

$\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{3^2+{\left(-2\right)}^2}\\ =\sqrt{13}\\ =3.6\end{array}$

The magnitude of vector $\overrightarrow{B}$ is

$\begin{array}{c}\left|\overrightarrow{B}\right|=\sqrt{4^2+{\left(-4\right)}^2}\\ =\sqrt{32}\\ =5.66\end{array}$

Thus from equation (I), the angle between them is

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{20}{3.6\times 5.66}\right)\\ ={\cos }^{-1}0.98\\ =11°\end{array}$

Thus the required angle is $11°$.

Determining the angle between vectors in the second case

The dot product of the two vectors is

$\begin{array}{c}\overrightarrow{A}·\overrightarrow{B}=\left(-2\hat{i}+4\hat{j}\right)·\left(3\hat{i}-4\hat{j}+2\hat{k}\right)\\ =-2\times 3+4\times \left(-4\right)+0\times 2\\ =-22\end{array}$

The magnitude of vector $\overrightarrow{A}$ is

$\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{{\left(-2\right)}^2+4^2}\\ =\sqrt{20}\\ =4.47\end{array}$

The magnitude of vector $\overrightarrow{\mathrm{B}}$ is

$\begin{array}{c}\left|\overrightarrow{B}\right|=\sqrt{3^2+{\left(-4\right)}^2+2^2}\\ =\sqrt{29}\\ =5.39\end{array}$

Thus from equation (I), the angle between them is

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-22}{4.47\times 5.39}\right)\\ ={\cos }^{-1}\left(-0.913\right)\\ =155.94°\end{array}$

Thus the required angle is $155.94°$.

Determining the angle between vectors in the third case

The dot product of the two vectors is

$\begin{array}{c}\overrightarrow{A}·\overrightarrow{B}=\left(\hat{i}-2\hat{j}+2\hat{k}\right)·\left(3\hat{j}+4\hat{k}\right)\\ =1\times 0-2\times 3+2\times 4\\ =2\end{array}$

The magnitude of vector $\overrightarrow{A}$ is

$\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{1^2+{\left(-2\right)}^2+2^2}\\ =\sqrt{9}\\ =3\end{array}$

The magnitude of vector $\overrightarrow{\mathrm{B}}$ is

$\begin{array}{c}\left|\overrightarrow{B}\right|=\sqrt{3^2+4^2}\\ =\sqrt{25}\\ =5\end{array}$

Thus from equation (I), the angle between them is

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{2}{3\times 5}\right)\\ ={\cos }^{-1}\left(0.133\right)\\ =82.34°\end{array}$

Thus the required angle is $82.34°$.