Question

Use the definition of scalar product, ${}^{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}\mathbf{.}}\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{=}\mathbf{a}\mathbf{b}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$and the fact that${}^{\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{=}{\mathbf{a}}_{\mathbf{x}}\mathbf{}{\mathbf{b}}_{\mathbf{x}}\mathbf{+}{\mathbf{a}}_{\mathbf{y}}\mathbf{}{\mathbf{b}}_{\mathbf{y}}\mathbf{+}{\mathbf{a}}_{\mathbf{z}}\mathbf{}{\mathbf{b}}_{\mathbf{z}}}$to calculate the angle between two vectors given bylocalid="1654245592038" ${}^{\overrightarrow{\mathbf{a}\mathbf{}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{k}}\mathbf{}\mathbf{}$and${}^{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{k}}$.

Short answer

Angle$\left(\mathrm{\theta }\right)$between $\stackrel{\rightharpoonup }{a}$and $\stackrel{\rightharpoonup }{b}$is ${22}^{°}$.$\stackrel{\rightharpoonup }{\mathrm{a}}\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{b}}$

Step-by-step solution

Given

$$\begin{gathered} \overrightarrow{a}=3.0\hat{i}+3.0\hat{j}+3.0\hat{k} \\ \overrightarrow{b}=2.0\hat{i}+1.0\hat{j}+3.0\hat{k} \end{gathered}$$

Understanding the concept

As we have given the unit vectors, first we calculate the magnitude for the individual vectors and dot product. Using the formula for dot product we can find the angle between them.

Formula:

The dot product is written as,

$\stackrel{\rightharpoonup }{a}.\stackrel{\rightharpoonup }{b}=ab\cos \theta$

The dot product is written as,

$a=\sqrt{{\left(a_x\right)}^2+{\left(a_y\right)}^2}$

Calculate the angle betweena→=3.00i^+3.00j^+3.00k^andb^=2.00i^+1.00j^+3.00k^ 

Use equation (ii) to calculate the magnitude of the vectors${}^{aandb}$. Therefore,

$$\begin{gathered} \left|\overrightarrow{a}\right|=\sqrt{3.0^2+3.0^2+3.0^2} \\ =5.2 \\ \left|\overrightarrow{b}\right|=\sqrt{2.0^2+1.0^2+3.0^2} \\ =3.74 \end{gathered}$$

The magnitude of the dot product is calculated as,

$$\begin{gathered} \left|\overrightarrow{a}.\overrightarrow{b}\right|=\left(3\times 2\right)+\left(3\times 1\right)+\left(3\times 3\right) \\ =6+3+9 \\ =18 \end{gathered}$$

Now, use the equation (i) to find the angle.

$$\begin{gathered} 18.0=5.2\times 3.74\times \cos \theta \\ \cos \theta =\frac{18.0}{19.24} \\ \theta ={\cos }^{-1}\left(\frac{18.0}{19.448}\right) \\ \theta =22° \end{gathered}$$

So, the angle between the two given vectors is${}^{22°}$
.