Question

Two vectors are given by these expressions: $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{3}\widehat{\mathbf{i}}\mathbf{+}\mathbf{7}\widehat{\mathbf{j}}\mathbf{-}\mathbf{4}\widehat{\mathbf{k}}$and $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{6}\widehat{\mathbf{i}}\mathbf{-}\mathbf{10}\widehat{\mathbf{j}}\mathbf{+}\mathbf{9}\widehat{\mathbf{k}}$. Evaluate the quantities

$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\left[\frac{{\displaystyle |\overrightarrow{A}\times \overrightarrow{B}|}}{{\displaystyle AB}}\right]$and

$\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left[\frac{{\displaystyle |\overrightarrow{A}\times \overrightarrow{B}|}}{{\displaystyle AB}}\right]$

(c) Which give(s) the angle between the vectors?

Short answer

$$\begin{gathered} a)co{s}^1\left[\frac{{\displaystyle \overrightarrow{A}\times \overrightarrow{B}}}{{\displaystyle AB}}\right]=168.2^o \\ b)si{n}^{-1}\left[\frac{{\displaystyle |\overrightarrow{A}\times \overrightarrow{B}|}}{{\displaystyle AB}}\right]=11.9^o \\ c)Bothgivestheanglevetweenthevectors \end{gathered}$$

Step-by-step solution

Define vector product:

In physics and astronomy, vector production and scalar production are two of the most widely used vector multiplication methods. Subtracting a product of vector size four times the angle between them reveals the product vector size of two vectors.

Given Information:

The vectors, $\overrightarrow{A}=-3\widehat{i}+7\widehat{j}-4\widehat{k}$

The vector, $\overrightarrow{B}=6\widehat{i}-10\widehat{j}+9\widehat{k}$

To find the two vectors:

$$\begin{gathered} \left|\overrightarrow{A}\right|=\sqrt{{(-3)}^2+7^2+{(-4)}^2} \\ \left|\overrightarrow{A}\right|=8.6 \end{gathered}$$

And

$\left|\overrightarrow{B}\right|=\sqrt{6^2+{(-10)}^2+9^2}$

$\left|\overrightarrow{B}\right|=14.731$

Define $(\overrightarrow{A}\times \overrightarrow{B})$ as follows.

$\overrightarrow{A}\times \overrightarrow{B}=(-3\times 6)+(-7\times 10)+(-4\times 9)$

$\overrightarrow{A}\times \overrightarrow{B}=-124$

Define $(\overrightarrow{A}\times \overrightarrow{B})$by metrics.

$\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ -3& \widehat{7}& -4\\ 6& -10& 9\end{array}\right|$

$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=\widehat{i}(63-40)+\widehat{j}(-24+27)+\widehat{k}(30-42) \\ \overrightarrow{A}\times \overrightarrow{B}=23\widehat{i}+3\widehat{j}-12 \end{gathered}$$

Take a magnitude, and you have

$$\begin{gathered} |\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{{23}^2+3^2+{(-12)}^2} \\ |\overrightarrow{A}\times \overrightarrow{B}|=26.115 \end{gathered}$$

Determine the given expression as below.

a.$\begin{array}{c}co{s}^1\left[\frac{\overrightarrow{A}\times \overrightarrow{B}}{AB}\right]=co{s}^{-1}\left[\frac{-124}{8.6\times 14.731}\right]\\ =168.2°\end{array}$

And

b.$\begin{array}{c}si{n}^{-1}\left[\frac{|\overrightarrow{A}\times \overrightarrow{B}|}{AB}\right]=si{n}^{-1}\left[\frac{26.115}{8.6\times 14.731}\right]\\ =11.9°\end{array}$

c.Both gives the angle between the vectors.