Show that$\mathbf{}\mathbf{2}\hat{\mathbf{i}}\mathbf{-}\hat{\mathbf{j}}\mathbf{+}\mathbf{4}\hat{\mathbf{k}}\mathbf{}$and $\mathbf{5}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\hat{\mathbf{j}}\mathbf{-}\mathbf{2}\hat{\mathbf{k}}$are orthogonal (perpendicular). Find a third vector perpendicular to both.

Write the expression of the angle between vectors.

$\mathbf{cos}\mathbf{}\mathbf{\theta }\mathbf{=}\left(\frac{\overrightarrow{A}-\overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)$ ….. (1)

Here, $\mathbf{\theta }$is the angle between vectors localid="1658990679832" $\overrightarrow{\mathbf{A}}\mathbf{}$and $\overrightarrow{\mathbf{B}}$.

Write the expression for perpendicular vector,

localid="1658989582602" $\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{=}\left|\begin{array}{ccc}\hat{i}& \hat{i}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$ ….. (2)

Here, $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$are vectors in one plane.

The dot product of vectors is calculated as follows,

$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=\left(2\hat{i}-\hat{j}+4\hat{k}\right)\times \left(5\hat{i}-2\hat{j}+2\hat{k}\right) \\ =10-2-8 \\ =0 \end{gathered}$$

Substituting 0 for$\overrightarrow{A}·\overrightarrow{B}$into equation (1).

$$\begin{gathered} \cos \theta =\left(\frac{0}{\left|A\right|\left|B\right|}\right) \\ =0 \\ \theta ={\cos }^{-1}\left(0\right) \\ =90° \end{gathered}$$

Hence, the angle between vectors is$90°.$Thus, the vectors are orthogonal.

The Perpendicular vector from equation (2) is calculated as follows,

$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{i}& \hat{i}\\ 2& -1& 4\\ 5& 2& -2\end{array}\right| \\ =\hat{i}\left[\left(-1\right)\left(-2\right)-\left(2\right)\left(4\right)\right]-\hat{j}\left[\left(2\right)\left(-2\right)-\left(5\right)\left(4\right)\right]+\hat{k}\left[\left(2\right)\left(2\right)-\left(5\right)\left(-1\right)\right] \\ =-6\hat{i}+24\hat{j}+9\hat{k} \end{gathered}$$

Vector can reduce as follows,

role="math" localid="1658990584400" $\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=3(-2\hat{\mathrm{i}}+8\mathrm{j}+3\hat{\mathrm{k}})$

The vector perpendicular to the given vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is $(-2\hat{\mathrm{i}}+8\hat{\mathrm{j}}+3\hat{\mathrm{k}})$.