Question

Find a unit vector perpendicular to the vectors a=4i-j+3k and b=-2i+j-2k

Answer 1

Please see the attachment

Answer 2

Answer:

The unit vector perpendicular to the given vectors a and b is given by

=$-\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$.

Step-by-step explanation:

We are given that two vectors are

$\vec{a}=4\hat{i}-\hat{j}+3\hat{k}$

$\vec{b}=-2\hat{i}+\hat{j}-2\hat{k}$

Let c is a unit vector which is perpendicular to the given vectors.

$\vec{c}=\vec{a}\times\vec{b}$

We have to find a vector which is perpendicular to both vectors a and b.

Therefore, we find $\vec{a}\times\vec{b}$ by determinant

$\vec{a}\times\vec{b}=\begin{vmatrix}i&j&k\\4&-1&3\\-2&1&-2\end{vmatrix}$

Expand along $R_1$

$\vec{a}\times\vec{b} = \hat{i}(2-3)-\hat{j}(-8+6)+\hat{k}(4-2)$

$\vec{a}\times\vec{b}=-\hat{i}+2\hat{j}+2\hat{k}$

$\mid{\vec{a}\times\vec{b}}\mid=\sqrt{1^1+2^2+2^2}$

$\mid{\vec{a}\times\vec{b}}\mid=\sqrt{1+4+4}$

$\mid{\vec{a}\times\vec{b}}\mid=\sqrt{9}=3$

The unit vector perpendicular to both vectors a and b is given by

$\hat{c}=\frac{\vec{c}}{\mid{c}\mid}$

The unit vector, $\hat{c}=\frac{-\hat{i}+2\hat{j}+2\hat{2}}{3}$

Hence, the unit vector c is perpendicular to the vectors a and b is given by

$\hat{c}=-\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$.