# What is the unit vector that is normal to the plane containing ( i +k ) and (2i+ j - 3k)?

May 27, 2016

+-(3hati-3hatj+hatk)/(sqrt19

#### Explanation:

If $\vec{A} = \hat{i} + \hat{j} \mathmr{and} \vec{B} = 2 \hat{i} + \hat{j} - 3 \hat{k}$
then vectors which will be normal to the plane containing $\vec{A} \mathmr{and} \vec{B}$ are either$\vec{A} \times \vec{B} \mathmr{and} \vec{B} \times \vec{A}$ .So we are to find out the unit vectors of these two vector . One is opposite to another.

Now $\vec{A} \times \vec{B} = \left(\hat{i} + \hat{j} + 0 \hat{k}\right) \times \left(2 \hat{i} + \hat{j} - 3 \hat{k}\right)$
$= \left(1 \cdot \left(- 3\right) - 0 \cdot 1\right) \hat{i} + \left(0 \cdot 2 - \left(- 3\right) \cdot 1\right) \hat{j} + \left(1 \cdot 1 - 1 \cdot 2\right) \hat{k}$
$= - 3 \hat{i} + 3 \hat{j} - \hat{k}$

So unit vector of $\vec{A} \times \vec{B} = \frac{\vec{A} \times \vec{B}}{|} \vec{A} \times \vec{B} |$
=-(3hati-3hatj+hatk)/(sqrt(3^2+3^2+1^2))=-(3hati-3hatj+hatk)/(sqrt19#

And unit vector of $\vec{B} \times \vec{A} = + \frac{3 \hat{i} - 3 \hat{j} + \hat{k}}{\sqrt{19}}$