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

Sep 23, 2017

The unit vector is =1/sqrt194〈7,-12,-1〉

#### Explanation:

The cross product of 2 vectors is calculated with the determinant

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(d , e , f\right) , \left(g , h , i\right) |$

where 〈d,e,f〉 and 〈g,h,i〉 are the 2 vectors

Here, we have veca=〈3,2,-3〉 and vecb=〈2,1,2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(3 , 2 , - 3\right) , \left(2 , 1 , 2\right) |$

$= \vec{i} | \left(2 , - 3\right) , \left(1 , 2\right) | - \vec{j} | \left(3 , - 3\right) , \left(2 , 2\right) | + \vec{k} | \left(3 , 2\right) , \left(2 , 1\right) |$

$= \vec{i} \left(2 \cdot 2 + 3 \cdot 1\right) - \vec{j} \left(3 \cdot 2 + 3 \cdot 2\right) + \vec{k} \left(3 \cdot 1 - 2 \cdot 2\right)$

=〈7,-12,-1〉=vecc

Verification by doing 2 dot products

〈7,-12,-1〉.〈3,2,-3〉=7*3-12*2+1*3=0

〈7,-12,-1〉.〈2,1,2〉=7*2-12*1-1*2=0

So,

$\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$

The modulus of $\vec{c}$ is

$| | \vec{c} | | = \sqrt{{7}^{2} + {\left(- 12\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{49 + 144 + 1} = \sqrt{194}$

Therefore,

The unit vector is

hatc=1/sqrt194〈7,-12,-1〉