# What is the unit vector that is orthogonal to the plane containing  (2i + 3j – 7k)  and  (-2i- 3j + 2k) ?

Jun 4, 2017

The unit vector is $= < - \frac{3}{\sqrt{13}} , \frac{2}{\sqrt{13}} , 0 >$

#### Explanation:

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

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

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

Here, we have veca=〈2,3,-7〉 and vecb=〈-2,-3,2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(2 , 3 , - 7\right) , \left(- 2 , - 3 , 2\right) |$

$= \vec{i} | \left(3 , - 7\right) , \left(- 3 , 2\right) | - \vec{j} | \left(2 , - 7\right) , \left(- 2 , 2\right) | + \vec{k} | \left(2 , 3\right) , \left(- 2 , - 3\right) |$

$= \vec{i} \left(3 \cdot 2 - 7 \cdot 3\right) - \vec{j} \left(2 \cdot 2 - 7 \cdot 2\right) + \vec{k} \left(- 2 \cdot 3 + 2 \cdot 3\right)$

=〈-15,10,0〉=vecc

Verification by doing 2 dot products

〈-15,10,0〉.〈2,3,-7〉=-15*2+10*3-7*0=0

〈-15,10,0〉.〈-2,-3,2〉=-15*-2+10*-3-0*2=0

So,

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

The modulus of $\vec{c}$ is $| | \vec{c} | | = \sqrt{{15}^{5} + {10}^{2}} = \sqrt{325}$

The unit vector is

$\hat{c} = \frac{\vec{c}}{|} | \vec{c} | | = \frac{1}{325} < - 15 , 10 , 0 >$

$= < - \frac{3}{\sqrt{13}} , \frac{2}{\sqrt{13}} , 0 >$