# Question 8bef5

Apr 21, 2017

The unit vector is =〈10/sqrt230,3/sqrt230,11/sqrt230〉

#### Explanation:

The vector perpendicular to the plane containing 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 〈d,e,f〉 and 〈g,h,i〉 are the 2 vectors

Here, we have veca=〈2,-3,-1〉 and vecb=〈1,4,-2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(2 , - 3 , - 1\right) , \left(1 , 4 , - 2\right) |$

$= \vec{i} | \left(- 3 , - 1\right) , \left(4 , - 2\right) | - \vec{j} | \left(2 , - 1\right) , \left(1 , - 2\right) | + \vec{k} | \left(2 , - 3\right) , \left(1 , 4\right) |$

$= \vec{i} \left(- 3 \cdot - 2 + 1 \cdot 4\right) - \vec{j} \left(- 2 \cdot 2 + 1 \cdot 1\right) + \vec{k} \left(2 \cdot 4 + 3 \cdot 1\right)$

=〈10,3,11〉=vecc

Verification by doing 2 dot products

〈10,3,11〉.〈2,-3,-1〉=10*2-3*3-11*1=0

〈10,3,11〉.〈1,4,-2〉=10*1+3*4-11*2=0

So,

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

The unit vector is

$= \frac{\vec{c}}{| | \vec{c} | |}$

=(〈10,3,11〉)/(||〈10,3,11〉||)

=1/sqrt(100+9+121)〈10,3,11〉

=1/sqrt230〈10,3,11〉#