What is the unit vector that is orthogonal to the plane containing  (3i - j - 2k)  and  (3i – 4j + 4k) ?

Jan 8, 2018

The unit vector is $= \frac{1}{\sqrt{549}} \left(- 12 i - 18 j - 9 k\right)$

Explanation:

A vector perpendicular to 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,-1,-2〉 and vecb=〈3,-4,4〉

Therefore,

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

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

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

=〈-12,-18,-9〉=vecc

Verification by doing 2 dot products

〈3,-1,-2〉.〈-12,-18,-9〉=-3*12+1*18+2*9=0

〈3,-4,4〉.〈-12,-18,-9〉=-3*12+4*18-4*9=0

So,

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

The unit vector $\hat{c}$ in the direction of $\vec{c}$ is

$\hat{c} = \frac{\vec{c}}{\sqrt{{\left(- 12\right)}^{2} + {\left(- 18\right)}^{2} + {\left(- 9\right)}^{2}}} = \frac{\vec{c}}{\sqrt{549}}$

$= \frac{1}{\sqrt{549}} \left(- 12 i - 18 j - 9 k\right)$