# What is the unit vector that is orthogonal to the plane containing  (29i-35j-17k)  and  (41j+31k) ?

Apr 2, 2017

The unit vector is =1/1540.3〈-388,-899,1189〉

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

Here, we have veca=〈29,-35,-17〉 and vecb=〈0,41,31〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(29 , - 35 , - 17\right) , \left(0 , 41 , 31\right) |$

$= \vec{i} | \left(- 35 , - 17\right) , \left(41 , 31\right) | - \vec{j} | \left(29 , - 17\right) , \left(0 , 31\right) | + \vec{k} | \left(29 , - 35\right) , \left(0 , 41\right) |$

$= \vec{i} \left(- 35 \cdot 31 + 17 \cdot 41\right) - \vec{j} \left(29 \cdot 31 + 17 \cdot 0\right) + \vec{k} \left(29 \cdot 41 + 35 \cdot 0\right)$

=〈-388,-899,1189〉=vecc

Verification by doing 2 dot products

〈-388,-899,1189〉.〈29,-35,-17〉=-388*29+899*35-17*1189=0

〈-388,-899,1189〉.〈0,41,31〉=-388*0-899*41+1189*31=0

So,

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

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

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

$| | \vec{c} | | = \sqrt{{388}^{2} + {899}^{2} + {1189}^{2}} = \sqrt{2372466}$

The unit vector is =1/1540.3〈-388,-899,1189〉