# What is the cross product of <-3 ,-6 ,-3 > and <0 ,1 , -7 >?

Apr 27, 2016

We will use determinants to calculate cross product.

#### Explanation:

First of all, let us rewrite both vectors, in terms of the vectors of the basis of $m a t h {\boldsymbol{R}}^{3}$: $\left\{\vec{{e}_{1}} , \vec{{e}_{2}} , \vec{{e}_{3}}\right\}$ (or you may use $\left\{\vec{i} , \vec{j} , \vec{k}\right\}$).

• $< - 3 , - 6 , - 3 > = - 3 \vec{{e}_{1}} - 6 \vec{{e}_{2}} - 3 \vec{{e}_{3}}$
• $< 0 , 1 , - 7 > = \vec{{e}_{2}} - 7 \vec{{e}_{3}}$

Now, cross product of two vectors $< x , y , z >$ and $< x ' , y ' , z ' >$ is given by:

$< x , y , z > \times < x ' , y ' , z ' > = \det \left(\begin{matrix}\vec{{e}_{1}} & \vec{{e}_{2}} & \vec{{e}_{3}} \\ x & y & z \\ x ' & y ' & z '\end{matrix}\right)$

In our case:

$< - 3 , - 6 , - 3 > \times < 0 , 1 , - 7 > =$

$= \det \left(\begin{matrix}\vec{{e}_{1}} & \vec{{e}_{2}} & \vec{{e}_{3}} \\ - 3 & - 6 & - 3 \\ 0 & 1 & - 7\end{matrix}\right) =$

$= \vec{{e}_{1}} \cdot \left[\left(- 6\right) \cdot \left(- 7\right) - \left(- 3\right) \cdot 1\right] -$
$- \vec{{e}_{2}} \cdot \left[\left(- 3\right) \cdot \left(- 7\right) - \left(- 3\right) \cdot 0\right] +$
$+ \vec{{e}_{3}} \cdot \left[\left(- 3\right) \cdot 1 - \left(- 6\right) \cdot 0\right] =$

$= 45 \vec{{e}_{1}} - 21 \vec{{e}_{2}} - 3 \vec{{e}_{3}} =$

$= < 45 , - 21 , - 3 >$