# What is the cross product of <-3,5,8 > and <6, -2, 7 >?

Apr 22, 2017

The vector is =〈51,69,-24〉=

#### Explanation:

The cross product of 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=〈-3,5,8〉 and vecb=〈6,-2,7〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(- 3 , 5 , 8\right) , \left(6 , - 2 , 7\right) |$

$= \vec{i} | \left(5 , 8\right) , \left(- 2 , 7\right) | - \vec{j} | \left(- 3 , 8\right) , \left(6 , 7\right) | + \vec{k} | \left(- 3 , 5\right) , \left(6 , - 2\right) |$

$= \vec{i} \left(5 \cdot 7 + 2 \cdot 8\right) - \vec{j} \left(- 3 \cdot 7 - 6 \cdot 8\right) + \vec{k} \left(3 \cdot 2 - 5 \cdot 6\right)$

=〈51,69,-24〉=vecc

Verification by doing 2 dot products

〈51,69,-24〉.〈-3,5,8〉=-51*3+69*5-24*8=0

〈51,69,-24〉.〈6,-2,7〉=51*6-69*2-24*7=0

So,

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

Apr 22, 2017

$= \left(51 , 69 , - 24\right) .$

#### Explanation:

Let, $\vec{x} = \left(- 3 , 5 , 8\right) = - 3 i + 5 j + 8 k ,$ and,

$\vec{y} = \left(6 , - 2 , 7\right) = 6 i - 2 j + 7 k .$

$\therefore \vec{x} \times \vec{y} = | \left(i , j , k\right) , \left(- 3 , 5 , 8\right) , \left(6 , - 2 , 7\right) |$

$= \left(35 - \left(- 16\right)\right) i - \left(- 21 - 48\right) j + \left(6 - 30\right) k$

$= 51 i + 69 j - 24 k$

$= \left(51 , 69 , - 24\right) .$