# What is the cross product of [1, 4, -2] and [3, 0, 5] ?

Jan 18, 2017

$20 \hat{\vec{i}} - 11 \hat{\vec{j}} - 12 \hat{\vec{k}}$

#### Explanation:

the cross product of two vectors

$\vec{a} = \left[{a}_{1} , {a}_{2} , {a}_{3}\right]$ and $\vec{b} = \left[{b}_{1} , {b}_{2} , {b}_{3}\right]$

is calculated by the determinate

$\vec{a} \times \vec{b} = | \left(\hat{\vec{i}} , \hat{\vec{j}} , \hat{\vec{k}}\right) , \left({a}_{1} , {a}_{2} , {a}_{3}\right) , \left({b}_{1} , {b}_{2} , {b}_{3}\right) |$

so we have here

$\vec{a} \times \vec{b} = | \left(\hat{\vec{i}} , \hat{\vec{j}} , \hat{\vec{k}}\right) , \left(1 , 4 , - 2\right) , \left(3 , 0 , 5\right) |$

expanding by Row 1

$= \hat{\vec{i}} | \left(4 , - 2\right) , \left(0 , 5\right) | - \hat{\vec{j}} | \left(1 , - 2\right) , \left(3 , 5\right) | + \hat{\vec{k}} | \left(1 , 4\right) , \left(3 , 0\right) |$

$= \left(4 \times 5 - 0 \times \left(- 2\right)\right) \hat{\vec{i}} - \left(1 \times 5 - 3 \times \left(- 2\right)\right) \hat{\vec{j}} + \left(1 \times 0 - 4 \times 3\right) \hat{\vec{k}}$

$= 20 \hat{\vec{i}} - 11 \hat{\vec{j}} - 12 \hat{\vec{k}}$