# How do you find the cross product and state whether the resulting vectors are perpendicular to the given vectors <-1,1,0>times<2,1,3>?

Jan 4, 2017

$\vec{a} \times \vec{b} = + 3 \hat{\vec{i}} + 3 \hat{\vec{j}} - 3 \hat{\vec{k}}$

#### Explanation:

The vector cross product of two vectors $\vec{a} \text{ }$&$\text{ "vecb" }$

is defined as

$\vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin \theta \hat{\vec{n}}$

where $\theta \text{ }$is the angle between the vectors & #" "hatvecn" "isa unit vector mutually perpendicular to

$\vec{a} \text{ }$&$\text{ "vecb" }$

If the vectors are in component form, in particular:

$\vec{a} = \left(\begin{matrix}{a}_{1} \\ {a}_{2} \\ {a}_{3}\end{matrix}\right)$and $\vec{b} = \left(\begin{matrix}{b}_{1} \\ {b}_{2} \\ {b}_{3}\end{matrix}\right)$

we can evaluate the cross product by the use of a determinate

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

In this case we have

$\vec{a} = \left(\begin{matrix}- 1 \\ 1 \\ 0\end{matrix}\right)$$\text{ } \vec{b} = \left(\begin{matrix}2 \\ 1 \\ 3\end{matrix}\right)$

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

expanding the determinant as usual

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

$\vec{a} \times \vec{b} = + 3 \hat{\vec{i}} + 3 \hat{\vec{j}} - 3 \hat{\vec{k}}$

By definition this is perpendicular tot eh original two vectors. A quick check using the dot product will confirm this.

$\vec{a} . \left(\begin{matrix}3 \\ 3 \\ - 3\end{matrix}\right) = \left(\begin{matrix}- 1 \\ 1 \\ 0\end{matrix}\right) . \left(\begin{matrix}3 \\ 3 \\ - 3\end{matrix}\right)$

$= - 3 + 3 + 0 = 0 \text{ }$perpendicular

$\vec{b} . \left(\begin{matrix}3 \\ 3 \\ - 3\end{matrix}\right) = \left(\begin{matrix}2 \\ 1 \\ 3\end{matrix}\right) . \left(\begin{matrix}3 \\ 3 \\ - 3\end{matrix}\right)$

$= 6 + 3 - 3 = 0 \text{ }$perpendicular