# How do you find the cross product <0,1,2>times<1,1,4> and verify that the resulting vectors are perpendicular to the given vectors?

Jul 7, 2016

To take the cross product, for some

$\vec{a} = \left\langle{a}_{1} , {a}_{2} , {a}_{3}\right\rangle$

and

$\vec{b} = \left\langle{b}_{1} , {b}_{2} , {b}_{3}\right\rangle ,$

you can use this formula:

$\setminus m a t h b f \left(\vec{a} \times \vec{b} = \left\langle{a}_{2} {b}_{3} - {a}_{3} {b}_{2} , {a}_{3} {b}_{1} - {a}_{1} {b}_{3} , {a}_{1} {b}_{2} - {a}_{2} {b}_{1}\right\rangle\right)$

If you notice, the subscripts are cyclic. They go $23 , 31 , 12$, and you subtract the "mirror-image" subscripts.

So, the cross product is:

$\textcolor{b l u e}{\vec{a} \times \vec{b}}$

$= \left\langle1 \cdot 4 - 2 \cdot 1 , 2 \cdot 1 - 0 \cdot 4 , 0 \cdot 1 - 1 \cdot 1\right\rangle$

$= \textcolor{b l u e}{\left\langle2 , 2 , - 1\right\rangle}$

An easy way to verify the perpendicularity of $\vec{a} \times \vec{b}$ is by performing the dot product with $\vec{a}$ and then with $\vec{b}$. If they are perpendicular, and neither of them is $\vec{0}$ (which would defeat the purpose), then the dot product should be $0$.

(It doesn't matter which direction you do it either, because dot products are commutative.)

$\textcolor{b l u e}{\left(\vec{a} \times \vec{b}\right) \cdot \vec{a}} = \vec{a} \cdot \left(\vec{a} \times \vec{b}\right)$

$= \left\langle2 , 2 , - 1\right\rangle \cdot \left\langle0 , 1 , 2\right\rangle$

$= 2 \cdot 0 + 2 \cdot 1 + \left(- 1\right) \cdot 2$

$= \textcolor{b l u e}{0}$

$\textcolor{b l u e}{\left(\vec{a} \times \vec{b}\right) \cdot \vec{b}} = \vec{b} \cdot \left(\vec{a} \times \vec{b}\right)$

$= \left\langle2 , 2 , - 1\right\rangle \cdot \left\langle1 , 1 , 4\right\rangle$

$= 2 \cdot 1 + 2 \cdot 1 + \left(- 1\right) \cdot 4$

$= \textcolor{b l u e}{0}$

And to further prove this, here is another formula you should know (modified to substitute $\vec{a} \times \vec{b}$ for one of the vectors):

$\setminus m a t h b f \left(\left(\vec{a} \times \vec{b}\right) \cdot \vec{a} = | | \left(\vec{a} \times \vec{b}\right) | | \cdot | | \vec{a} | | \cos \theta\right)$
$\setminus m a t h b f \left(\left(\vec{a} \times \vec{b}\right) \cdot \vec{b} = | | \left(\vec{a} \times \vec{b}\right) | | \cdot | | \vec{b} | | \cos \theta\right)$

Since both results were $0$, and since neither vector was $\vec{0}$ (hence their norms are nonzero), it must be that $\cos \theta = 0$. Therefore, $\theta = {90}^{\circ}$ or ${270}^{\circ}$, and either way, CW or CCW, that indicates perpendicularity.

Hence, $\setminus m a t h b f \left(\vec{a} \times \vec{b}\right)$ is perpendicular to $\setminus m a t h b f \left(\vec{a}\right)$ and is also perpendicular to $\setminus m a t h b f \left(\vec{b}\right)$.