3.7: The Cross Product- Geometry

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The Cross Product of two Vectors and the Right-Hand Rule

The cross product of the two vectors $\overrightarrow{u}$ and $\overrightarrow{v}$, is itself a vector. Where is this vector $\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{?}$ The cross product of two vectors is a vector perpendicular to the plane formed by the two vectors. What if there are two perpendicular directions? Does this perpendicular vector lie above or below the plane formed by the two vectors? Let’s use the right-hand rule.

$Illustration of a hand showing direction of fingers representing right-hand rule and the cross product of two vectors. The index finger is u, the middle finger v, and the thumb is the cross product u x v.$

The Geometry of the Cross Product

Definition: Term

If $\theta$ is the angle between the two vectors $\overrightarrow{u}\boldsymbol{=}\left\langle u_x,\left.u_y,\ u_z\right\rangle \right.$ and $\overrightarrow{v}\boldsymbol{=}\left\langle v_x,\left.v_y,\ v_z\right\rangle \right.$, then the length (magnitude) of the cross product $\overrightarrow{u}\times \overrightarrow{v}$ is

\[\|\vec{u} \times \vec{v}\|=\|\vec{u}\|\|\vec{v}\| \sin \theta \nonumber \]

\[\|\vec{u}\|=\sqrt{u_x^2+u_y{ }^2+u_z^2} \text { and }\|\vec{v}\|=\sqrt{v_x^2+v_y^2+v_z^2} \nonumber \]

$Image showing the angle between two vectors u and v and the length of the cross product shown as u x v. U x v is perpendicular to u and v.$

Example $\PageIndex{1}$

Length of the vector $\overrightarrow{u}\times \overrightarrow{v}$, where $\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.$ and $\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.$

Solution

The length of the vector $\overrightarrow{u}\times \overrightarrow{v}$, where $\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.$ and $\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.$ is

\[\left\|\overrightarrow{u}\times \overrightarrow{v}\right\|=\ \left\|\overrightarrow{u}\right\|\left\|\overrightarrow{v}\right\|\mathrm{sin}\theta \nonumber \]

\[\sqrt{{u_x}^2+{u_y}^2+\ {u_z}^2}\bullet \sqrt{{v_x}^2+{v_y}^2+\ {v_z}^2}\bullet \mathrm{sin}\theta \nonumber \]

We now need to get $\mathrm{sin}\theta$. We’ll use the formula for the angle between two vectors.

\[\begin{gathered}
\theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \\
\theta=\cos ^{-1} \frac{\langle 5,2,4\rangle \cdot\langle 3,4,-7\rangle}{\sqrt{5^2+2^2+4^2} \cdot \sqrt{3^2+4^2+(-7)^2}} \\
\theta=\cos ^{-1} \frac{5 \cdot 3+2 \cdot 4+4 \cdot(-7)}{\sqrt{45} \cdot \sqrt{74}} \\
\theta=\cos ^{-1} \frac{-5}{\sqrt{45} \cdot \sqrt{74}} \\
\theta=94.97^{\circ}
\end{gathered} \nonumber \]

Now we compute $\left\|\overrightarrow{u}\times \overrightarrow{v}\right\|=\sqrt{45}\bullet \sqrt{74}{\mathrm{sin} \left(94.97{}^\circ \right)\ }=57.49$ units.

Using Technology

We can use technology to find the magnitude of the cross product of two vectors.

Go to www.wolframalpha.com.

To find the length of the cross product of the vectors $\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.2,\ 4\right\rangle \right.$ and $\overrightarrow{v}\boldsymbol{=}\left\langle 3,\left.4,\ -7\right\rangle \right.$ enter magnitude of $\mathrm{<}$5, 2, 4$\mathrm{>}$ cross $\mathrm{<}$3, 4, -7$\mathrm{>}$ in the entry field. Wolframalpha tells you what it thinks you entered and its answer. In this case it shows you result of $\sqrt{3305}$. Click on the approximate form button to get the result in decimal form as $57.49$.

$This screenshot from WolframAlpha shows the result as square root of 3305, with an arrow pointing to the Approximate form button.$

This screenshot from WolframAlpha shows the calculation of the magnitude of <5,2,4 x <3,4, negative 7>. The result is 57.4891." src="/@api/deki/files/97458/clipboard_e13da1df53b00e241350a56220a04e76b.png">

Area of a Parallelogram

Geometrically, $\left\|\overrightarrow{u}\times \overrightarrow{v}\right\|$ produces the area of a parallelogram determined by $\overrightarrow{u}$ and $\overrightarrow{v}$.

\[\text { Area }=\|\vec{u} \times \vec{v}\|=\|\vec{u}\|\|\vec{v}\| \sin \theta \nonumber \]

$Image showing that area of a parallelogram is determined by vectors v and u.$

The area of the parallelogram determined by the vectors $\vec{u}=\langle 5,2,4\rangle$ and $\vec{v}=\langle 3,4,-7\rangle$ from the above example is 57.49 square units.

The Cross Product of Perpendicular and Parallel Vectors

If vectors $\vec{u}$ and $\vec{v}$ are perpendicular to each other, then the angle between them is 90° and sin(90°) = 1, so that \[\vec{u} \times \vec{v}=\|\vec{u}\|\|\vec{v}\| \nonumber \]

If vectors $\vec{u}$ and $\vec{v}$ are parallel to each other, then the angle between them is $0^{\circ}$ and $\sin \left(0^{\circ}\right)=0$.

It makes sense then to define the cross product of parallel vectors to be the zero vector, $\overrightarrow{0}$. Also, if at least one of the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ is the zero vector $\overrightarrow{0}$, then the cross product $\overrightarrow{u}\times \overrightarrow{v}$ is defined to be the zero vector. We can say that if the cross product of two vectors is zero, then the vectors are parallel to each other. Also, if two vectors are parallel to each other, then their cross product is zero. We combine these statements together in an if-and-only-if statement.

Nonzero vectors $\vec{u}$ and $\vec{v}$ are parallel to each other if and only if $\vec{u} \times \vec{v}=0$.

Properties of the Cross Product

$\vec{u} \times \vec{v}=-\vec{v} \times \vec{u}$, the cross product is anti-commutative
$k(\vec{u} \times \vec{v})=k \vec{u} \times \vec{v}=\vec{u} \times k \vec{v}$, multiplication by a scalar
$\vec{u} \times(\vec{v}+\vec{w})=\vec{u} \times \vec{v}+\vec{u} \times \vec{w}$, the cross product distributes over vector addition
$\vec{u} \times \overrightarrow{0}=\overrightarrow{0}$, the cross product with the zero vector $\vec{0}$, is the zero vector $\vec{0}$

Example $\PageIndex{2}$

Use WolframAlpha to verify that \[\overrightarrow{u}\times \left(\overrightarrow{v}+\overrightarrow{w}\right)=\overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{u}\times \overrightarrow{w} \nonumber \]

where $\overrightarrow{u}\boldsymbol{=}\left\langle 5,\left.-2,\ -3\right\rangle \right.$, $\overrightarrow{v}\boldsymbol{=}\left\langle 6,\left.4,\ 1\right\rangle \right.$, and $\overrightarrow{w}\boldsymbol{=}\left\langle -3,\left.7,\ 2\right\rangle \right.$.

Solution

Use WA to first compute $\overrightarrow{u}\times \left(\overrightarrow{v}+\overrightarrow{w}\right)\boldsymbol{=}\overrightarrow{u}$ and then $\overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{u}\times \overrightarrow{w}$. Determine if the results do or do not match. We can do this in one step by entering

$\mathrm{<}$5,-2,-3$\mathrm{>}$ x ($\mathrm{<}$6,4,1$\mathrm{>}$ + $\mathrm{<}$-3,7,2$\mathrm{>}$) = $\mathrm{<}$5,-2,-3$\mathrm{>}$ x $\mathrm{<}$6,4,1$\mathrm{>}$ + $\mathrm{<}$5,-2,-3$\mathrm{>}$ x $\mathrm{<}$-3,7,2$\mathrm{>}$

If the statement on the left of the = equals the statement on the right, WA responds with True.

If the statement on the left of the $\mathrm{\neq}$ equals the statement on the right, WA responds with False.

In this case, we get a True response and have verified the truth of the statement.

This screenshot from WolframAlpha shows if a calculation is true or not. The calculation is <5,negative 2,3 x (<6,4,1>+) =<5,negative 2,negative 3> x <6,4,1> + <5,negative 2,negative 3> x . The result is True." src="/@api/deki/files/97460/clipboard_edbe1ddbbffad25bcefc5fe83c7b2ec61.png">

Try These

Exercise $\PageIndex{1}$

Find the cross product of the vectors $\vec{u}=\langle-4,3,5\rangle$ and $\vec{v}=\langle 5,-1,2\rangle$.

Answer: $\overrightarrow{u}\times \overrightarrow{v}\boldsymbol{=\ }\left\langle 11,\left.33,-11\right\rangle \right.$

Exercise $\PageIndex{2}$

Find the cross product of the vectors $\vec{u}=\langle-2,3,-9\rangle$ and $\vec{v}=\langle 6,-9,27\rangle$.

Answer: $\overrightarrow{u}\times \overrightarrow{v}=\left.\boldsymbol{\langle }0,\ 0,\ 0\right\rangle = \overrightarrow{0}$

Exercise $\PageIndex{3}$

Find the length of the vector formed by the cross product of the vectors $\vec{u}=\langle 3,-5,4\rangle$ and $\vec{v}=\langle 2,-4,1\rangle$.

Answer: $5\sqrt{6}$ = 12.2 units

Exercise $\PageIndex{4}$

Find the angle between the vectors $\vec{u}=\langle 4,-7,-6\rangle$ and $\vec{v}=\langle 5,-1,2\rangle$.

Answer: $\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{‖\overrightarrow{u}‖∙‖\overrightarrow{v}‖}$$\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\left\langle 4,\left.-7,6\right\rangle \right.\boldsymbol{\bullet }\left\langle 5,\left.-1,\ 2\right\rangle \right.}{\sqrt{4^2+{(-7)}^2+\ 6^2}\bullet \sqrt{5^2+{(-1)}^2+\ 2^2}}$$\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{4∙5+(-7)∙(-1)+6∙2}{\sqrt{91}\bullet \sqrt{30}}$$\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{25}{\sqrt{101}\bullet \sqrt{30}}$$\theta =\ \mathrm{74.19}\mathrm{{}^\circ }$

Exercise $\PageIndex{5}$

Determine if the vectors $\vec{u}=\langle 3,-2,1\rangle$ and $\vec{v}=\langle 0,2,4\rangle$ are perpendicular or parallel to each other.

Answer: Perpendicular since $\overrightarrow{u}\bullet \overrightarrow{v}\boldsymbol{=}0$

Exercise $\PageIndex{6}$

Find the area of the parallelogram and the triangle formed by the vectors $\vec{u}=\langle 1,-2,-4\rangle$ and $\vec{v}=\langle 4,3,-5\rangle$.

Answer: Parallelogram is $26.94$ square units. Triangle is (½ of $26.94)$ = $13.47$ square units.

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Definition: Term

Example \(\PageIndex{1}\)

Solution

Properties of the Cross Product

Example \(\PageIndex{2}\)

Solution

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)

Exercise \(\PageIndex{4}\)

Exercise \(\PageIndex{5}\)

Exercise \(\PageIndex{6}\)