2.4 The cross product  (Page 4/16)

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Using expansion along the first row to compute a $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ Determinant

Evaluate the determinant $|\begin{array}{ccc}\hfill 2& \hfill 5& \hfill -1\\ \hfill -1& \hfill 1& \hfill 3\\ \hfill -2& \hfill 3& \hfill 4\end{array}|.$

We have

$\begin{array}{cc}\hfill |\begin{array}{ccc}\hfill 2& \hfill 5& \hfill -1\\ \hfill -1& \hfill 1& \hfill 3\\ \hfill -2& \hfill 3& \hfill 4\end{array}|& =2|\begin{array}{cc}1\hfill & 3\hfill \\ 3\hfill & 4\hfill \end{array}|-5|\begin{array}{cc}-1\hfill & 3\hfill \\ -2\hfill & 4\hfill \end{array}|-1|\begin{array}{cc}-1\hfill & 1\hfill \\ -2\hfill & 3\hfill \end{array}|\hfill \\ & =2\left(4-9\right)-5\left(-4+6\right)-1\left(-3+2\right)\hfill \\ & =2\left(-5\right)-5\left(2\right)-1\left(-1\right)=-10-10+1\hfill \\ & =-19.\hfill \end{array}$

Evaluate the determinant $|\begin{array}{ccc}\hfill 1& \hfill -2& \hfill -1\\ \hfill 3& \hfill 2& \hfill -3\\ \hfill 1& \hfill 5& \hfill 4\end{array}|.$

$40$

Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.

Rule: cross product calculated by a determinant

Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ be vectors. Then the cross product $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ is given by

$\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=|\begin{array}{ccc}\text{i}\hfill & \text{j}\hfill & \text{k}\hfill \\ {u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \end{array}|=|\begin{array}{cc}{u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{2}\hfill & {v}_{3}\hfill \end{array}|\text{i}-|\begin{array}{cc}{u}_{1}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{3}\hfill \end{array}|\text{j}+|\begin{array}{cc}{u}_{1}\hfill & {u}_{2}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill \end{array}|\text{k}.$

Using determinant notation to find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}$

Let $\text{p}=⟨-1,2,5⟩$ and $\text{q}=⟨4,0,-3⟩.$ Find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}.$

We set up our determinant by putting the standard unit vectors across the first row, the components of $\text{u}$ in the second row, and the components of $\text{v}$ in the third row. Then, we have

$\begin{array}{cc}\hfill \text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -1& \hfill 2& \hfill 5\\ \hfill 4& \hfill 0& \hfill -3\end{array}|=|\begin{array}{cc}\hfill 2& \hfill 5\\ \hfill 0& \hfill -3\end{array}|\text{i}-|\begin{array}{cc}\hfill -1& \hfill 5\\ \hfill 4& \hfill -3\end{array}|\text{j}+|\begin{array}{cc}\hfill -1& \hfill 2\\ \hfill 4& \hfill 0\end{array}|\text{k}\hfill \\ & =\left(-6-0\right)\text{i}-\left(3-20\right)\text{j}+\left(0-8\right)\text{k}\hfill \\ & =-6\text{i}+17\text{j}-8\text{k}.\hfill \end{array}$

Notice that this answer confirms the calculation of the cross product in [link] .

Use determinant notation to find $\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b},$ where $\text{a}=⟨8,2,3⟩$ and $\text{b}=⟨-1,0,4⟩.$

$8\text{i}-35\text{j}+2\text{k}$

Using the cross product

The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped . The following examples illustrate these calculations.

Finding a unit vector orthogonal to two given vectors

Let $\text{a}=⟨5,2,-1⟩$ and $\text{b}=⟨0,-1,4⟩.$ Find a unit vector orthogonal to both $\text{a}$ and $\text{b}.$

The cross product $\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}$ is orthogonal to both vectors $\text{a}$ and $\text{b}.$ We can calculate it with a determinant:

$\begin{array}{cc}\hfill \text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill 5& \hfill 2& \hfill -1\\ \hfill 0& \hfill -1& \hfill 4\end{array}|=|\begin{array}{cc}\hfill 2& \hfill -1\\ \hfill -1& \hfill 4\end{array}|\text{i}-|\begin{array}{cc}\hfill 5& \hfill -1\\ \hfill 0& \hfill 4\end{array}|\text{j}+|\begin{array}{cc}\hfill 5& \hfill 2\\ \hfill 0& \hfill -1\end{array}|\text{k}\hfill \\ & =\left(8-1\right)\text{i}-\left(20-0\right)\text{j}+\left(-5-0\right)\text{k}\hfill \\ & =7\text{i}-20\text{j}-5\text{k}.\hfill \end{array}$

Normalize this vector to find a unit vector in the same direction:

$‖\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}‖=\sqrt{{\left(7\right)}^{2}+{\left(-20\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{474}.$

Thus, $⟨\frac{7}{\sqrt{474}},\frac{-20}{\sqrt{474}},\frac{-5}{\sqrt{474}}⟩$ is a unit vector orthogonal to $\text{a}$ and $\text{b}.$

Find a unit vector orthogonal to both $\text{a}$ and $\text{b},$ where $\text{a}=⟨4,0,3⟩$ and $\text{b}=⟨1,1,4⟩.$

$⟨\frac{-3}{\sqrt{194}},\frac{-13}{\sqrt{194}},\frac{4}{\sqrt{194}}⟩$

To use the cross product for calculating areas, we state and prove the following theorem.

Area of a parallelogram

If we locate vectors $\text{u}$ and $\text{v}$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖$ ( [link] ).

Proof

We show that the magnitude of the cross product is equal to the base times height of the parallelogram.

$\begin{array}{cc}\hfill \text{Area of a parallelogram}& =\text{base}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{height}\hfill \\ & =‖\text{u}‖\left(‖\text{v}‖\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)\hfill \\ & =‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖\hfill \end{array}$

Finding the area of a triangle

Let $P=\left(1,0,0\right),Q=\left(0,1,0\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}R=\left(0,0,1\right)$ be the vertices of a triangle ( [link] ). Find its area.

We have $\stackrel{\to }{PQ}=⟨0-1,1-0,0-0⟩=⟨-1,1,0⟩$ and $\stackrel{\to }{PR}=⟨0-1,0-0,1-0⟩=⟨-1,0,1⟩.$ The area of the parallelogram with adjacent sides $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$ is given by $‖\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}‖\text{:}$

$\begin{array}{ccc}\hfill \stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}& =\hfill & |\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}|=\left(1-0\right)\text{i}-\left(-1-0\right)\text{j}+\left(0-\left(-1\right)\right)\text{k}=\text{i}+\text{j}+\text{k}\hfill \\ \hfill ‖\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}‖& =\hfill & ‖⟨1,1,1⟩‖=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3}.\hfill \end{array}$

The area of $\text{Δ}PQR$ is half the area of the parallelogram, or $\sqrt{3}\text{/}2.$

Find the area of the parallelogram $PQRS$ with vertices $P\left(1,1,0\right),Q\left(7,1,0\right),R\left(9,4,2\right),$ and $S\left(3,4,2\right).$

$6\sqrt{13}$

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