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1.6 The vector product in ${\mathbb{R}}^{3}$

Definition 1.21

Let ${\bf u}=(u_{1}\;\;u_{2}\;\;u_{3})$ , ${\bf v}=(v_{1}\;\;v_{2}\;\;v_{3})$ be vectors in ${\mathbb{R}}^{3}$ . The vector (or cross) product of ${\bf u}$ and ${\bf v}$ :

{\bf u}\times{\bf v}=(u_{2}v_{3}-u_{3}v_{2}\;\;u_{3}v_{1}-u_{1}v_{3}\;\;u_{1}v% _{2}-u_{2}v_{1})

Remark 1.22

i) If you know about determinants of $2\times 2$ matrices, then one way to remember the entries of ${\bf u}\times{\bf v}$ is to write ${\bf u}$ and ${\bf v}$ as the first two rows of a $3\times 3$ matrix:

\left(\begin{array}[]{ccc}u_{1}&u_{2}&u_{3}\\ v_{1}&v_{2}&v_{3}\\ **&*&*\end{array}\right)

Now ${\bf u}\times{\bf v}$ can be entered in the third row of the matrix: to find the first entry in ${\bf u}\times{\bf v}$ , simply cover up the first column and the third row, and calculate the determinant of the matrix you then obtain:

\left|\begin{array}[]{cc}u_{2}&u_{3}\\ v_{2}&v_{3}\end{array}\right|=u_{2}v_{3}-u_{3}v_{2}.

To find the second entry in ${\bf u}\times{\bf v}$ , do the same thing with the second column and the third row covered up, but then multiply by $(-1)$ . So the second entry is:

-\left|\begin{array}[]{cc}u_{1}&u_{3}\\ v_{1}&v_{3}\end{array}\right|=u_{3}v_{1}-u_{1}v_{3}.

Finally, the third entry can be found by covering up the third row and column, and calculating the determinant of the matrix you thus obtain.

ii) Be careful of signs! The $y$ -coordinate of ${\bf u}\times{\bf v}$ is $u_{3}v_{1}-u_{1}v_{3}$ , not $u_{1}v_{3}-u_{3}v_{1}$ .

Theorem 1.23

The vector product has the following properties:

(i) ${\bf u}\times{\bf v}=-{\bf v}\times{\bf u}$ for any ${\bf u},{\bf v}\in{\mathbb{R}}^{3}$ ,

(ii) ${\bf u}\times({\bf v}+{\bf w})={\bf u}\times{\bf v}+{\bf u}\times{\bf w}$ for any ${\bf u},{\bf v},{\bf w}\in{\mathbb{R}}^{3}$ ,

(iii) ${\bf u}\times(\lambda{\bf v})=\lambda({\bf u}\times{\bf v})=(\lambda{\bf u})% \times{\bf v}$ for any ${\bf u},{\bf v}\in{\mathbb{R}}^{3}$ and $\lambda\in{\mathbb{R}}$ ,

(iv) $({\bf u}\times{\bf v})\cdot{\bf u}=({\bf u}\times{\bf v})\cdot{\bf v}=0$ for any ${\bf u},{\bf v}\in{\mathbb{R}}^{3}$ ,

(v) ${\bf u}\times{\bf v}={\bf 0}$ if ${\bf u}=\lambda{\bf v}$ for some $\lambda\in{\mathbb{R}}$ .

Proof

(i) follows immediately from inspecting the coordinate values of ${\bf u}\times{\bf v}$ and ${\bf v}\times{\bf u}$ . (ii) and (iii) can be proved in a similar way to Thm. 1.9. For (iv), it will suffice to show that $({\bf u}\times{\bf v})\cdot{\bf u}=0$ , since then $({\bf u}\times{\bf v})\cdot{\bf v}=0$ can be shown by swapping ${\bf u}$ and ${\bf v}$ . Now we calculate directly: $({\bf u}\times{\bf v})\cdot{\bf u}=(u_{2}v_{3}-u_{3}v_{2})u_{1}+(u_{3}v_{1}-u_% {1}v_{3})u_{2}+(u_{1}v_{2}-u_{2}v_{1})u_{3}=u_{1}u_{2}v_{3}-u_{1}u_{3}v_{2}+u_% {2}u_{3}v_{1}-u_{1}u_{2}v_{3}+u_{1}u_{3}v_{2}-u_{2}u_{3}v_{1}=0$ . For (v), if ${\bf u}=\lambda{\bf v}$ then ${\bf u}\times{\bf v}=\lambda({\bf u}\times{\bf u})$ , so it will suffice to show that ${\bf u}\times{\bf u}={\bf 0}$ . But

{\bf u}\times{\bf u}=(u_{2}u_{3}-u_{3}u_{2}\;\;u_{3}u_{1}-u_{1}u_{3}\;\;u_{1}u% _{2}-u_{2}u_{1})=(0\;\;0\;\;0)

so the proof is complete. $\square$

Thus we have the following important principle. We say that two vectors ${\bf u}$ , ${\bf v}$ are collinear (i.e. on the same line) if either ${\bf u}=\lambda{\bf v}$ or ${\bf v}=\lambda{\bf u}$ for some $\lambda\in{\mathbb{R}}$ .

Principle 1.24

If ${\bf u}$ and ${\bf v}$ are not collinear vectors, then ${\bf u}\times{\bf v}$ is a non-zero vector which is orthogonal to both ${\bf u}$ and ${\bf v}$ .

Theorem 1.25

Let $\theta$ be the angle between ${\bf u}$ and ${\bf v}$ . Then $|{\bf u}\times{\bf v}|=|{\bf u}|\,|{\bf v}|\,|\sin\theta|$ .

Proof

Since $|{\bf u}\times{\bf v}|\geq 0$ , it will suffice to show that $|{\bf u}\times{\bf v}|^{2}=|{\bf u}|^{2}|{\bf v}|^{2}\sin^{2}\theta$ . But ${\bf u}\cdot{\bf v}=|{\bf u}|\,|{\bf v}|\cos\theta$ , so we have to show that:

|{\bf u}\times{\bf v}|^{2}+({\bf u}\cdot{\bf v})^{2}=|{\bf u}|^{2}|{\bf v}|^{2}

We prove this statement by directly calculating the left-hand side. Since ${\bf u}\times{\bf v}=(u_{2}v_{3}-u_{3}v_{2}\;\;u_{3}v_{1}-u_{1}v_{3}\;\;u_{1}v% _{2}-u_{2}v_{1})$ we have:

|{\bf u}\times{\bf v}|^{2}=u_{1}^{2}v_{2}^{2}+u_{1}^{2}v_{3}^{2}+u_{2}^{2}v_{1% }^{2}+u_{2}v_{3}^{2}+u_{3}^{2}v_{1}^{2}+u_{3}^{2}v_{2}^{2}-2(u_{1}u_{2}v_{1}v_% {2}+u_{1}u_{3}v_{1}v_{3}+u_{2}u_{3}v_{2}v_{3})

On the other hand, ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}$ so

({\bf u}\cdot{\bf v})^{2}=u_{1}^{2}v_{1}^{2}+u_{2}^{2}v_{2}^{2}+u_{3}^{2}v_{3}% ^{2}+2(u_{1}u_{2}v_{1}v_{2}+u_{1}u_{3}v_{1}v_{3}+u_{2}u_{3}v_{2}v_{3})

Adding these two, we obtain:

|{\bf u}\times{\bf v}|^{2}+({\bf u}\cdot{\bf v})^{2}=\sum_{i,j=1}^{3}u_{i}^{2}% v_{j}^{2}=(u_{1}^{2}+u_{2}^{2}+u_{3}^{2})(v_{1}^{2}+v_{1}^{2}+v_{3}^{2})=|{\bf u% }|^{2}|{\bf v}|^{2}

as required. $\square$

Using the vector product to find equations of planes in ${\mathbb{R}}^{3}$

Suppose we are given two vectors ${\bf u}$ , ${\bf v}$ and we want to find the equation of the plane passing through the origin which contains ${\bf u}$ and ${\bf v}$ . We require ${\bf u}$ and ${\bf v}$ to be non-collinear, i.e. linearly independent. To describe the plane $\pi$ containing ${\bf u}$ and ${\bf v}$ we just have to find a normal vector to $\pi$ . But as we saw above, ${\bf u}\times{\bf v}$ is orthogonal to both ${\bf u}$ and ${\bf v}$ , so we can set ${\bf n}={\bf u}\times{\bf v}$ . This principle is best demonstrated with some examples.

Example 1.26

Find the equation of the plane passing through the origin in ${\mathbb{R}}^{3}$ which includes ${\bf u}=(1\;\;-1\;\;2)$ and ${\bf v}=(0\;\;3\;\;-1)$ .

$\quad$

What about the equation of a plane which doesn’t pass through the origin? We can usually describe a plane $\pi$ if we know three points $A, B, C$ in $\pi$ .

Example 1.27

Find the equation of the plane in ${\mathbb{R}}^{3}$ which passes through the points $A=(1,2,-1)$ , $B=(2,-3,1)$ and $C=(0,3,-4)$ .

Using the vector product to calculate areas of parallelograms and triangles

Let $O A C B$ be a parallelogram in ${\mathbb{R}}^{3}$ , where $\overrightarrow{OA}={\bf a}$ and $\overrightarrow{OB}={\bf b}$ .

We can use the vector product ${\bf a}\times{\bf b}$ to calculate the area of the parallelogram $O A C B$ .

To see how, recall the well-known formula for the area of a parallelogram:

A=\ell h

where $\ell$ is the length of the base, and $h$ is the (perpendicular) height. We can choose $O A$ to be the base, so that $\ell=|{\bf a}|$ . Moreover, if $\theta$ is the angle between ${\bf a}$ and ${\bf b}$ then the height of the parallelogram is $|{\bf b}|\,\sin\theta$ . Thus $A=|{\bf a}|\,|{\bf b}|\sin\theta=|{\bf a}\times{\bf b}|$ .

Example 1.28

Calculate the area of the parallelogram $O A C B$ in ${\mathbb{R}}^{2}$ with corners $A=(1\;\;2)$ , $B=(-1\;\;1)$ , $C=(0\;\;3)$ .

The question does not ask us to verify that $O A C B$ is a parallelogram, but we might check: $\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OB}$ . Though $O A C B$ is a parallelogram in ${\mathbb{R}}^{2}$ , we can consider it as being a parallelogram in ${\mathbb{R}}^{3}$ just by adding a third coordinate equal to zero: $A=(1\;\;2\;\;0)$ , $B=(-1\;\;1\;\;0)$ , $C=(0\;\;3\;\;0)$ . Now the area of the parallelogram is $|\overrightarrow{OA}\times\overrightarrow{OB}|=|(0\;\;0\;\;3)|=3$ .

Similarly to parallelograms, we can calculate the area of a triangle with sides $\overrightarrow{OA}$ and $\overrightarrow{OB}$ : it is just $\frac{1}{2}bh=\frac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}|$ .

Example 1.29

Find the area of the triangle in ${\mathbb{R}}^{3}$ with corners $A=(1\;\;1\;\;-1)$ , $B=(2\;\;-1\;\;0)$ and $C=(0\;\;2\;\;7)$ .

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1.6 The vector product in ℝ3

Definition 1.21

Remark 1.22

Theorem 1.23

Proof

Principle 1.24

Theorem 1.25

Proof

Example 1.26

Example 1.27

Example 1.28

Example 1.29

1.6 The vector product in ${\mathbb{R}}^{3}$