Vector product, mixed product, areas and volumes – Vector Space ℝ³ of Real Numbers (Senior 5 Mathematics) | Ganzaa

Vector Space ℝ³ of Real Numbers

Lessons

Senior 5 · Mathematics

Vector Space ℝ³ of Real Numbers

4 lessons

Lesson 4 of 4

Vector product, mixed product, areas and volumes

45 min

A product that points

Tighten a bolt with a spanner: you push on the handle, sideways, yet the bolt moves along its own axis, at right angles both to the handle and to your push. The turning effect is a vector perpendicular to the two vectors that create it. The vector product builds exactly such a vector, and its length turns out to be an area. Combined with the scalar product of lesson 3 it gives volumes too, and a test for whether four points lie in one plane. Unlike the scalar product, it exists only in three dimensions.

Definition

The vector product (cross product) of $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ is the vector

u \times v = \begin{vmatrix} i & j & k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2v_3 - u_3v_2)\,i - (u_1v_3 - u_3v_1)\,j + (u_1v_2 - u_2v_1)\,k ,

where the "determinant" is expanded along its first row. Example: $i \times j = k$ , $j \times k = i$ , $k \times i = j$ .

Important

Properties of the vector product

Anti-commutative: $v \times u = -(u \times v)$ (swapping two rows of the determinant changes its sign); hence $u \times u = 0$ .
$u \times (v + w) = u \times v + u \times w$ and $(\lambda u) \times v = \lambda(u \times v)$ .
$u \times v$ is orthogonal to both $u$ and $v$ : $u \cdot (u \times v)$ is a determinant with two equal rows, so it is 0.
$|u \times v| = |u|\,|v|\sin\theta$ , where $\theta$ is the angle between $u$ and $v$ . (Expanding both sides proves Lagrange's identity $|u \times v|^2 = |u|^2|v|^2 - (u \cdot v)^2$ , and $1 - \cos^2\theta = \sin^2\theta$ .)
$u \times v = 0$ if and only if $u$ and $v$ are parallel (or one is 0).
$u, v, u \times v$ form a right-handed set: curl the fingers of your right hand from $u$ towards $v$ ; your thumb points along $u \times v$ .

Worked example 1: computing and checking a vector product

Find $u \times v$ for $u = (2, - 1, 3)$ and $v = (1, 4, - 2)$ .

Solution.

u \times v = \begin{vmatrix} i & j & k \\ 2 & -1 & 3 \\ 1 & 4 & -2 \end{vmatrix} = (2 - 12)\,i - (-4 - 3)\,j + (8 + 1)\,k = (-10, 7, 9) .

Check orthogonality: $(-10, 7, 9) \cdot (2, -1, 3) = -20 - 7 + 27 = 0$ and $(-10, 7, 9) \cdot (1, 4, -2) = -10 + 28 - 18 = 0$ . ✓

Important

Areas. The parallelogram built on $u$ and $v$ has base $∣ u ∣$ and height $|v|\sin\theta$ , so

\text{area of parallelogram} = |u \times v|, \qquad \text{area of triangle } ABC = \tfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| .

Worked example 2: the area of a triangle in space

Find the area of the triangle with vertices $A (1, 0, 2)$ , $B (3, 1, 1)$ , $C (0, 2, 3)$ .

Solution. $\overrightarrow{AB} = (2, 1, -1)$ , $\overrightarrow{AC} = (-1, 2, 1)$ , and

\overrightarrow{AB} \times \overrightarrow{AC} = (1 \cdot 1 - (-1) \cdot 2,\ -(2 \cdot 1 - (-1)(-1)),\ 2 \cdot 2 - 1 \cdot (-1)) = (3, -1, 5) .

Area $= \frac{1}{2}\sqrt{9 + 1 + 25} = \frac{\sqrt{35}}{2} \approx 2.96$ square units.

Definition

The mixed product (scalar triple product) of $u, v, w$ is the number

[u, v, w] = u \cdot (v \times w) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} .

By the properties of determinants it is unchanged by a cyclic shift, $[u, v, w] = [v, w, u] = [w, u, v]$ , and changes sign when two vectors are swapped.

Volume of a parallelepiped

The box built on $u, v, w$ has base area $|u \times v|$ . Its height $h$ is the part of $w$ along the direction of $u \times v$ , perpendicular to the base: $h = \frac{|w \cdot (u \times v)|}{|u \times v|}$ . Multiplying base by height:

V_{\text{parallelepiped}} = \left|[u, v, w]\right|, \qquad V_{\text{tetrahedron } ABCD} = \frac{1}{6}\left|\left[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}\right]\right| .

A tetrahedron is $\frac{1}{3} \times$ base $\times$ height, with half the parallelogram as its base, hence $\frac{1}{6}$ . And $[u, v, w] = 0$ exactly when the box is flat: the three vectors are coplanar (linearly dependent), which matches the determinant test of lesson 2.

Worked example 3: a box and a tetrahedron

(a) Find the volume of the parallelepiped built on $(1, 2, 0)$ , $(0, 1, 3)$ , $(2, 0, 1)$ . (b) Find the volume of the tetrahedron $A B C D$ with $A (1, 0, 2)$ , $B (3, 1, 1)$ , $C (0, 2, 3)$ , $D (2, 3, 4)$ .

Solution. (a) $\begin{vmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{vmatrix} = 1(1 - 0) - 2(0 - 6) + 0 = 13$ , so the volume is 13 cubic units.

(b) With $\overrightarrow{AB} \times \overrightarrow{AC} = (3, -1, 5)$ from worked example 2 and $\overrightarrow{AD} = (1, 3, 2)$ : $[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}] = (3, -1, 5) \cdot (1, 3, 2) = 3 - 3 + 10 = 10$ . The volume is $\frac{10}{6} = \frac{5}{3}$ cubic units.

Worked example 4: when are three vectors coplanar?

Find $k$ so that $(1, 2, k)$ , $(2, 1, 1)$ and $(1, - 1, 2)$ are coplanar.

Solution.

\begin{vmatrix} 1 & 2 & k \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = 1(2 + 1) - 2(4 - 1) + k(-2 - 1) = -3 - 3k .

This is 0 for $k = - 1$ . Check: with $k = - 1$ , $(1, 2, - 1) = a (2, 1, 1) + b (1, - 1, 2)$ needs $2 a + b = 1$ and $a - b = 2$ , so $a = 1$ , $b = - 1$ , and the third component $1 - 2 = - 1$ agrees. ✓ So $(1, 2, - 1) = (2, 1, 1) - (1, - 1, 2)$ .

Warning

Mind the minus sign on the $j$ component of $u \times v$ : it is $- (u_{1} v_{3} - u_{3} v_{1})$ .
$u \times v \ne v \times u$ : they are opposite vectors. Their lengths, and hence areas, agree.
Volumes are absolute values. A negative mixed product only means $u, v, w$ are left-handed; the volume is positive.
Don't forget the $\frac{1}{2}$ for a triangle and the $\frac{1}{6}$ for a tetrahedron.

Tip

Check every cross product with one scalar product: $(u \times v) \cdot u$ must be 0. If it is not, a sign has slipped.

3Blue1Brown: Cross products, Chapter 10 of Essence of Linear Algebra (about 9 min). Watch the length of u × v appear as the area of a parallelogram and its direction come from the right-hand rule.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

For $u = (1, 2, - 1)$ and $v = (3, 0, 2)$ , find $u \times v$ and $v \times u$ .
Find a unit vector perpendicular to both $(1, 1, 0)$ and $(0, 1, 1)$ .
Find the area of the parallelogram built on $(2, 1, 0)$ and $(1, - 1, 2)$ .
Find the area of the triangle $P (1, 1, 1)$ , $Q (2, 3, 1)$ , $R (1, 2, 4)$ .
Find the volume of the parallelepiped built on $(1, 0, 1)$ , $(1, 1, 0)$ , $(0, 1, 1)$ , and the volume of the tetrahedron $O (0, 0, 0)$ , $A (2, 0, 0)$ , $B (0, 3, 0)$ , $C (1, 1, 4)$ .
Show that $(1, - 1, 2)$ , $(2, 1, 1)$ and $(1, 2, - 1)$ are coplanar, and write one of them as a combination of the other two.
Explain why $v \cdot (u \times v) = 0$ for all $u, v$ , and what this means geometrically.

Show answersHide answers

$u \times v = (4, -5, -6)$ ; $v \times u = (-4, 5, 6)$ .
$(1, 1, 0) \times (0, 1, 1) = (1, -1, 1)$ , so $\pm\frac{1}{\sqrt{3}}(1, -1, 1)$ .
$(2, 1, 0) \times (1, -1, 2) = (2, -4, -3)$ ; area $\sqrt{29} \approx 5.39$ .
$\overrightarrow{PQ} \times \overrightarrow{PR} = (1, 2, 0) \times (0, 1, 3) = (6, -3, 1)$ ; area $\frac{\sqrt{46}}{2} \approx 3.39$ .
Mixed product $= 2$ , volume 2; tetrahedron: $[O A, O B, O C] = 24$ , volume $\frac{24}{6} = 4$ .
The mixed product is $1 (- 1 - 2) + 1 (- 2 - 1) + 2 (4 - 1) = 0$ ; $(2, 1, 1) = (1, - 1, 2) + (1, 2, - 1)$ .
It is a determinant with rows $v, u, v$ : two equal rows, so 0. Geometrically, $u \times v$ is perpendicular to $v$ (and to $u$ ).

1A product that points

2Volume of a parallelepiped

A product that points

Volume of a parallelepiped