IB Analysis and Approaches HL · Geometry in space · Section 3 of 7

Vector product, areas and volumes

3 worked examples · By Álvaro López

Key idea

The vector (cross) product takes two vectors in space and returns another vector, perpendicular to both. Its magnitude is the area of the parallelogram they form, and combined with a third vector it gives volumes. In the IB exam it is used to find a normal vector to a plane, to calculate areas of triangles in $R^{3}$ , and to find volumes of parallelepipeds and tetrahedrons.

Definition and calculation

The vector product only exists in three dimensions. It is written $u \times v$ and read "u cross v".

Memorising three long expressions is awkward. An easier rule is to expand, along the first row, a determinant whose first row is the base vectors $i, j, k$ :

$u \times v = i u_{1} v_{1} j u_{2} v_{2} k u_{3} v_{3} = i u_{2} v_{2} u_{3} v_{3} - j u_{1} v_{1} u_{3} v_{3} + k u_{1} v_{1} u_{2} v_{2}$

Look at the minus sign in front of the $j$ component: it is the most common mistake (see the common mistakes below). It is a "symbolic" determinant, because the first row is not numbers, but it works as a calculation rule.

Properties

To see why the result is perpendicular, take the dot product of $u$ with $u \times v$ : the six terms that appear cancel in pairs, and the same happens with $v$ . The property for parallel vectors also follows from this: if $v = λ u$ , each component has the form $λ (u_{2} u_{3} - u_{3} u_{2}) = 0$ .

The magnitude formula comes from the identity $∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2} - (u \cdot v)^{2}$ , which you can check by expanding the components. If you substitute $u \cdot v = ∣ u ∣ ∣ v ∣ cos θ$ , you get

$∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2} (1 - cos^{2} θ) = ∣ u ∣^{2} ∣ v ∣^{2} sin^{2} θ$

and since $sin θ \geq 0$ between $0^{\circ}$ and $18 0^{\circ}$ , the square root can be taken without ambiguity. Compare this with the scalar (dot) product: there $cos θ$ appears and gives a number; here $sin θ$ appears and gives the magnitude of a vector.

A vector perpendicular to two others

This is the most frequent application. If $u$ and $v$ are not parallel, any vector perpendicular to both is a multiple of $u \times v$ . To get one of magnitude $1$ , divide by its magnitude:

$n^= \pm \frac{u \times v}{∣ u \times v ∣}$

Given three points $A$ , $B$ , $C$ that are not collinear, the vector $A B \times A C$ is perpendicular to the plane that contains them. It will be the basis for finding the equation of the plane in the next section.

Area of a parallelogram and of a triangle

Parallelogram ABCD with sides u and v, angle theta at A and height h equal to the magnitude of v times sine of theta

A parallelogram with sides $u$ and $v$ has base $∣ u ∣$ and height $h = ∣ v ∣ sin θ$ . Its area is base times height, that is, exactly the magnitude of the vector product.

The triangle is half the parallelogram, because the diagonal splits it into two equal triangles. The advantage of this formula is that it works in space, where you cannot use "base times height divided by two" without first finding the height. You can also compare it with the trigonometric formula $\frac{1}{2} ab sin C$ for the area of a triangle: it is the same formula, written with vectors.

Scalar triple product and volumes

With three vectors $u$ , $v$ , $w$ that start from the same point you form a parallelepiped (a box with parallelogram faces). Its base is the parallelogram of $u$ and $v$ , with area $∣ u \times v ∣$ ; its height is the component of $w$ in the direction perpendicular to the base, that is, $∣ w ∣ ∣ cos φ ∣$ , where $φ$ is the angle between $w$ and $u \times v$ .

The absolute value is needed: the scalar triple product can come out negative depending on the order of the vectors, but a volume cannot. The one sixth comes from the fact that a pyramid occupies a third of the prism with the same base and height, and the triangular base is half the parallelogram: $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ .

A very useful special case: if $(u \times v) \cdot w = 0$ , the volume is zero and the three vectors are coplanar (they lie in the same plane).

Worked examples

Example 1 Basic

Calculating a vector product and checking it

Let $u = 123$ and $v = 4 - 1 2$ .

a) Calculate $u \times v$ .

b) Calculate $(u \times v) \cdot u$ and check that the result is the one you expect.

c) Write down $v \times u$ without repeating the calculation.

Solution

a) Using the determinant:

$u \times v = i 14 j 2 - 1 k 32 = i (2 \cdot 2 - 3 \cdot (- 1)) - j (1 \cdot 2 - 3 \cdot 4) + k (1 \cdot (- 1) - 2 \cdot 4)$

$= 7 i + 10 j - 9 k = 710 - 9$

b) $(u \times v) \cdot u = 7 \cdot 1 + 10 \cdot 2 + (- 9) \cdot 3 = 7 + 20 - 27 = 0$ . It is zero, as it must be: the vector product is perpendicular to $u$ .

c) Swapping the order changes the sign: $v \times u = - (u \times v) = - 7 - 10 9$ .

Example 2 Medium

A triangle in space: normal vector, area and unit vector

The points $A (1, 0, 2)$ , $B (3, 1, 0)$ and $C (2, 3, 1)$ are the vertices of a triangle.

a) Find $A B \times A C$ .

b) Find the area of triangle $A B C$ .

c) Find the vector of magnitude $1$ perpendicular to the plane of the triangle whose first component is positive.

Solution

First, the vectors that start at $A$ , subtracting end minus start:

$A B = 21 - 2, A C = 13 - 1$

a) Using the determinant:

$A B \times A C = i (1 \cdot (- 1) - (- 2) \cdot 3) - j (2 \cdot (- 1) - (- 2) \cdot 1) + k (2 \cdot 3 - 1 \cdot 1)$

$= 5 i - 0 j + 5 k = 505$

b) The magnitude is $25 + 0 + 25 = 52$ , and the area is half of it:

$A = \frac{1}{2} \cdot 52 = \frac{5 2}{2}$

c) The vector $505$ is perpendicular to the plane. Divide it by its magnitude $52$ :

$n^= \frac{1}{5 2} 505 = \frac{1}{2} 0 \frac{1}{2} = \frac{2}{2} 0 \frac{2}{2}$

The first component is positive, so the sign does not need to change.

Example 3 Hard

A box with a parameter and coplanar vectors

Consider the vectors $u = 210$ , $v = 031$ and $w = 1 k 2$ , where $k \in R$ .

a) Find $u \times v$ and, from it, the scalar triple product $(u \times v) \cdot w$ as a function of $k$ .

b) For $k = 0$ , calculate the volume of the parallelepiped and the volume of the tetrahedron determined by the three vectors.

c) Find the values of $k$ for which the parallelepiped has volume $5$ .

d) Find the value of $k$ for which the three vectors are coplanar.

Solution

a) We expand the symbolic determinant:

$u \times v = i (1 \cdot 1 - 0 \cdot 3) - j (2 \cdot 1 - 0 \cdot 0) + k (2 \cdot 3 - 1 \cdot 0) = 1 - 2 6$

We multiply by $w$ using the dot product:

$(u \times v) \cdot w = 1 \cdot 1 + (- 2) \cdot k + 6 \cdot 2 = 13 - 2 k$

b) With $k = 0$ the scalar triple product is $13$ . The parallelepiped has volume $∣ 13 ∣ = 13$ and the tetrahedron $\frac{13}{6}$ .

c) The volume is $∣ 13 - 2 k ∣$ , and it must equal $5$ . There are two cases:

$13 - 2 k = 5 \Rightarrow k = 4 or 13 - 2 k = - 5 \Rightarrow k = 9$

The absolute value gives two solutions; the two boxes have the same volume, although with opposite orientations.

d) Coplanar means zero volume: $13 - 2 k = 0$ , so $k = \frac{13}{2}$ .

In the exam: when a volume is set equal to a number, do not forget the absolute value, which gives two values of the parameter. And if you are asked for the volume of the tetrahedron, divide by $6$ : it is the most common confusion with the parallelepiped.

Common mistakes

Summary

Vector product: $u \times v = u_{2} v_{3} - u_{3} v_{2} u_{3} v_{1} - u_{1} v_{3} u_{1} v_{2} - u_{2} v_{1}$ , found from the symbolic determinant of $i, j, k$ (minus sign on the $j$ component). It is a vector.
Properties: $v \times u = - (u \times v)$ ; it is perpendicular to $u$ and to $v$ ; it equals $0$ if and only if they are parallel; $∣ u \times v ∣ = ∣ u ∣ ∣ v ∣ sin θ$ .
Vector perpendicular to $u$ and $v$ : $u \times v$ (or a multiple of it); unit vector: divide by its magnitude.
Areas: parallelogram $∣ u \times v ∣$ ; triangle $\frac{1}{2} ∣ u \times v ∣$ , for example $\frac{1}{2} ∣ A B \times A C ∣$ .
Scalar triple product $(u \times v) \cdot w$ = determinant of the three vectors. Volume of the parallelepiped $= ∣(u \times v) \cdot w ∣$ ; of the tetrahedron, one sixth of that. If it is $0$ , the vectors are coplanar.

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