If *three* mutually perpendicular copies of the real line intersect at their origins, any point in the resulting space is specified by an ordered *triple* of real numbers (*x* _{1}, *x* _{2}, *x* _{3}). The set of all ordered triples of real numbers is called **3‐space**, denoted **R** ^{3} (“R three”). See Figure .

**Figure 1**

The operations of addition and scalar multiplication difined on **R** ^{2} carry over to **R** ^{3}:

Vectors in **R** ^{3} are called **3‐vectors** (because there are 3 components), and the geometric descriptions of addition and scalar multiplication given for 2‐vectors also carry over to 3‐vectors.

**Example 1**: If **x** = (3, 0, 4) and **y** = (2, 1, −1), then

**Standard basis vectors in R **^{3} . Since for any vector **x** = (*x* _{1}, *x* _{2}, *x* _{3}) in **R** ^{3},

the standard basis vectors in **R** ^{3} are

Any vector **x** in **R** ^{3} may therefore be written as

See Figure .

**Figure 2**

**Example 2**: What vector must be added to **a** = (1, 3, 1) to yield **b** = (3, 1, 5)?

Let **c** be the required vector; then **a + c = b**. Therefore,

Note that **c** is the vector **ab**; see Figure .

**Figure 3**

**The cross product**. So far, you have seen how two vectors can be added (or subtracted) and how a vector is multiplied by a scalar. Is it possible to somehow “multiply” two vectors? One way to define the product of two vectors—which is done only with vectors in **R** ^{3}—is to form their *cross product*. Let **x** = (*x* _{1}, *x* _{2}, *x* _{3}) and **y** =(*y* _{1}, *y* _{2}, *y* _{3}) be two vectors in **R** ^{3}. The **cross product** (or **vector product**) of **x** and **y** is defined as follows:

The cross product of two vectors is a vector, and perhaps the most important characteristic of this vector product is that *it is perpendicular to both factors*. (This will be demonstrated when the dot product is introduced.) That is, the vector **x** x **y** will be perpendicular to both **x** and **y**; see Figure . [There is an ambiguity here: the plane in Figure , which contains the vectors **x** and **y**, has two perpendicular directions: “up” and “down.” Which one does the cross product choose? The answer is given by the *right‐hand rule*: Place the wrist of your right hand at the common initial point of **x** and **y**, with your fingers pointing along **x**; as you curl your fingers toward **y**, your thumb will point in the direction of **x** x **y**. This shows that the cross product is anticommutative: **y** x **x** = − (**x** x **y**).]

**Figure 4**

The **length** of a vector **x** = (*x* _{1}, *x* _{2}, *x* _{3}) in **R** ^{3}, which is denoted ∥ **x**∥, is given by the equation

a result which follows from the Pythagorean Theorem (see the discussion preceding Figure below). While the direction of the cross product of **x** and **y** is determined by orthogonality and the right‐hand rule, the **magnitude** (that is, the length) of **x** x **y** is equal to the area of the parallelogram spanned by the vectors **x** and **y**.

Since the area of the parallelogram in Figure is

the following equation holds:

where θ is the angle between **x** and **y**.

**Figure 5**

**Example 3**: Let **x** = (2, 3, 0) and **y** = (−1, 1, 4) be position vectors in **R** ^{3}. Compute the area of the triangle whose vertices are the origin and the endpoints of **x** and **y** and determine the angle between the vectors **x** and **y**.

Since the area of the triangle is half the area of the parallelogram spanned by **x** and **y**,

Now, since ‖ **x** x **y**‖ = ‖ **x**‖·‖ **y**‖ sin θ, the angle between **x** and **y** is given by

Therefore, θ = sin ^{−1} √233/234.