Vectors in ℝ³ and their operations – 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 1 of 4

Vectors in ℝ³ and their operations

35 min

Three numbers for a position in space

A drone hovers above a football pitch. To say where it is, "40 m east and 25 m north of the corner flag" is not enough: you must also say how high it is. Positions in space need three numbers, and so do the moves between them. In S4 you worked with vectors of the plane, pairs $(x, y)$ . This unit extends everything to triples $(x, y, z)$ , the vector space $\mathbb{R}^3$ . Engineers use it for forces on a roof truss, pilots for wind and flight paths, and game designers for every object on a screen.

Definition

$\mathbb{R}^3 = \{ (x, y, z) : x, y, z \in \mathbb{R} \}$ is the set of ordered triples of real numbers. An element $u = (u_{1}, u_{2}, u_{3})$ is a vector of $\mathbb{R}^3$ , and $u_{1}, u_{2}, u_{3}$ are its components. With the standard unit vectors $i = (1, 0, 0)$ , $j = (0, 1, 0)$ , $k = (0, 0, 1)$ ,

u = (u_{1}, u_{2}, u_{3}) = u_{1} i + u_{2} j + u_{3} k .

In a right-handed system of axes, the position vector of the point $P (x, y, z)$ is $\overrightarrow{OP} = (x, y, z)$ . For two points $A$ and $B$ with position vectors $a$ and $b$ , the vector from $A$ to $B$ is $\overrightarrow{AB} = b - a$ .

Locating a point

To reach $P (3, 4, 2)$ from the origin, move 3 units along the $x$ -axis, 4 units parallel to the $y$ -axis, and 2 units up. The arrow $\overrightarrow{OP} = 3i + 4j + 2k$ does all three moves at once.

Definition

Operations. For $u = (u_{1}, u_{2}, u_{3})$ , $v = (v_{1}, v_{2}, v_{3})$ and a real number (a scalar) $\lambda$ :

u + v = (u_1 + v_1,\ u_2 + v_2,\ u_3 + v_3), \qquad \lambda u = (\lambda u_1,\ \lambda u_2,\ \lambda u_3), \qquad u - v = u + (-1)v .

The zero vector is $0 = (0, 0, 0)$ . Two vectors are equal when their three components are equal. Two non-zero vectors are parallel (collinear) when one is a scalar multiple of the other: $v = \lambda u$ .

Important

$\mathbb{R}^3$ is a real vector space. For all $u, v, w \in \mathbb{R}^3$ and $\lambda, \mu \in \mathbb{R}$ :

For addition	For multiplication by a scalar
$u + v \in \mathbb{R}^3$ (closure)	$\lambda u \in \mathbb{R}^3$ (closure)
$u + v = v + u$	$\lambda(u + v) = \lambda u + \lambda v$
$(u + v) + w = u + (v + w)$	$(\lambda + \mu)u = \lambda u + \mu u$
$u + 0 = u$	$\lambda(\mu u) = (\lambda\mu)u$
$u + (- u) = 0$	$1 u = u$

Each rule holds because it holds for real numbers, component by component. Any set with an addition and a scalar multiplication obeying these ten rules is a vector space: $\mathbb{R}^2$ , $\mathbb{R}^3$ and $M_3(\mathbb{R})$ of unit 6 are all examples.

Worked example 1: a vector between two points

For $A (2, - 1, 3)$ and $B (5, 1, - 1)$ , find $\overrightarrow{AB}$ and $\overrightarrow{BA}$ , and write $\overrightarrow{AB}$ with $i, j, k$ .

Solution. $\overrightarrow{AB} = b - a = (5 - 2,\ 1 - (-1),\ -1 - 3) = (3, 2, -4) = 3i + 2j - 4k$ , and $\overrightarrow{BA} = -\overrightarrow{AB} = (-3, -2, 4)$ .

Worked example 2: combining and solving

Let $u = (2, - 1, 3)$ , $v = (- 1, 4, 0)$ , $w = (3, 0, - 2)$ . (a) Find $2 u - v + 3 w$ . (b) Find the vector $x$ with $3 x + u = 2 (v - x)$ .

Solution. (a) $2u - v + 3w = (4 + 1 + 9,\ -2 - 4 + 0,\ 6 - 0 - 6) = (14, -6, 0)$ .

(b) The vector-space rules let you solve as in algebra: $3 x + 2 x = 2 v - u$ , so $5x = (-2 - 2,\ 8 + 1,\ 0 - 3) = (-4, 9, -3)$ and $x = \left(-\frac{4}{5}, \frac{9}{5}, -\frac{3}{5}\right)$ .

Worked example 3: parallel vectors

Find $m$ and $n$ so that $(2, m, - 6)$ is parallel to $(- 1, 3, n)$ .

Solution. We need $(2, m, -6) = \lambda(-1, 3, n)$ . The first components give $\lambda = -2$ . Then $m = 3\lambda = -6$ and $- 6 = - 2 n$ , so $n = 3$ .

Worked example 4: the fourth vertex of a parallelogram

$A B C D$ is a parallelogram with $A (1, 0, 2)$ , $B (3, - 1, 4)$ and $C (2, 5, 1)$ . Find $D$ .

Solution. In a parallelogram $\overrightarrow{AD} = \overrightarrow{BC}$ , so $d - a = c - b$ and

d = a + c - b = (1 + 2 - 3,\ 0 + 5 + 1,\ 2 + 1 - 4) = (0, 6, -1) .

Check: $\overrightarrow{AD} = (-1, 6, -3)$ and $\overrightarrow{BC} = (-1, 6, -3)$ . ✓ So $D (0, 6, - 1)$ .

Warning

$\overrightarrow{AB} = b - a$ , "end minus start", not $a - b$ .
A point is not a vector. $P (3, 4, 2)$ is a place; $\overrightarrow{OP} = (3, 4, 2)$ is the move from $O$ to it. Adding two points has no meaning; adding a vector to a point gives another point.
For parallel vectors, one scalar $\lambda$ must work for all three components. $(2, 4, 5)$ and $(1, 2, 3)$ are not parallel: $\lambda = 2$ fits the first two components but not the third.

Tip

To sketch a point in 3-D, draw its "shadow" $(x, y, 0)$ in the floor first, then go straight up by $z$ . The dashed box in the figure is exactly this route.

3Blue1Brown: Vectors, Chapter 1 of Essence of Linear Algebra (about 10 min). Watch the three views of a vector (an arrow, a list of numbers, and an object you add and scale) and how they fit together.

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Practice

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

For $A (1, - 2, 4)$ and $B (- 3, 0, 5)$ , find $\overrightarrow{AB}$ in component form and with $i, j, k$ .
With $u = (1, 2, - 1)$ and $v = (3, - 1, 2)$ , find $u + v$ , $3 u - 2 v$ and $-\frac{1}{2}v$ .
Which of $(6, - 3, 9)$ , $(- 4, 2, - 6)$ and $(2, - 1, 4)$ are parallel to $(2, - 1, 3)$ ?
Find $p$ and $q$ such that $(p, - 2, 4)$ and $(3, q, - 6)$ are parallel.
Solve $2 (x - u) = x + 3 v$ for the vector $x$ , with $u = (1, 0, 2)$ and $v = (0, - 1, 1)$ .
$A (2, 1, 0)$ , $B (4, 3, - 2)$ and $C (1, 5, 3)$ are three vertices of the parallelogram $A B C D$ . Find $D$ , and the point where its diagonals meet.
Prove, using components, that $\lambda(u + v) = \lambda u + \lambda v$ for all $u, v \in \mathbb{R}^3$ and $\lambda \in \mathbb{R}$ .

Show answersHide answers

$(- 4, 2, 1) = - 4 i + 2 j + k$ .
$(4, 1, 1)$ ; $(3 - 6, 6 + 2, - 3 - 4) = (- 3, 8, - 7)$ ; $\left(-\frac{3}{2}, \frac{1}{2}, -1\right)$ .
$(6, - 3, 9) = 3 (2, - 1, 3)$ and $(- 4, 2, - 6) = - 2 (2, - 1, 3)$ are parallel; $(2, - 1, 4)$ is not.
$\lambda = \frac{-6}{4} = -\frac{3}{2}$ from the third components, so $3 = -\frac{3}{2}p$ gives $p = - 2$ , and $q = -\frac{3}{2}(-2) = 3$ .
$x = 2 u + 3 v = (2, - 3, 7)$ .
$d = a + c - b = (- 1, 3, 5)$ . The diagonals meet at the midpoint of $A C$ : $\left(\frac{3}{2}, 3, \frac{3}{2}\right)$ .
The first component of $\lambda(u + v)$ is $\lambda(u_1 + v_1) = \lambda u_1 + \lambda v_1$ , the first component of $\lambda u + \lambda v$ ; the same holds for the other two components.

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every vector was checked by computer.
Video: Vectors | Chapter 1, Essence of linear algebra, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1Three numbers for a position in space

2Locating a point

Three numbers for a position in space

Locating a point