Vector Representation

Representation of vectors in three-dimensional space

A three-dimensional vector space is described using coordinates similar to a two-dimensional space. Imagine three number lines intersecting perpendicularly at the origin. One line, extending out of the horizontal plane, represents depth or height. The other two represent length and width. These are the x, y, and z axes.

The unit vectors in the x, y, and z directions are $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$, respectively. A three-dimensional vector, representing width, length, and depth, is defined in this space.

Vectors are denoted by two capital letters with an arrow above ($\overrightarrow{OA}$) or a single small boldface letter ($\mathbf{a}$). Sometimes an underlined or tilde-marked small letter ($\underline{a}$ or $\tilde{a}$) is used.

A vector $\mathbf{a}$ in three-dimensional space is an ordered triplet of real numbers (components) $a_1$, $a_2$, and $a_3$ such that $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ (a position vector), where $a_1\mathbf{i}$, $a_2\mathbf{j}$, and $a_3\mathbf{k}$ lie along the coordinate axes and have a common initial point at the origin.

In three-dimensional space, every point is identified by three numerical values. The z-axis is perpendicular to the x and y axes.

The positive z-axis is above the xy-plane, and the negative z-axis is below. The three axes follow the right-hand rule: if you align your right-hand fingers with the positive x-axis and curl them towards the positive y-axis, your thumb points in the positive z-axis direction.

If the figure is viewed from above along the z-axis, the positive z-axis points directly towards the viewer, with the positive x and y axes pointing left and right, respectively. Coordinates are written as $(x, y, z)$. The x and y coordinates locate a point in the xy-plane, and the z-coordinate shows the vertical position.

A vector $\mathbf{a}$ with three components is represented as $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$, where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are unit vectors in the x, y, and z directions, respectively.

A vector $\overrightarrow{OP} = (a, b, c)$ is represented in three-dimensional space.

For example, the vector $\overrightarrow{OA} = (2, -3, 4)$ can be written as $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$.

Representation of vectors in three-dimensional space follows the same principles as in two-dimensional space.

Magnitude of a vector in three-dimensional space

The magnitude of a vector $\mathbf{a} = (a_1, a_2, a_3)$, denoted by $|\mathbf{a}|$, is:

$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

The unit vector $\hat{\mathbf{a}}$ of vector $\mathbf{a}$ is:

$\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}}{\sqrt{a_1^2 + a_2^2 + a_3^2}}$

The distance between two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is:

$r = |P_2 - P_1| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Consider the figures

In Figure 1, the distance between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is given by:

$|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

(This is from the Pythagorean theorem.)

In Figure 2, the distance between points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is given by $r = |P_2 - P_1|$. The length $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ is parallel to the xy-plane and represents the distance between the projections of $P_1$ and $P_2$ onto that plane.

Thus, the distance between points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is:

$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

In Figure 2, the positive x, y, and z-axes are labeled x, y, and z, respectively. Their negative counterparts are shown with dotted lines. The origin is the intersection of all three axes.

The most convenient way to express a vector analytically is using its components.

Example 1

Find a unit vector in the direction opposite to $\overrightarrow{PQ}$, where $P(1, 3, 2)$ and $Q(-1, 0, 8)$.

Solution:

$\overrightarrow{PQ} = (-1 - 1)\mathbf{i} + (0 - 3)\mathbf{j} + (8 - 2)\mathbf{k} = -2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$

$\overrightarrow{QP} = -\overrightarrow{PQ} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$

$|\overrightarrow{QP}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

The unit vector in the direction opposite to $\overrightarrow{PQ}$ (which is the direction of $\overrightarrow{QP}$) is:

$\frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} = \frac{2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}}{7} = \frac{2}{7}\mathbf{i} + \frac{3}{7}\mathbf{j} - \frac{6}{7}\mathbf{k}$

Example 2

Write the ordered triplet that represents the vector from $R(-2, 0, 8)$ to $S(5, 4, -1)$, and then find the magnitude of $\overrightarrow{RS}$.

Solution:

$\overrightarrow{RS} = (5 - (-2), 4 - 0, -1 - 8) = (7, 4, -9)$

$|\overrightarrow{RS}| = \sqrt{7^2 + 4^2 + (-9)^2} = \sqrt{49 + 16 + 81} = \sqrt{146}$

Example 3

Write $\overrightarrow{AB}$ in terms of $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ if $A(-5, 10, 3)$ and $B(1, 4, -2)$.

Solution:

$\overrightarrow{AB} = (1 - (-5), 4 - 10, -2 - 3) = (6, -6, -5)$

$\overrightarrow{AB} = 6\mathbf{i} - 6\mathbf{j} - 5\mathbf{k}$

Example 4

Find a unit vector in the direction of $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} + \mathbf{k}$.

Solution:

$|\mathbf{a}| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$

$\hat{\mathbf{a}} = \frac{3\mathbf{i} + 4\mathbf{j} + \mathbf{k}}{\sqrt{26}} = \frac{3}{\sqrt{26}}\mathbf{i} + \frac{4}{\sqrt{26}}\mathbf{j} + \frac{1}{\sqrt{26}}\mathbf{k}$