Vectors in 3D - Coordinates in Space

Everything built for vectors in the plane carries over to space with one addition: a third coordinate. This chapter extends components, magnitude, addition, subtraction and scalar multiplication to three dimensions, and introduces the extra geometric intuition needed to "see" a point or a vector in space. The algebra barely changes; the payoff is that the whole of the rest of this block — lines, planes, angles and distances in space — depends on being fluent with it.

To follow the chapter you need everything from the 2D vector chapter: components, magnitude, addition, scalar multiplication and parallel vectors. Have a clear mental (or physical) model of three mutually perpendicular axes ready, since sketches in 3D are harder to draw accurately than in 2D and intuition has to compensate.

A third axis, a third coordinate. Space is described by three mutually perpendicular axes, $x$, $y$ and $z$, meeting at the origin $O$. Every point $P$ in space corresponds to exactly one ordered triple $(x,y,z)$, exactly as a point in the plane corresponds to an ordered pair.

1. Coordinates in Three-Dimensional Space

The three coordinate axes are conventionally drawn so that $x$, $y$ and $z$ obey the right-hand rule: point the fingers of the right hand along $x$ and curl them toward $y$; the thumb points along $z$. A point $P(x,y,z)$ is located by moving $x$ units along the $x$-axis, then $y$ units parallel to the $y$-axis, then $z$ units parallel to the $z$-axis.

The three coordinate planes are $xy$ (where $z=0$), $xz$ (where $y=0$) and $yz$ (where $x=0$); together they divide space into eight regions called octants, the 3D analogue of the plane's four quadrants.

2. Vectors in Component Form

A vector in space has three components:

where $\vec{i}=\begin{pmatrix}1\\0\\0\end{pmatrix}$, $\vec{j}=\begin{pmatrix}0\\1\\0\end{pmatrix}$, $\vec{k}=\begin{pmatrix}0\\0\\1\end{pmatrix}$ are the three standard unit vectors along the axes. Given $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$,

exactly as in two dimensions, just with one extra row. A position vector $\overrightarrow{OP} = \begin{pmatrix}x\\y\\z\end{pmatrix}$ again lets any point be treated as a vector from the origin.

3. Magnitude in Three Dimensions

The magnitude of $\vec{v}=\begin{pmatrix}a\\b\\c\end{pmatrix}$ requires two applications of Pythagoras' theorem, not one. First, the diagonal across the base rectangle (sides $a$ and $b$) has length $\sqrt{a^2+b^2}$. This diagonal and the vertical side $c$ are then the two legs of a second right triangle, whose hypotenuse is $\vec v$ itself:

The same two-step argument applied to two points $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$ gives the distance formula in space:

Worked example. Find $|\vec v|$ for $\vec v = \begin{pmatrix}2\\-3\\6\end{pmatrix}$.

4. Addition, Subtraction and Scalar Multiplication

All three operations extend component-wise, exactly as in the plane, now with three rows instead of two:

The geometric pictures (triangle rule, parallelogram rule, direction reversal under a negative scalar) are unchanged — they simply now live in three dimensions rather than two, and the parallel vectors condition $\vec u = k\vec v$ (proportional components in all three coordinates) extends identically.

Worked example. Let $\vec u=\begin{pmatrix}1\\4\\-2\end{pmatrix}$, $\vec v=\begin{pmatrix}-3\\0\\5\end{pmatrix}$. Find $2\vec u - \vec v$.

5. Unit Vectors in Space

The unit vector in the direction of a non-zero $\vec v$ is still $\hat v = \dfrac{\vec v}{|\vec v|}$, and a unit vector still satisfies $|\hat v|=1$. The three axis unit vectors $\vec i, \vec j, \vec k$ themselves have magnitude 1, and are mutually perpendicular — a fact this chapter states and the dot-product chapter will shortly prove algebraically.

Worked example. Find the unit vector in the direction of $\vec v=\begin{pmatrix}2\\-3\\6\end{pmatrix}$ (the vector from the earlier worked example, with $|\vec v|=7$).

6. Midpoint and Section Formula in Space

The midpoint and section formulas extend with one more component each, by exactly the same averaging argument as in the plane:

Worked example — the harder case. $A(2,-1,5)$, $B(-4,7,2)$. Find the point $P$ dividing $AB$ in the ratio $2:1$, and verify it lies strictly between $A$ and $B$ by checking that each coordinate of $P$ lies between the corresponding coordinates of $A$ and $B$.

Checking: $x$-coordinate $-2$ lies between $2$ and $-4$ ✓; $y$-coordinate $\tfrac{13}{3}\approx4.33$ lies between $-1$ and $7$ ✓; $z$-coordinate $3$ lies between $5$ and $2$ ✓. The edge case worth noting: a student who only checks one coordinate (commonly just $x$) can be fooled if an arithmetic slip happened in another — checking all three is the only reliable verification in three dimensions.

Common Exam Pitfalls

• Forgetting the third component entirely. Under exam pressure, a student used to 2D habitually drops the $z$-row when adding, subtracting or finding a magnitude — always count that every vector has exactly three rows before starting arithmetic.
• Applying the 2D distance formula with only two terms. The 3D magnitude needs three squared terms under the root, not two; a common error is to use the familiar two-term formula out of habit.
• Misreading the right-hand rule / axis orientation, which affects how a sketch is drawn (though not any algebraic answer) — algebraic results do not depend on how the axes are drawn, only on consistent labelling of coordinates.
• Assuming the three coordinate planes behave like three separate 2D planes with no relationship. The $xy$-, $xz$- and $yz$-planes are literally defined by one coordinate being zero; a point's membership in a plane is checked by substitution, not by inspecting a drawing.
• Treating a 3D unit vector's components as if they must be "nice" fractions. Unlike many textbook 2D examples, 3D unit vector components are frequently irrational; don't assume an error just because $\sqrt{a^2+b^2+c^2}$ isn't a perfect square.
• Checking only one coordinate when verifying that a point lies on a segment or between two others. All three coordinates must be checked; an error affecting only the $y$- or $z$-coordinate can otherwise go unnoticed.

Summary

• A point in space is $(x,y,z)$; a vector in space is $\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = v_1\vec i+v_2\vec j+v_3\vec k$.
• $\overrightarrow{AB} = \begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}$, exactly as in 2D with an extra row.
• Magnitude: $|\vec v|=\sqrt{a^2+b^2+c^2}$, from two applications of Pythagoras; distance formula: $|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$.
• Addition, subtraction, scalar multiplication and the parallel-vectors condition all extend component-wise, unchanged in structure from 2D.
• Unit vector: $\hat v = \vec v/|\vec v|$; the axis unit vectors $\vec i,\vec j,\vec k$ are mutually perpendicular unit vectors.
• Midpoint: average each coordinate; section formula: $\vec p = \dfrac{n\vec a+m\vec b}{m+n}$, identical in form to 2D.