Components of a vector

Pick up any vector in three-dimensional space and there is a completely mechanical way to describe it with three numbers: how far it reaches along $x$ , how far along $y$ , how far along $z$ . Those three numbers are its components, and writing the vector as $x\hat i+y\hat j+z\hat k$ turns every geometric question about it — its length, its direction, what happens when you add it to another vector — into ordinary arithmetic. This is the tool that makes the rest of vector algebra computable rather than just pictorial. Once a vector is in component form, magnitude is a square root of a sum of squares, addition is done term by term, and direction cosines fall out by dividing by the magnitude. This article builds the idea from the standard basis vectors $\hat i,\hat j,\hat k$ , derives the component and magnitude formulas, and works through the same numeric examples used throughout the Vector Algebra chapter.

The core insight

Any vector can be written as $a_1\hat i+a_2\hat j+a_3\hat k$ , where $a_1,a_2,a_3$ are its scalar components along the axes, and this component form converts magnitude, addition, and scalar multiplication into simple arithmetic on those three numbers.

The core idea

Set up axes $OX$ , $OY$ , $OZ$ through a common origin $O$ , mutually perpendicular. Along the positive $x$ , $y$ , $z$ directions there are three fixed unit vectors, written $\hat i,\hat j,\hat k$ , satisfying $|\hat i|=|\hat j|=|\hat k|=1$ . These are not arbitrary — $\hat i$ is literally the position vector of the point $(1,0,0)$ , $\hat j$ of $(0,1,0)$ , and $\hat k$ of $(0,0,1)$ . Every other vector in space is built as a combination of these three.

Position vector of a point. For a point $P(x,y,z)$ , the position vector $\overrightarrow{OP}$ is

\overrightarrow{OP} = x\hat i + y\hat j + z\hat k,

where: $x,y,z$ are the scalar components of $\overrightarrow{OP}$ along the three axes, and $x\hat i,\ y\hat j,\ z\hat k$ are its vector components. The two terms sound similar but mean different things — a scalar component is a bare number, a vector component still carries a direction.

Components of a general vector. A vector does not have to start at the origin. For $\overrightarrow{AB}$ running from $A(x_1,y_1,z_1)$ to $B(x_2,y_2,z_2)$ , the components come from subtracting the initial point's coordinates from the terminal point's:

\overrightarrow{AB} = (x_2-x_1)\hat i + (y_2-y_1)\hat j + (z_2-z_1)\hat k.

where: the order matters — it is always terminal minus initial, i.e. $B$ minus $A$ , never the other way round.

Magnitude from components. Once a vector is in the form $\vec a = a_1\hat i+a_2\hat j+a_3\hat k$ , its length follows from a three-dimensional Pythagorean-style sum:

|\vec a| = \sqrt{a_1^2+a_2^2+a_3^2}.

where: $a_1,a_2,a_3$ are the scalar components of $\vec a$ along $x,y,z$ . Component form is what makes vector algebra genuinely computational. Geometric constructions like the triangle law or the parallelogram law are excellent for building intuition about why vector addition works, but they do not scale to an arbitrary vector in three dimensions when an exact numeric answer is needed. Component form fixes that: addition, subtraction, and scalar multiplication all reduce to doing the same operation on matching components, one axis at a time. It also feeds directly forward into direction cosines, $l=\dfrac{a_1}{|\vec a|}$ and so on, and later into the dot and cross products, both of which are defined entirely in terms of components.

Figure

A three axis diagram showing a vector from the origin to point x y z resolved into three perpendicular component arrows along the x y and z axes

Derivation AMagnitude of a vector from its components

This derivation shows why $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$ follows from ordinary two-dimensional geometry applied twice, once in a horizontal plane and once vertically, rather than being an assumed extra rule bolted onto vector algebra.

Step 1. Take $\vec a = \overrightarrow{OP} = a_1\hat i+a_2\hat j+a_3\hat k$ for a point $P(a_1,a_2,a_3)$ . Drop a perpendicular from $P$ to the point $Q(a_1,a_2,0)$ lying directly below it in the $xy$ -plane. The segment $OQ$ lies entirely in that plane, and $QP$ is vertical, so triangle $OQP$ has a right angle at $Q$ .

Step 2. Inside the $xy$ -plane, $Q(a_1,a_2,0)$ is reached from $O$ by moving $a_1$ along $x$ and $a_2$ along $y$ , which are themselves perpendicular directions. Ordinary two-dimensional Pythagoras on that right triangle gives $OQ^2 = a_1^2+a_2^2$ , so $OQ=\sqrt{a_1^2+a_2^2}$ .

Step 3. Now apply Pythagoras to the right triangle $OQP$ , whose legs are $OQ$ and $QP=a_3$ and whose hypotenuse is $OP=|\vec a|$ . That gives $OP^2 = OQ^2+QP^2 = (a_1^2+a_2^2)+a_3^2$ .

Step 4. Take the square root of both sides: $|\vec a| = \sqrt{a_1^2+a_2^2+a_3^2}$ . The same argument applies unchanged to $\overrightarrow{AB}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k$ , since it is just a position vector once the axes are shifted so that $A$ is the origin — so the magnitude of any vector between two points is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ .

Worked example

Find the scalar components and the magnitude of $\overrightarrow{AB}$ , where $A(2,-1,3)$ and $B(-4,5,1)$ . Then use $\vec a = 2\hat i+3\hat j-\hat k$ and $\vec b=-\hat i+4\hat j+5\hat k$ to find the components and magnitude of $2\vec a-3\vec b$ .

Step 1. Subtract $A$ 's coordinates from $B$ 's coordinates, terminal minus initial:

\overrightarrow{AB} = (-4-2)\hat i + (5-(-1))\hat j + (1-3)\hat k = -6\hat i + 6\hat j - 2\hat k.

So the scalar components of $\overrightarrow{AB}$ are $-6$ , $6$ , and $-2$ along $x$ , $y$ , $z$ .

Step 2. Find the magnitude using $|\overrightarrow{AB}|=\sqrt{a_1^2+a_2^2+a_3^2}$ :

|\overrightarrow{AB}| = \sqrt{(-6)^2+6^2+(-2)^2} = \sqrt{36+36+4} = \sqrt{76} = 2\sqrt{19}.

Step 3. For the second part, scale each vector componentwise: $2\vec a = 4\hat i+6\hat j-2\hat k$ and $3\vec b=-3\hat i+12\hat j+15\hat k$ .

Step 4. Subtract componentwise, matching $x$ with $x$ , $y$ with $y$ , $z$ with $z$ :

2\vec a-3\vec b = (4-(-3))\hat i + (6-12)\hat j + (-2-15)\hat k = 7\hat i - 6\hat j - 17\hat k.

Step 5. Apply the magnitude formula one more time: $|2\vec a-3\vec b| = \sqrt{7^2+(-6)^2+(-17)^2} = \sqrt{49+36+289} = \sqrt{374}$ . So $\overrightarrow{AB}=-6\hat i+6\hat j-2\hat k$ has magnitude $\mathbf{2\sqrt{19}}$ , and $2\vec a-3\vec b = 7\hat i-6\hat j-17\hat k$ has magnitude $\mathbf{\sqrt{374}}$ .

Competitive stretch

Once $\vec a=2\hat i+3\hat j-\hat k$ is known, its direction cosines follow immediately: $|\vec a|=\sqrt{4+9+1}=\sqrt{14}$ , so the direction cosines are $\left(\dfrac{2}{\sqrt{14}},\dfrac{3}{\sqrt{14}},\dfrac{-1}{\sqrt{14}}\right)$ . This is exactly the position-vector direction-cosine formula from earlier in the chapter, just applied to a general vector instead of only to $\overrightarrow{OP}$ — component form is what lets the same formula transfer over unchanged.

Common traps

Subtracting in the wrong order. $\overrightarrow{AB}$ is terminal minus initial ( $B$ minus $A$ ); doing it the other way silently computes $\overrightarrow{BA}=-\overrightarrow{AB}$ , flipping every sign.
Mixing up scalar and vector components. For $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ , the scalar components are the plain numbers $a_1,a_2,a_3$ ; the vector components are $a_1\hat i,\,a_2\hat j,\,a_3\hat k$ . Both terms show up in exam questions, so read carefully which one is being asked for.
Squaring negative components incorrectly. $(-6)^2=36$ , not $-36$ ; dropping the sign before squaring is one of the most common arithmetic slips in magnitude questions.
Treating raw components as direction cosines. The components $x,y,z$ equal the direction cosines only when $|\vec r|=1$ ; for any other vector you must divide each component by the magnitude first.

Ask the tutor: stuck on whether to subtract A from B or B from A, or want a few more practice problems on finding components and magnitude? Ask the AI tutor.

Why do we bother writing vectors with i-hat, j-hat, k-hat instead of just listing three numbers like (a1, a2, a3)?

Both notations carry the same information, and many textbooks use them interchangeably. Writing $a_1\hat i+a_2\hat j+a_3\hat k$ makes explicit that a vector is a sum of pieces along fixed reference directions, which matters once you start adding vectors, taking dot products, or computing cross products — those operations are defined in terms of the $\hat i,\hat j,\hat k$ pieces, so keeping the basis vectors visible in the notation is a reminder of what is actually being combined.

Do the components of AB depend on where A and B are, or only on the vector's length and direction?

Only on length and direction. Two different segments with the same length and the same direction — for example $\overrightarrow{AB}$ from $(0,0,0)$ to $(3,4,0)$ and $\overrightarrow{CD}$ from $(1,1,1)$ to $(4,5,1)$ — give identical component vectors $3\hat i+4\hat j$ , because only the coordinate differences matter, not the starting point. This is exactly why we can freely slide a "free vector" around without changing what it represents.

If two vectors have the same magnitude, must they have the same components?

No. Magnitude only fixes $\sqrt{a_1^2+a_2^2+a_3^2}$ , and many different triples $(a_1,a_2,a_3)$ can give the same value under a square root — for instance $3\hat i+4\hat j$ and $4\hat i+3\hat j$ both have magnitude $5$ but are clearly different vectors pointing in different directions. Magnitude alone never determines a vector; you also need its direction, which is exactly what the individual components (or the direction cosines built from them) supply.

Components of a vector

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads