A vector is a magnitude and a direction bundled into one object, written in component form as \langle x, y\rangle or in bold as \mathbf{v}. Everything about vectors from here forward is arithmetic on components: addition, scaling, and the dot and cross products. Every one of those operations works the same way with three components instead of two. Getting the two-component version automatic now is what makes the three-component version feel like nothing new.
Think of a vector as walking directions, not a fixed address. "3 blocks east, 4 blocks north" is the same instructions no matter where you start. Adding two vectors just means following one set of directions, then the other. Scaling by k means walking the same directions k times as far, turning around first if k is negative. A unit vector is what's left once you strip out how far you walked, keeping only which way. That is why dividing by the length works: it shrinks the "how far" part down to exactly 1, leaving "which way" untouched.
A vector is a quantity with both a magnitude (a length) and a direction, written in component form as \langle x,y\rangle or in bold as \mathbf{v}; a plane vector is just the two-component case, the one this section reviews. Two vectors are equal exactly when their components match. The arrow's starting point does not matter, since a vector carries no fixed position of its own; only its length and direction are meaningful.
Addition and scalar multiplication both work component-by-component: \langle u_1,u_2\rangle + \langle v_1,v_2\rangle = \langle u_1+v_1,\ u_2+v_2\rangle and k\langle x,y\rangle = \langle kx,ky\rangle. Here k scales the magnitude by |k| and reverses direction exactly when k<0.
A unit vector is any vector of magnitude 1. Dividing a nonzero vector by its own magnitude, \hat{\mathbf{v}} = \mathbf{v}/|\mathbf{v}|, always produces one, since that division is scalar multiplication by the positive constant 1/|\mathbf{v}|. The magnitude itself is |\mathbf{v}| = \sqrt{x^2+y^2}, the Pythagorean distance from the origin to the point (x,y).
Formula
Vector Components, Magnitude, and Direction Angle
Recalled from College Algebra · Geometric and component vectors — first taught there.
For a vector with magnitude |\mathbf{v}| and direction angle \theta, x = |\mathbf{v}|\cos\theta, \qquad y = |\mathbf{v}|\sin\theta. Conversely, for a vector with components \langle x, y\rangle, |\mathbf{v}| = \sqrt{x^2+y^2}, \qquad \theta = \tan^{-1}\!\left(\frac{y}{x}\right)\ \text{(adjusted for the quadrant of $\langle x,y\rangle$).}
Tip-to-tail addition, live
Drag u or v — the parallelogram's diagonal always lands on u + v, no matter how either vector is dragged.
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Preview: projecting one vector onto another
Drag u or v — the dashed shadow is how much of u points along v, the exact question the dot product answers a few sections from now.
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Worked example
Normalize a vector
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1
Find the unit vector pointing in the same direction as \mathbf{v} = \langle 3, 4\rangle.
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2never skip this step. Dividing by the wrong number is the most common error here
Magnitude first: |\mathbf{v}| = \sqrt{3^2+4^2} = \sqrt{25} = 5.
never skip this step. Dividing by the wrong number is the most common error here
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Divide every component by 5: \hat{\mathbf{v}} = \left\langle \dfrac{3}{5},\ \dfrac{4}{5}\right\rangle = \langle 0.6,\ 0.8\rangle.
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Check: |\hat{\mathbf{v}}| = \sqrt{0.6^2+0.8^2} = \sqrt{0.36+0.64} = \sqrt{1} = 1. \boxed{\hat{\mathbf{v}} = \langle 0.6,\ 0.8\rangle}
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Where this goes next
Chapter 1: Vectors and the geometry of space
Every operation reviewed here is about to gain a third coordinate and nothing else: components, magnitude, addition, and unit vectors all carry over unchanged. A 3D vector \langle x,y,z\rangle adds, scales, and normalizes by the same rules. The geometry just moves off the page and into a room.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Why is a vector's magnitude always non-negative?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
It's not a notational convention papering over negative values — length genuinely has no negative version to convert.
Direction (left, down, or any angle) is captured separately, by the vector's angle — it never leaks into the magnitude as a negative sign.
Magnitude is quadrant-independent; \sqrt{x^2+y^2} comes out non-negative no matter which quadrant \langle x,y\rangle sits in.
Why
Magnitude is a length — the size of the arrow — and length is fundamentally a non-negative quantity. Direction (which can point any which way) is handled entirely separately, by the vector's angle.
Magnitude is a length — can a length itself be negative?
From College Algebra
A vector has magnitude 9 and direction angle 240°. Find its exact y-component.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
y = 9\sin240° = 9\left(-\frac{\sqrt3}{2}\right) = -\frac{9\sqrt3}{2}.
Correct answer
Also accepted:
Why
240° is a Quadrant III reference angle of 60° — both components should come out negative.
From College Algebra
A GPS app reports 'distance to destination: 3.2 miles.' Is this a scalar or a vector statement?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The app could add a direction (a compass heading), but as stated, this particular readout is a bare number with no direction attached.
A single quantity is classified by what's actually reported, not by what could theoretically accompany it.
It's a real numeric quantity, just without a direction — that makes it a scalar, not neither.
Why
As stated, only a magnitude (3.2 miles) is given — no direction. That makes it a scalar, the same way plain 'speed' is a scalar until a heading is attached.
Check whether a direction is actually stated alongside the distance.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
With \mathbf{u}=\langle7,3\rangle and \mathbf{v}=\langle-2,-9\rangle, what is the x-component of \mathbf{u}-\mathbf{v}?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
7-(-2)=9.
Correct answer
Also accepted:
Why
From College Algebra
Write v = ⟨7, -3⟩ using standard basis vectors, then find the coefficient of i minus the coefficient of j.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
v = 7i - 3j, so the coefficient of i is 7 and the coefficient of j is -3. Their difference is 7 - (-3) = 10.
Correct answer
Also accepted:
Why
Write v in ai+bj form first, then subtract the coefficient of j from the coefficient of i.
From College Algebra
What is the magnitude of the zero vector ⟨0, 0⟩?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
|⟨0,0⟩| = √(0²+0²) = 0. The zero vector is the only vector with zero length.
Correct answer
Also accepted:
Why