Difficulty: Introductory | Prerequisites: Part 1 of these notes (Vectors in 3D Space) and comfort with square roots
Once you can write a vector in component form, the next questions are: how long is it, and how do I scale it to length 1? These are the magnitude and unit vector operations. You will also learn to express any vector as a weighted sum of the standard basis vectors i, j and k. These tools recur throughout Calculus III, from computing dot and cross products to finding tangent and normal vectors along curves.
The magnitude of a vector is its length, found by taking the square root of the sum of its squared components. A unit vector points in the same direction but has length 1, obtained by dividing each component by the magnitude. Any 3D vector can be written as a linear combination of the standard unit vectors i, j and k.
Magnitude (norm, length) of a vector
The scalar quantity representing a vector's length, written ‖v⃗‖. Computed as the square root of the sum of the squares of its components. Think of it as "how far the arrow reaches."
Unit vector
A vector with magnitude exactly 1. It preserves direction while stripping away length. In simple terms, it answers "which way?" without "how far?"
Standard unit vectors (i, j, k)
The three vectors of length 1 pointing along the positive x-, y- and z-axes respectively: i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩. They form the standard basis for 3D space.
Linear combination
A way of writing a vector as a sum of scalar multiples of other vectors. For example, ⟨−5, 11, 5⟩ = −5i + 11j + 5k. Each component becomes the coefficient (weight) on its corresponding basis vector.
Distance formula (3D)
The magnitude of the vector between two points, giving the straight-line distance between them in 3D space. It is the three-dimensional version of the Pythagorean theorem.
Given P = (4, −5, 2), Q = (−1, 6, 7) and R = (1, −2, 5):
PQ = ⟨−1 − 4, 6 − (−5), 7 − 2⟩ = ⟨−5, 11, 5⟩
PR = ⟨1 − 4, −2 − (−5), 5 − 2⟩ = ⟨−3, 3, 3⟩
Replace each component with a scalar multiple of its corresponding unit vector.
u⃗ = ⟨−5, 11, 5⟩ becomes −5i + 11j + 5k.
v⃗ = ⟨−3, 3, 3⟩ becomes −3i + 3j + 3k.
This notation is interchangeable with angle-bracket notation. Professors use whichever suits the context.
The magnitude of PQ = ⟨−5, 11, 5⟩ is: ‖PQ‖ = sqrt(25 + 121 + 25) = sqrt(171) ≈ 13.1
The magnitude of PR = ⟨−3, 3, 3⟩ is: ‖PR‖ = sqrt(9 + 9 + 9) = sqrt(27) ≈ 5.2
The distance between two points equals the magnitude of the vector connecting them.
Divide each component by the magnitude of the vector.
Unit vector in the direction of PQ: ⟨−5/sqrt(171), 11/sqrt(171), 5/sqrt(171)⟩
Unit vector in the direction of PR: ⟨−3/sqrt(27), 3/sqrt(27), 3/sqrt(27)⟩
A quick check: the magnitude of any unit vector should be 1. If you square the components, add them and get 1, you have done it correctly.
Magnitude of a vector:
\|\vec{v}\| = \sqrt{a^2 + b^2 + c^2} \quad \text{where } \vec{v} = \langle a, b, c \rangleUnit vector in the direction of v⃗:
\hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \left\langle \frac{a}{\|\vec{v}\|},\; \frac{b}{\|\vec{v}\|},\; \frac{c}{\|\vec{v}\|} \right\rangleLinear combination form:
\vec{v} = a\,\vec{i} + b\,\vec{j} + c\,\vec{k}Distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
d(P,Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}Unit vectors are how GPS systems and robotics encode direction independently of speed or force. A drone's controller, for example, computes a unit vector toward a waypoint, then scales it by the desired speed to get the velocity vector. The magnitude formula is the same calculation your phone uses to find straight-line distance between two map pins in 3D (accounting for altitude).
Students sometimes forget to square each component before adding when computing magnitude. You must square first, add, then take the square root.
When finding a unit vector, students occasionally divide the magnitude by the vector instead of the vector by the magnitude. The vector goes in the numerator.
The notation ⟨−5, 11, 5⟩ and −5i + 11j + 5k represent exactly the same vector. They are not different vectors or different "types" of vectors.
A unit vector is not ⟨1, 1, 1⟩. That vector has magnitude sqrt(3), not 1. A unit vector has magnitude 1, which usually means irrational components.
⚠️ "Find the magnitude" and "find the unit vector" are standard exam questions. Know the formulas by heart and show the intermediate steps.
⚠️ Rounding: the assignment specifies rounding to the tenths place. On exams, follow the stated precision or leave answers in exact radical form (e.g. sqrt(171)) unless told otherwise.
⚠️ Writing vectors as linear combinations of i, j, k may appear as a conversion question. It is mechanical but easy to lose marks on if you mix up signs.
The magnitude of ⟨3, 4, 0⟩ is ____. (5)
True or false: the unit vector of ⟨0, 0, 6⟩ is ⟨0, 0, 1⟩. (True)
Write ⟨2, −7, 1⟩ as a linear combination of i, j, k: ____. (2i − 7j + k)
True or false: ‖unit vector‖ = 1 always. (True)
Fill in the blank: to find the unit vector, divide the vector by its ____. (magnitude)
Q: Given P = (4, −5, 2) and Q = (−1, 6, 7), find ‖PQ‖. Round to the tenths place.
A: PQ = ⟨−5, 11, 5⟩. ‖PQ‖ = sqrt(25 + 121 + 25) = sqrt(171) ≈ 13.1.
Q: Find the unit vector in the direction of ⟨−3, 3, 3⟩.
A: ‖v⃗‖ = sqrt(9 + 9 + 9) = sqrt(27) = 3sqrt(3). Unit vector = ⟨−3/(3sqrt(3)), 3/(3sqrt(3)), 3/(3sqrt(3))⟩ = ⟨−1/sqrt(3), 1/sqrt(3), 1/sqrt(3)⟩.
Q: Write the vector ⟨−5, 11, 5⟩ as a linear combination of i, j and k.
A: −5i + 11j + 5k.
Q: A student computes the unit vector of ⟨4, 0, 3⟩ as ⟨4, 0, 3⟩/7. What is wrong?
A: The magnitude is sqrt(16 + 0 + 9) = sqrt(25) = 5, not 7. The student likely added the components (4 + 3) instead of squaring them. The correct unit vector is ⟨4/5, 0, 3/5⟩.
Q: Verify that ⟨−5/sqrt(171), 11/sqrt(171), 5/sqrt(171)⟩ is a unit vector.
A: Sum of squared components = 25/171 + 121/171 + 25/171 = 171/171 = 1. The magnitude is sqrt(1) = 1, confirming it is a unit vector.
Magnitude is the denominator in the dot product's angle formula (cos θ = (u⃗ · v⃗) / (‖u⃗‖ ‖v⃗‖)), so fluency here saves time later. Unit vectors also appear as tangent, normal and binormal vectors when studying curves in 3D. The i, j, k notation is essential for computing cross products using the determinant method in the next lesson.
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