Source: Homework Set 1, §12.2–12.3
Tags: projection, orthogonal projection, orthogonal decomposition, dot product, vector projection, comp, proj, orth, Gram-Schmidt
Difficulty: Intermediate | Prerequisites: §12.1 (vectors, dot product basics). You should be comfortable computing dot products and know what it means for two vectors to be orthogonal.
Big Picture
Orthogonal projection is one of the central tools in §12.2 and §12.3 of Calculus III. The idea is simple: given two vectors, break one of them into a piece that runs along the other and a piece that runs perpendicular to it. This "orthogonal decomposition" turns up everywhere in multivariable calculus, from computing distances between points and lines to understanding work and force components. It also forms the backbone of Gram-Schmidt orthogonalisation, which you will meet again in linear algebra. If you can project comfortably, the later material on planes and rotations (§12.4, §12.5) becomes much more manageable.
Projection splits a vector v into a component along u (proj) and a component perpendicular to u (orth). Together they give you the orthogonal decomposition v = proj_u v + orth_u v. The formulas lean entirely on the dot product, and verifying orthogonality is just checking that one dot product equals zero.
Projection of v onto u (proj_u v)
The vector component of v in the direction of u, given by ((v · u) / (u · u)) u.
In simple terms, this is the "shadow" of v cast along the line of u.
Orthogonal projection (orth_u v)
The component of v that is perpendicular to u, defined as v - proj_u v.
Think of it as: everything in v that the projection leaves behind.
Orthogonal decomposition
The statement that any vector v can be written as the sum of its projection onto u and its orthogonal projection: v = proj_u v + orth_u v. The two pieces are at right angles.
In simple terms, you are splitting one vector into two perpendicular parts that add back to the original.
Dot product (v · u)
The scalar quantity v₁u₁ + v₂u₂ + ... + vₙuₙ. Two vectors are orthogonal precisely when their dot product is zero.
Think of it as: a single number that measures how much two vectors "agree" in direction.
The projection formula
proj_u v = ((v · u) / (u · u)) u
The scalar coefficient (v · u) / (u · u) is sometimes called comp_u v when divided by |u| instead, but the vector projection is the one you will use most often.
Notice that proj_u v always points in the direction of u (or opposite to u). It is a scalar multiple of u.
The orthogonal projection formula
orth_u v = v - proj_u v
This gives you whatever is "left over" once you subtract the part of v lying along u.
Proving orth_u v is orthogonal to u
Compute orth_u v · u = (v - proj_u v) · u.
Expand: v · u - proj_u v · u.
Substitute proj_u v = ((v · u)/(u · u)) u, so proj_u v · u = ((v · u)/(u · u))(u · u) = v · u.
The result is v · u - v · u = 0. Since the dot product is zero, orth_u v is perpendicular to u.
Worked example: R² (u = ⟨2, -3⟩, v = ⟨3, 1⟩)
v · u = (3)(2) + (1)(-3) = 6 - 3 = 3
u · u = 4 + 9 = 13
proj_u v = (3/13)⟨2, -3⟩ = ⟨6/13, -9/13⟩
orth_u v = ⟨3, 1⟩ - ⟨6/13, -9/13⟩ = ⟨33/13, 22/13⟩
Check: ⟨33/13, 22/13⟩ · ⟨2, -3⟩ = 66/13 - 66/13 = 0. Confirmed orthogonal.
Worked example: R³ (u = ⟨2, -6, -3⟩, v = ⟨8, 4, -6⟩)
v · u = 16 - 24 + 18 = 10
u · u = 4 + 36 + 9 = 49
proj_u v = (10/49)⟨2, -6, -3⟩ = ⟨20/49, -60/49, -30/49⟩
orth_u v = ⟨8, 4, -6⟩ - ⟨20/49, -60/49, -30/49⟩ = ⟨372/49, 256/49, -264/49⟩
Check: dot this with u and confirm it equals zero.
\text{proj}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}}\,\mathbf{u}\text{orth}_{\mathbf{u}}\mathbf{v} = \mathbf{v} - \text{proj}_{\mathbf{u}}\mathbf{v}\mathbf{v} = \text{proj}_{\mathbf{u}}\mathbf{v} + \text{orth}_{\mathbf{u}}\mathbf{v} \quad (\text{orthogonal decomposition})The scalar projection (component) is sometimes written separately:
\text{comp}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|}Note: proj_u v = (comp_u v / |u|) u, so the vector projection is the scalar projection times the unit vector in the direction of u.
Projection is how engineers resolve forces into components. When a crate sits on a ramp, the component of gravity along the surface (projection) tells you the sliding force, and the perpendicular component (orthogonal projection) tells you the normal force pressing the crate into the ramp.
In computer graphics, projections determine how light reflects off surfaces, since the reflected ray depends on decomposing the incoming light vector relative to the surface normal.
Students often divide by |u| instead of u · u in the projection formula. The denominator is the dot product of u with itself (which equals |u|²), not the magnitude.
Confusing the scalar projection (comp) with the vector projection (proj). The scalar projection is a number; the vector projection is a vector in the direction of u.
Forgetting that orth_u v is defined relative to a particular u. Changing u changes both the projection and the orthogonal part.
Assuming proj_u v and proj_v u are the same. They point in different directions and generally have different magnitudes.
⚠️ The orthogonality proof (showing orth_u v · u = 0) is a classic exam question. Be able to do it from scratch.
⚠️ Expect at least one problem where you compute proj and orth in R² and another in R³, then verify with a dot-product check.
⚠️ This decomposition is reused directly in Question 5 of this homework (vector rotation) and in later sections on distance from a point to a line.
True or false: proj_u v always has the same magnitude as v.
False. It has magnitude ≤ |v|, with equality only when v is parallel to u.
Fill in the blank: orth_u v · u = ___.
0
True or false: proj_u v + orth_u v = v.
True. That is the orthogonal decomposition.
True or false: if v is perpendicular to u, then proj_u v = 0.
True. The dot product v · u is zero, so the formula yields the zero vector.
Fill in the blank: the denominator in the projection formula is ___, not |u|.
u · u (equivalently, |u|²)
Q: Let u = ⟨1, 2⟩ and v = ⟨4, 0⟩. Compute proj_u v and orth_u v.
A: v · u = 4. u · u = 5. proj_u v = (4/5)⟨1, 2⟩ = ⟨4/5, 8/5⟩. orth_u v = ⟨4, 0⟩ - ⟨4/5, 8/5⟩ = ⟨16/5, -8/5⟩.
Q: Prove that orth_u v is orthogonal to u for any nonzero vectors u and v.
A: orth_u v · u = (v - ((v · u)/(u · u))u) · u = v · u - ((v · u)/(u · u))(u · u) = v · u - v · u = 0.
Q: If u = ⟨1, 0, 1⟩ and v = ⟨2, 3, 0⟩, what is proj_u v?
A: v · u = 2. u · u = 2. proj_u v = (2/2)⟨1, 0, 1⟩ = ⟨1, 0, 1⟩.
Q: Suppose proj_u v = 0. What does this tell you about the relationship between u and v?
A: It means v · u = 0, so v and u are orthogonal. The entire vector v lies in the direction perpendicular to u.
Q: Can proj_u v ever be longer than v? Why or why not?
A: No. By the Cauchy-Schwarz inequality, |v · u| ≤ |v||u|, so the scalar factor (v · u)/(u · u) has absolute value at most |v|/|u|, making |proj_u v| ≤ |v|. Equality holds only when v is parallel to u.
This material connects directly to the cross product (§12.4), because the area formula for parallelograms uses the cross product of two vectors, and understanding which component is "along" versus "across" relies on the same decomposition idea.
It also connects to the parametric description of planes (§12.5), where you navigate a plane using two direction vectors, and to Question 5 of this homework, which rotates a vector by first projecting it into a plane orthogonal to the axis of rotation.
In linear algebra, this process generalises to Gram-Schmidt orthogonalisation, which builds an orthonormal basis by repeatedly subtracting projections.
vector projection, scalar projection, component of v along u, orthogonal complement, perpendicular component, Gram-Schmidt, dot product, inner product, proj formula, orth formula, orthogonal decomposition, shadow of a vector, force decomposition, normal component, tangential component, Calculus III §12.2, MATH 215