Source: Math 126 Midterm 1 Practice Exam
Difficulty: Intermediate | Prerequisites: Familiarity with basic vector operations (addition, scalar multiplication, magnitude) from Math 125 or equivalent.
Vector projections and components are central to the first midterm in Math 126. They build directly on the dot product and connect to nearly everything that follows: equations of planes, work done by a force, and decomposing motion into tangential and normal parts. If you can compute a dot product and know what a unit vector is, you have the prerequisites. If you skipped the dot product lecture, read those notes first.
The dot product lets you measure how much one vector points in the direction of another. The scalar projection (comp) gives you a signed length along that direction, while the vector projection (proj) gives you the actual vector component. Master the formulas for both, know the difference, and most midterm projection problems become routine algebra.
Dot product (scalar product)
For vectors u = ⟨u₁, u₂, u₃⟩ and v = ⟨v₁, v₂, v₃⟩, the dot product is u · v = u₁v₁ + u₂v₂ + u₃v₃. Equivalently, u · v = |u| |v| cos θ, where θ is the angle between them. Think of it as a single number that tells you how aligned two vectors are.
Scalar projection (component)
comp_v u = (u · v) / |v|. This is the signed length of the shadow u casts onto the line through v. In simple terms, it answers: how far does u reach in the direction of v?
Vector projection
proj_v u = ((u · v) / |v|²) v. This is the vector component of u that lies along v. Think of it as the shadow of u onto v, returned as an actual vector.
Unit vector
A vector of magnitude 1. The unit vector in the direction of v is v̂ = v / |v|. In simple terms, it keeps the direction but scales the length to exactly 1.
Angle between vectors
cos θ = (u · v) / (|u| |v|). This is the standard way to find the angle. If θ is acute, the dot product is positive; if obtuse, negative; if 90°, zero.
Orthogonal decomposition
Any vector u can be written as the sum of its projection onto v and the component perpendicular to v: u = proj_v u + u_⊥. In simple terms, you split a vector into a "parallel" part and a "leftover" part.
The scalar projection (comp) is a number. The vector projection (proj) is a vector.
They are related by: proj_v u = (comp_v u) · (v / |v|)
Equivalently: comp_v u = |proj_v u| when the projection is in the same direction as v, and comp_v u = -|proj_v u| when it points opposite.
Projection is linear: proj_v(u + w) = proj_v u + proj_v w
This is used constantly in exam problems. If you know proj_v u and proj_v(u + v), you can find proj_v v by subtraction.
Special case: proj_v v = v always. The projection of any vector onto itself is itself.
Given:
The angle θ between u and v is acute with sin θ = 1/2
proj_v u = ⟨1, 1, 2⟩
proj_v(u + v) = ⟨3, 3, 6⟩
Find comp_u v.
Step 1: Use linearity to find proj_v v
proj_v(u + v) = proj_v u + proj_v v
⟨3, 3, 6⟩ = ⟨1, 1, 2⟩ + proj_v v
So proj_v v = ⟨2, 2, 4⟩.
But proj_v v = v always (any vector projected onto itself is itself).
Therefore v = ⟨2, 2, 4⟩.
Step 2: Find |u| using the projection formula
proj_v u = ((u · v) / |v|²) v
Since proj_v u = ⟨1, 1, 2⟩ = (1/2) ⟨2, 2, 4⟩ = (1/2) v, the scalar factor is 1/2.
So (u · v) / |v|² = 1/2, which means u · v = |v|²/2.
|v| = √(4 + 4 + 16) = √24 = 2√6, so |v|² = 24.
Therefore u · v = 12.
Step 3: Find |u| using sin θ
We know sin θ = 1/2 and θ is acute, so θ = π/6 and cos θ = √3/2.
From u · v = |u||v| cos θ:
12 = |u| · 2√6 · (√3/2) = |u| · √18 = |u| · 3√2
So |u| = 12 / (3√2) = 4/√2 = 2√2.
Step 4: Compute comp_u v
comp_u v = (u · v) / |u| = 12 / (2√2) = 6/√2 = 3√2.
\text{comp}_{\mathbf{v}}\,\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|}\text{proj}_{\mathbf{v}}\,\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\,\mathbf{v}\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}\mathbf{u} = \text{proj}_{\mathbf{v}}\,\mathbf{u} + \mathbf{u}_{\perp}Vector projection is how engineers decompose forces. When you push a lawnmower at an angle, only the horizontal component of your force moves it forward; the rest pushes it into the ground. That horizontal component is the projection of your force vector onto the ground direction. The same idea appears in physics (work = F · d, which is |F| times the component of F along d) and in computer graphics (lighting calculations project light direction onto surface normals).
Students often confuse comp_v u (a scalar) with proj_v u (a vector). The component is a number; the projection is a vector. If the answer should be a vector and you wrote a number, you mixed them up.
Students often divide by the wrong magnitude. comp_v u divides by |v|, not |u|. The subscript tells you which vector you are projecting onto, and that is the one whose magnitude appears in the denominator.
Students sometimes forget that proj_v v = v. This is the identity that unlocks many exam problems where you are given proj_v(u + v).
Students often assume comp_v u is always positive. It can be negative when the angle between the vectors is obtuse (greater than 90°). The sign tells you direction: positive means u has a component in the same direction as v, negative means opposite.
⚠️ Projection problems that give you proj_v(u + v) are testing whether you know proj_v v = v. This appeared on the Autumn 2024 midterm.
⚠️ Problems that give sin θ expect you to recover cos θ (or vice versa) using sin²θ + cos²θ = 1. Watch the sign: if θ is acute, cos θ is positive.
⚠️ Answers must be in exact form. Leave square roots, not decimals. 3√2 is correct; 4.2426 is not.
⚠️ Know the difference between comp_u v and comp_v u. The subscript and argument swap, and they give different numbers.
True or False: proj_v u is always a scalar. (False, it is a vector.)
True or False: comp_v u can be negative. (True, when the angle is obtuse.)
Fill in the blank: proj_v v = ___. (v itself.)
True or False: proj_v(u + w) = proj_v u + proj_v w. (True, projection is linear.)
Fill in the blank: If u · v = 0, then comp_v u = ___. (0, the vectors are orthogonal.)
Q: Let u = ⟨3, 0, 4⟩ and v = ⟨1, 0, 0⟩. Find proj_v u and comp_v u.
A: u · v = 3. |v| = 1. So comp_v u = 3/1 = 3. proj_v u = (3/1) ⟨1, 0, 0⟩ = ⟨3, 0, 0⟩.
Q: If proj_v u = ⟨2, -1, 3⟩ and proj_v(u + v) = ⟨5, -2.5, 7.5⟩, find v.
A: proj_v v = proj_v(u + v) - proj_v u = ⟨3, -1.5, 4.5⟩. Since proj_v v = v, we have v = ⟨3, -1.5, 4.5⟩.
Q: Given |u| = 5, |v| = 3, and u · v = -6, find comp_u v and comp_v u.
A: comp_u v = (u · v) / |u| = -6/5. comp_v u = (u · v) / |v| = -6/3 = -2. Note they are different numbers.
Q: Vectors a and b satisfy |a| = 4, |b| = 6, and the angle between them is π/3. Find comp_b a.
A: a · b = |a||b| cos(π/3) = 4 · 6 · (1/2) = 12. comp_b a = 12/6 = 2.
Vector projections connect directly to equations of planes (the normal vector determines distance via projection), to work in physics (W = F · d = |F| comp_d F · |d|), and to the tangent/normal decomposition of acceleration that appears later in Math 126. The cross product, covered in the next set of notes, gives you the perpendicular component that projection leaves behind.
Tags: vector projection, scalar projection, component of a vector, comp, proj, dot product, scalar product, inner product, angle between vectors, orthogonal decomposition, unit vector, Math 126, Calculus III, UW, midterm 1, vector operations, linearity of projection