Scalar and Vector Projections- Practice Problems

What Are Scalar and Vector Projections?

Projections sound complicated until you realize they're just measuring how much of one vector goes in the direction of another. That's it. No fancy geometry required.

A scalar projection (also called the component) tells you the length of one vector when projected onto another. It gives you a number — positive, negative, or zero.

A vector projection does the same thing but keeps the direction. It gives you a vector that points along the second vector.

The Formulas You Need to Memorize

Scalar projection of u onto v:

compv(u) = (u · v) / ||v||

Vector projection of u onto v:

projv(u) = [(u · v) / (v · v)] * v

That's it. Two formulas. Commit them to memory right now.

Practice Problems

Problem 1: Basic Scalar Projection

Find the scalar projection of u = ⟨3, 4⟩ onto v = ⟨1, 0⟩.

Step 1: Calculate the dot product.

u · v = (3)(1) + (4)(0) = 3

Step 2: Find ||v||.

||v|| = √(1² + 0²) = √1 = 1

Step 3: Divide.

comp_v(u) = 3 / 1 = 3

This makes sense. Projecting ⟨3, 4⟩ onto the x-axis gives you 3 — the x-component of the vector.

Problem 2: Vector Projection

Find the vector projection of u = ⟨2, 3⟩ onto v = ⟨1, 1⟩.

Step 1: Calculate u · v.

u · v = (2)(1) + (3)(1) = 5

Step 2: Calculate v · v.

v · v = (1)(1) + (1)(1) = 2

Step 3: Apply the formula.

proj_v(u) = (5/2) * ⟨1, 1⟩ = ⟨2.5, 2.5⟩

The projection of ⟨2, 3⟩ onto the line y = x is ⟨2.5, 2.5⟩. The scalar component would be 5/√2 ≈ 3.54.

Problem 3: Projection With Negative Result

Find the scalar projection of u = ⟨-2, 4⟩ onto v = ⟨1, 2⟩.

u · v = (-2)(1) + (4)(2) = -2 + 8 = 6

||v|| = √(1 + 4) = √5

comp_v(u) = 6 / √5 = (6√5) / 5 ≈ 2.68

Positive result. The vectors point generally in the same direction.

Problem 4: Negative Scalar Projection

Find the scalar projection of u = ⟨3, 2⟩ onto v = ⟨-1, 0⟩.

u · v = (3)(-1) + (2)(0) = -3

||v|| = 1

comp_v(u) = -3 / 1 = -3

The negative sign tells you the projection points opposite to v. This is why scalar projections can be negative — they're measuring direction, not just length.

Problem 5: 3D Projection

Find the vector projection of u = ⟨1, 2, 2⟩ onto v = ⟨0, 3, 4⟩.

u · v = (1)(0) + (2)(3) + (2)(4) = 0 + 6 + 8 = 14

v · v = 0 + 9 + 16 = 25

proj_v(u) = (14/25) * ⟨0, 3, 4⟩ = ⟨0, 42/25, 56/25⟩ = ⟨0, 1.68, 2.24⟩

3D problems work the same way. More coordinates, same process.

Scalar vs. Vector Projection: When to Use Which

Use this table to decide:

Common Mistakes That Will Cost You Points

• Forgetting to find ||v|| in the scalar formula. The denominator is the magnitude of v, not just 1.
• Using the wrong dot product in the vector formula. Some students use ||v||² instead of v · v. They're the same thing, so it's fine — but v · v is faster to compute.
• Getting a negative result wrong. A negative scalar projection isn't an error. It means the vectors point in roughly opposite directions.
• Confusing u onto v with v onto u. The order matters. u onto v uses u as the source and v as the target direction.

Quick Reference

Scalar projection:

compv(u) = (u · v) / ||v||

Vector projection:

projv(u) = [(u · v) / (v · v)] * v

Remember:

• Scalar projection = the length of the vector projection
• Vector projection = scalar projection × unit vector in direction of v
• If v is a unit vector, the formulas simplify: scalar = u · v, vector = (u · v)v

Practice these problems until you can do them without looking at the formulas. The test won't let you use notes.