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Vector Projections

Formulas & Definitions
Core Definition

Vector projection decomposes a vector u⃗\vec{u} into two perpendicular components: one that lies parallel to a reference vector v⃗\vec{v}, and one that lies orthogonal to v⃗\vec{v}.

  1. Scalar Projection (Component of u⃗\vec{u} along v⃗\vec{v}): A scalar number representing the signed magnitude of the shadow of u⃗\vec{u} on v⃗\vec{v}:
    compv⃗u⃗=u⃗⋅v⃗∣v⃗∣=∣u⃗∣cos⁡θ\text{comp}_{\vec{v}}\vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|} = |\vec{u}|\cos\theta
  2. Vector Projection (projv⃗u⃗\text{proj}_{\vec{v}}\vec{u}): A vector pointing along the line of v⃗\vec{v} with magnitude equal to the scalar projection:
    projv⃗u⃗=(compv⃗u⃗)v⃗∣v⃗∣=(u⃗⋅v⃗∣v⃗∣2)v⃗\text{proj}_{\vec{v}}\vec{u} = \left(\text{comp}_{\vec{v}}\vec{u}\right) \frac{\vec{v}}{|\vec{v}|} = \left(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v}
  3. Orthogonal Component (u⃗⊥\vec{u}_{\perp}): The part of u⃗\vec{u} perpendicular to v⃗\vec{v}:
    u⃗⊥=u⃗−projv⃗u⃗\vec{u}_{\perp} = \vec{u} - \text{proj}_{\vec{v}}\vec{u}

Concept Explanation

In engineering modeling, vectors rarely align perfectly with standard Cartesian x,y,zx, y, z coordinate axes. Vector projections allow engineers to project forces, velocities, or displacements onto structural members, inclined ramps, or flow streamlines.

Scalar Projection vs. Vector Projection
  • compv⃗u⃗\text{comp}_{\vec{v}}\vec{u} is a SCALAR: It is a single number (can be positive, zero, or negative depending on cos⁡θ\cos\theta).
  • projv⃗u⃗\text{proj}_{\vec{v}}\vec{u} is a VECTOR: It is the scalar projection multiplied by the unit vector in the direction of v⃗\vec{v}.

Engineering Applications

Example Problems

Example 1: Calculating Projections

Let u⃗=⟨1,2,3⟩\vec{u} = \langle 1, 2, 3 \rangle and v⃗=⟨−2,0,4⟩\vec{v} = \langle -2, 0, 4 \rangle. Compute the scalar projection compv⃗u⃗\text{comp}_{\vec{v}}\vec{u} and the vector projection projv⃗u⃗\text{proj}_{\vec{v}}\vec{u}.

Solution
  1. Calculate Dot Product and Magnitude Squared:
    u⃗⋅v⃗=(1)(−2)+(2)(0)+(3)(4)=−2+0+12=10\vec{u} \cdot \vec{v} = (1)(-2) + (2)(0) + (3)(4) = -2 + 0 + 12 = 10
    ∣v⃗∣2=(−2)2+02+42=4+0+16=20  ⟹  ∣v⃗∣=20=25|\vec{v}|^2 = (-2)^2 + 0^2 + 4^2 = 4 + 0 + 16 = 20 \implies |\vec{v}| = \sqrt{20} = 2\sqrt{5}
  2. Calculate Scalar Projection (Scalar):
    compv⃗u⃗=1020=1025=5≈2.236\text{comp}_{\vec{v}}\vec{u} = \frac{10}{\sqrt{20}} = \frac{10}{2\sqrt{5}} = \sqrt{5} \approx 2.236
  3. Calculate Vector Projection (Vector):
    projv⃗u⃗=(1020)⟨−2,0,4⟩=12⟨−2,0,4⟩=⟨−1,0,2⟩\text{proj}_{\vec{v}}\vec{u} = \left(\frac{10}{20}\right) \langle -2, 0, 4 \rangle = \frac{1}{2} \langle -2, 0, 4 \rangle = \langle -1, 0, 2 \rangle

Connections & References