Dot Products and Projections

Develop dot product and orthogonality, vector projection.

Lesson 4 of 42 parts Practice Flashcards

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Dot Product and Orthogonality

Objective: Explain dot product and orthogonality, apply the rule, and justify relevant restrictions.

For u=⟨u₁,u₂⟩ and v=⟨v₁,v₂⟩, u·v=u₁v₁+u₂v₂=|u||v| cos θ. The result is a scalar. For two nonzero vectors, zero dot product means they are perpendicular. To find an angle, divide by the product of nonzero magnitudes and apply principal arccos. A dot product can be negative without either magnitude being negative.

Formulas

$u·v=u₁v₁+u₂v₂=|u||v| cos θ$

Worked examples

Try itFor u=⟨3,7⟩ and v=⟨5,−1⟩, find u·v.Show answer

Step 1: Multiply corresponding components and add. Step 2: Compute (3)(5)+(7)(−1). Step 3: The dot product is 8.

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Dot Product and Orthogonality

Objective: Explain dot product and orthogonality, apply the rule, and justify relevant restrictions.

Formulas

$u·v=u₁v₁+u₂v₂=|u||v| cos θ$

Worked examples

Try itFor u=⟨3,7⟩ and v=⟨5,−1⟩, find u·v.Show answer

Step 1: Multiply corresponding components and add. Step 2: Compute (3)(5)+(7)(−1). Step 3: The dot product is 8.

Vector Projection

Objective: Explain vector projection, apply the rule, and justify relevant restrictions.

For v≠0, the vector projection of u onto v is ((u·v)/(v·v))v. It isolates the component of u parallel to the line in direction v. The remainder u−projᵥu is perpendicular to v. Vector projection is distinct from the scalar component (u·v)/|v|. A projection can point opposite v when the scalar coefficient is negative.

Formulas

$projᵥu=((u·v)/(v·v))v$

Worked examples

Try itFor u=⟨3,7⟩ and v=⟨5,0⟩ with 5>0, find the x-component of projᵥ u.Show answer

Step 1: Use projᵥ u=((u·v)/(v·v))v. Step 2: The scalar is (15)/(25)=3/5. Step 3: The projection is ⟨3,0⟩, so its x-component is 3.

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