Advanced Partial Derivatives: Background You’ll Need 2

  • Calculate the dot product of two given vectors

The Dot Product

The dot product is a new way to “multiply” vectors, but unlike vector addition and scalar multiplication, the result is a scalar (a number), not a vector.

dot product

For vectors [latex]\mathbf{u} = \langle u_1, u_2, u_3 \rangle[/latex] and [latex]\mathbf{v} = \langle v_1, v_2, v_3 \rangle[/latex]:

[latex]\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[/latex]

For 2D vectors [latex]\mathbf{u} = \langle u_1, u_2 \rangle[/latex] and [latex]\mathbf{v} = \langle v_1, v_2 \rangle[/latex]:

[latex]\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2[/latex]

The dot product is also called the scalar product (because the result is a scalar) or the inner product.

Multiply corresponding components and add the results. For [latex]\langle 2,3,1 \rangle \cdot \langle 4,-1,5 \rangle[/latex]:

[latex](2)(4) + (3)(-1) + (1)(5) = 8 - 3 + 5 = 10[/latex]
  1. Find the dot product of [latex]\mathbf{u} = \langle 3,5,2 \rangle[/latex] and [latex]\mathbf{v} =\langle -1,3,0 \rangle[/latex].
  2. Find the scalar product of [latex]\mathbf{p} = 10\mathbf{i} - 4\mathbf{j} +7 \mathbf{k}[/latex] and [latex]\mathbf{q} = -2\mathbf{i} + \mathbf{j} + 6\mathbf{k}[/latex].

Properties of the Dot Product

The dot product follows familiar algebraic rules, making it easy to work with in calculations.

properties of the dot product

Let [latex]\mathbf{u}[/latex], [latex]\mathbf{v}[/latex], and [latex]\mathbf{w}[/latex] be vectors, and let [latex]c[/latex] be a scalar.

  1. Commutative: [latex]\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}[/latex]
  2. Distributive: [latex]\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}[/latex]
  3. Associative with scalars: [latex]c(\mathbf{u} \cdot \mathbf{v}) = (c\mathbf{u}) \cdot \mathbf{v} = \mathbf{u} \cdot (c\mathbf{v})[/latex]
  4. Magnitude relationship: [latex]\mathbf{v} \cdot \mathbf{v} = ||\mathbf{v}||^2[/latex]

From the definition and properties, we can make several key observations:

  • [latex]\mathbf{0} \cdot \mathbf{v} = 0[/latex] for any vector [latex]\mathbf{v}[/latex]
  • If [latex]\mathbf{v} \cdot \mathbf{v} = 0[/latex], then [latex]\mathbf{v} = \mathbf{0}[/latex] (only the zero vector has dot product zero with itself)
  • The dot product of a vector with itself gives the square of its magnitude
Remember that [latex]||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2}[/latex], so [latex]||\mathbf{v}||^2 = v_1^2 + v_2^2 + v_3^2 = \mathbf{v} \cdot \mathbf{v}[/latex].

Let [latex]\mathbf{a} = \langle 1,2,-3 \rangle[/latex], [latex]\mathbf{b} = \langle 0,2,4 \rangle[/latex], and [latex]\mathbf{c} = \langle 5,-1,3 \rangle[/latex]. Find each of the following products.

  1. [latex](\mathbf{a} \cdot \mathbf{b})\mathbf{c}[/latex]
  2. [latex]\mathbf{a}\cdot (2\mathbf{c} )[/latex]
  3. [latex]||\mathbf{b}||^2[/latex]