Vector Operations in Space: Learn It 1

  • Calculate the dot product of two given vectors
  • Find angles between vectors and determine when vectors are perpendicular using dot products
  • Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it
  • Calculate the work done by a given force

The Dot Product

When you apply a force to move an object, you do work on that object. In earlier calculus courses, you calculated work as force times distance—but only when the force acted in the same direction as the motion.

In reality, force and motion are vectors that often point in different directions. The dot product allows us to calculate work when the force vector and motion vector have different directions. It tells us how much of the force vector acts in the direction of the motion vector.

Beyond calculating work, the dot product helps us measure angles between vectors, determine if vectors are perpendicular, and analyze how vectors relate to coordinate axes.

What is 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]

Proof

Let [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]. Then

[latex]\begin{align*} \mathbf{u \cdot v} &= \langle u_1, u_2, u_3 \rangle \cdot \langle v_1, v_2, v_3 \rangle\\ &= u_1v_1 + u_2v_2 + u_3v_3\\ &= v_1u_1 + v_2u_2 + v_3u_3\\ &= \langle v_1, v_2, v_3 \rangle \cdot \langle u_1, u_2, u_3 \rangle\\ &= \mathbf{v \cdot u}.\\ \end{align*}[/latex]

The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:

[latex]\begin{align*} c (\mathbf{u \cdot v}) &= c(u_1v_1 + u_2v_2 + u_3v_3)\\ &= c(u_1v_1) + c(u_2v_2) + c(u_3v_3)\\ &= (cu_1)v_1 + (cu_2)v_2 + (cu_3)v_3\\ &= \langle cu_1, cu_2, cu_3 \rangle \cdot \langle v_1, v_2, v_3 \rangle\\ &= c\langle u_1, u_2,u_3 \rangle \cdot \langle v_1, v_2, v_3 \rangle\\ &= (c\mathbf{u}) \cdot \mathbf{v}.\\ \end{align*}[/latex]

The proof that [latex]c (\mathbf{u \cdot v}) = \mathbf{u}\cdot (c\mathbf{v})[/latex] is similar.

The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:

[latex]\begin{align*} \mathbf{v} \cdot \mathbf{v} &= \langle v_1, v_2, v_3 \rangle \cdot \langle v_1, v_2, v_3 \rangle\\ &= (v_1)^2 + (v_2)^2 + (v_3)^2\\ &= [\sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}]^2\\ &= ||\mathbf{v}||^2.\\ \end{align*}[/latex]

[latex]_\blacksquare[/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]