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Vector Algebra and Analytic Geometry of Space

The Scalar Product of Vectors (the Dot Product)

The Definition of the Scalar Product

Scalar product

Let a⃗=(a1,a2,a3)\vec{a} = (a_1, a_2, a_3) and b⃗=(b1,b2,b3)\vec{b} = (b_1, b_2, b_3). The scalar product is defined as:

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

The result of the scalar product is a scalar, and it is also called the dot product.

The Geometric Meaning

The geometric meaning of the scalar product
a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta

where θ\theta is the angle between the two vectors.

A Geometric Explanation

  • when θ=0∘\theta = 0^\circ, cos⁡θ=1\cos\theta = 1 and a⃗⋅b⃗=∣a⃗∣∣b⃗∣\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|
  • when θ=90∘\theta = 90^\circ, cos⁡θ=0\cos\theta = 0 and a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0
  • when θ=180∘\theta = 180^\circ, cos⁡θ=−1\cos\theta = -1 and a⃗⋅b⃗=−∣a⃗∣∣b⃗∣\vec{a} \cdot \vec{b} = -|\vec{a}| |\vec{b}|

Properties

The basic properties of the scalar product

The commutative law: a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

Distributivity: a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}

Associativity with scalar multiplication: (ka⃗)⋅b⃗=k(a⃗⋅b⃗)(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b})

The inner product of a vector with itself: a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2

Applications

Testing for Perpendicularity

The condition for two vectors to be perpendicular
a⃗⊥b⃗⇔a⃗⋅b⃗=0\vec{a} \perp \vec{b} \Leftrightarrow \vec{a} \cdot \vec{b} = 0

Computing an Angle

The formula for the angle between two vectors
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}

Vector Projection

Vector projection

The projection of the vector a⃗\vec{a} in the direction of b⃗\vec{b} is:

a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

and the projection in the direction of the unit vector is:

a⃗⋅b⃗∣b⃗∣2b⃗\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}

Exercises

Exercise 1

Given a⃗=(1,2,3)\vec{a} = (1, 2, 3) and b⃗=(4,0,−1)\vec{b} = (4, 0, -1), find a⃗⋅b⃗\vec{a} \cdot \vec{b}.

the scalar productcoordinate computationvector operations

Approach:

Use the coordinate formula for the scalar product.

Steps:

  1. a⃗⋅b⃗=1×4+2×0+3×(−1)\vec{a} \cdot \vec{b} = 1 \times 4 + 2 \times 0 + 3 \times (-1)
  2. =4+0−3=1= 4 + 0 - 3 = 1

Answer: a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1

Exercise 2

Given a⃗=(1,2,3)\vec{a} = (1, 2, 3) and b⃗=(4,0,−1)\vec{b} = (4, 0, -1), find the angle between the two vectors.

the angle between vectorsthe scalar productgeometric meaning

Approach:

Use the scalar product to compute the angle.

Steps:

  1. ∣a⃗∣=12+22+32=14|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}
  2. ∣b⃗∣=42+02+(−1)2=17|\vec{b}| = \sqrt{4^2 + 0^2 + (-1)^2} = \sqrt{17}
  3. a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1
  4. cos⁡θ=114×17=1238\cos\theta = \frac{1}{\sqrt{14} \times \sqrt{17}} = \frac{1}{\sqrt{238}}
  5. θ=arccos⁡1238≈80.2∘\theta = \arccos\frac{1}{\sqrt{238}} \approx 80.2^\circ

Answer: the angle is approximately 80.2∘80.2^\circ

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
a⃗⋅b⃗\vec{a} \cdot \vec{b}Scalar productdot productThe scalar product of the vectors a\mathbf{a} and b\mathbf{b}
cos⁡θ\cos\thetaCosinecosine thetaThe cosine of the angle
⊥\perpPerpendicular symbolperpendicularThe relation of perpendicularity
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