Scalar (or Dot) Product of Two Vectors

We have already studied about the addition and subtraction of vectors. Vectors can be multiplied in two ways, scalar or dot product where the result is a scalar and vector or cross product where is the result is a vector. In this article, we will look at the scalar or dot product of two vectors.

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Definition

The scalar or dot product of two non-zero vectors \( \vec{a} \) and \( \vec{b} \), denoted by \( \vec{a} \).\( \vec{b} \) is

\( \vec{a} \).\( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)| \( \cos \theta \)

where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \) and 0 ≤ \( \theta \) ≤ \( \pi \) as shown in the figure below.

It is important to note that if either \( \vec{a} \) = \( \vec{0} \) or \( \vec{b} \) = \( \vec{0} \), then \( \theta \) is not defined, and in this case

\( \vec{a} \).\( \vec{b} \) = 0

Browse more Topics under Vector Algebra

• Basic Concepts of Vectors
• Components of a Vector
• Types of Vectors
• Addition of Vectors
• Vector (or Cross) Product of Two Vectors
• Section Formula
• Projection of a Vector on a Line

Important Observations

• \( \vec{a} \).\( \vec{b} \) is a real number
• If \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors, then \( \vec{a} \).\( \vec{b} \) = 0 if and only if \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other.

\( \vec{a} \).\( \vec{b} \) = 0 \( \Longleftrightarrow \) \( \vec{a} \) ⊥ \( \vec{b} \)

• If \( \theta \) = 0, then \( \vec{a} \).\( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)|. In particular,

\( \vec{a} \).\( \vec{a} \) = \( \vec{|a|}^2 \), as in this case \( \theta \) is 0.

• If \( \theta \) = \( \pi \), then \( \vec{a} \).\( \vec{b} \) = – |\( \vec{a} \)| |\( \vec{b} \)|. In particular,

\( \vec{a} \).\( \vec{(- a)} \) = – \( \vec{|a|}^2 \), as in this case \( \theta \) is \( \pi \).

• Following from observations 2 and 3, for mutually perpendicular vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \), we have

\( \hat{i} \).\( \hat{i} \) = \( \hat{j} \).\( \hat{j} \) = \( \hat{k} \).\( \hat{k} \) = 1

And, \( \hat{i} \).\( \hat{j} \) = \( \hat{j} \).\( \hat{k} \) = \( \hat{k} \).\( \hat{i} \) = 0

• The angle between two non-zero vectors is given by,

\( \cos \theta \) = \( \frac{\vec{a}.\vec{b}}{|\vec{a}| |\vec{b}|} \)

Or, \( \theta \) =  cos^-1 [\( \frac{\vec{a}.\vec{b}}{|\vec{a}| |\vec{b}|} \)]

• The scalar product is commutative, i.e.

\( \vec{a} \).\( \vec{b} \) = \( \vec{b} \).\( \vec{a} \)

Two Important Properties of Dot Product

Property 1: Distributivity of a scalar or dot product over addition

If \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) are any three vectors, then

\( \vec{a} \).(\( \vec{b} \) + \( \vec{c} \)) = \( \vec{a} \).\( \vec{b} \) + \( \vec{a} \).\( \vec{c} \)

Property 2:

If \( \vec{a} \) and \( \vec{b} \) are any two vectors and \( \lambda \) is a scalar, then

(\( \lambda \)\( \vec{a} \)).\( \vec{b} \) = \( \lambda \)(\( \vec{a} \).\( \vec{b} \)) = \( \vec{a} \).(\( \lambda \)\( \vec{b} \))

If two vectors \( \vec{a} \) and \( \vec{b} \) are given in the component form as a₁\( \vec{i} \) + a₂\( \vec{j} \) + a₃\( \vec{k} \) and b₁\( \vec{i} \) + b₂\( \vec{j} \) + b₃\( \vec{k} \), then their scalar or dot product is as follows:

\( \vec{a} \).\( \vec{b} \) = (a₁\( \vec{i} \) + a₂\( \vec{j} \) + a₃\( \vec{k} \)).(b₁\( \vec{i} \) + b₂\( \vec{j} \) + b₃\( \vec{k} \))

Or, \( \vec{a} \).\( \vec{b} \) = a₁\( \vec{i} \)(b₁\( \vec{i} \) + b₂\( \vec{j} \) + b₃\( \vec{k} \)) + a₂\( \vec{j} \)(b₁\( \vec{i} \) + b₂\( \vec{j} \) + b₃\( \vec{k} \)) + a₃\( \vec{k} \)(b₁\( \vec{i} \) + b₂\( \vec{j} \) + b₃\( \vec{k} \))

On solving further, and using the properties 1 and 2 above, we get

\( \vec{a} \).\( \vec{b} \) = a₁b₁(\( \vec{i} \).\( \vec{i} \)) + a₁b₂(\( \vec{i} \).\( \vec{j} \)) + a₁b₃(\( \vec{i} \).\( \vec{k} \)) + a₂b₁(\( \vec{j} \).\( \vec{i} \)) + a₂b₂(\( \vec{j} \).\( \vec{j} \)) + a₂b₃(\( \vec{j} \).\( \vec{k} \)) + a₃b₁(\( \vec{k} \).\( \vec{i} \)) + a₃b₂(\( \vec{k} \).\( \vec{j} \)) + a₃b₃(\( \vec{k} \).\( \vec{k} \))

Using Observation 5 mentioned above, we get

\( \vec{a} \).\( \vec{b} \) = a₁b₁ + a₂b₂ + a₃b₃

Solved Problems for You

Question: Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes \( \sqrt{3} \) and 2 ,
respectively having \( \vec{a} \).\( \vec{b} \) = \( \sqrt{6} \).

Solution: By definition of the scalar or dot product of vectors, we know that

\( \vec{a} \).\( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)| \( \cos \theta \)

where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). In this question, we have,

|\( \vec{a} \)| = \( \sqrt{3} \), |\( \vec{b} \)| = 2, and \( \vec{a} \).\( \vec{b} \) = \( \sqrt{6} \)

Replacing these values in the formula, we get

\( \vec{a} \).\( \vec{b} \) = |\( \vec{a} \)| |\( \vec{b} \)| \( \cos \theta \)
\( \Longleftrightarrow \) \( \sqrt{6} \) = \( \sqrt{3} \) x 2 x \( \cos \theta \)
\( \Longleftrightarrow \) \( \sqrt{3} \) x \( \sqrt{2} \) = \( \sqrt{3} \) x \( \sqrt{2} \) x \( \sqrt{2} \) x \( \cos \theta \)

Cancelling the common terms on both sides, we get

1 = \( \sqrt{2} \) x \( \cos \theta \)
Or, \( \cos \theta \) = \( \frac{1}{ \sqrt{2}} \)
Therefore, \( \theta \) = \( \frac{\pi}{4} \)
Hence, the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \).