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Maths > Vector Algebra > Scalar (or Dot) Product of Two Vectors
Vector Algebra

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.

dot product

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

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Scalar (or Dot) Product of Two Vectors

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 \( \frac{\pi}{4} \).

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