Vector Dot Product

The vector dot product is also called a scalar product because the product of vectors gives a scalar quantity. Sometimes, a dot product is also named as an inner product. In vector algebra, the dot product is an operation applied to vectors.

The scalar product or dot product is commutative. When two vectors are operated under a dot product, the answer is only a number. A brief explanation of dot products is given below.

Dot Product of Two Vectors

If we have two vectors, a = a_x+a_y and b = b_x+b_y, then the dot product or scalar product between them is defined as

a.b = a_xb_x+ a_yb_y

The Formula for Vectors Dot Product

\begin{array}{l} Let \vec{A} and \vec{B} be two non zero vectors. \end{array}

Then, the scalar product is denoted by

\begin{array}{l} \vec{A} . \vec{B} \end{array}

and is defined as the scalar;

\begin{array}{l} | \vec{A} | | \vec{B} | \cos θ \end{array}

\begin{array}{l} \vec{A} . \vec{B} = | \vec{A} | | \vec{B} | \cos θ = A B \cos θ \end{array}

Double Dot Product

To perform a double inner product on vectors, consider the result of the single inner product, then repeat the process of setting the nearest pair of indices equal and summing over them. Only tensors of 2nd rank or higher can participate in double inner products.

Properties

\begin{array}{l} Let \vec{A}, \vec{B} and \vec{C} be three non zero vectors. \end{array}

Commutative Property

The scalar product of two vectors is commutative, i.e.,

\begin{array}{l} \vec{A} . \vec{B} = \vec{B} . \vec{A} \end{array}

Distributive Property

The scalar product is distributive over addition.

\begin{array}{l} \vec{A} . (\vec{B} + \vec{C}) = \vec{A} . \vec{B} + \vec{A} . \vec{C} \end{array}

Scalar Property

\begin{array}{l} m \vec{A} . n \vec{B} = m n (\vec{A} . \vec{B}) = (m n \vec{A}) . \vec{B} = \vec{A} . (m n \vec{B}) \end{array}

Where m and n are scalars.

Not associative because the dot product between a scalar (a ⋅ b) and a vector (c) is not defined, which means that the expressions involved in the associative property, (a ⋅ b) ⋅ c or a ⋅ (b ⋅ c), are both ill-defined. Note that the previously mentioned scalar multiplication property is sometimes called the “associative law for scalar and dot product”, or one can say that “the dot product is associative with respect to scalar multiplication” because c (a ⋅ b) = (c a) ⋅ b = a ⋅ (c b).

Orthogonal

Two non-zero vectors, a and b, are orthogonal if and only if a ⋅ b = 0.

No Cancellation

Unlike the multiplication of ordinary numbers, where if ab = ac, then b always equals c unless a is zero, the dot product does not obey the cancellation law.

If a ⋅ b = a ⋅ c and a ≠ 0, then we can write a ⋅ (b − c) = 0 by the distributive law; the result above says that a is perpendicular to (b − c), which still allows (b − c) ≠ 0, and therefore, b ≠ c.

Product Rule

If a and b are functions, then the derivative (denoted by a prime ′) of a ⋅ b is a′ ⋅ b + a ⋅ b′.

Bilinear

\begin{array}{l} a \cdot (r b + c) = r (a \cdot b) + (a \cdot c) . \end{array}

Dot Product of Two Parallel Vectors

If two vectors have the same direction or two vectors are parallel to each other, then the dot product of two vectors is the product of their magnitude.

Here, θ = 0 degree

so, cos 0 = 1

Therefore,

\begin{array}{l} \vec{A} . \vec{B} = | \vec{A} | | \vec{B} | \cos θ \end{array}

= AB

Dot Product of Opposite Vectors

If the two vectors are opposite in direction, then

\begin{array}{l} θ = π \end{array}

Here,

\begin{array}{l} \cos π = - 1 \end{array}

Now,

\begin{array}{l} \vec{A} . \vec{B} = | \vec{A} | | \vec{B} | \cos π \end{array}

= – AB

Derivative of Dot Product

If we have A(x) = A₁(x), A₂(x), …., A_n(x) and B(x) = B₁(x), B₂(x), …., B_n(x) are two vectors, then the derivative of the dot product is given by

\begin{array}{l} \frac{d}{d x} [A (x) . B (x)] = A^{'} (x) . B (x) + A (x) . B^{'} (x) \end{array}

Proof

We can prove this with the help of the definition of dot product and product of real number property as

\begin{array}{l} \frac{d}{d x} [A (x) . B (x)] = \frac{d}{d x} [\sum_{j = 1}^{n} A_{j} (x) B_{j} (x)] \end{array}

\begin{array}{l} = \sum_{j = 1}^{n} \frac{d}{d x} [A_{j} (x) B_{j} (x)] \end{array}

\begin{array}{l} = \sum_{j = 1}^{n} A_{j}^{'} (x) B_{j} (x) + \sum_{j = 1}^{n} A_{j} (x) B_{j}^{'} (x) \end{array}

\begin{array}{l} = A^{'} (x) . B (x) + A (x) . B^{'} (x) \end{array}

Hence proved.

Geometric Interpretation of Dot Product

If we have two vectors, A and B, and let θ be the angle between them, then the dot product is given by

\begin{array}{l} A . B = | A | . | B | \cos θ \end{array}

We can prove this result with the help of the cosine formula, as explained below.

We know that the formula for the dot product of two vectors, A and B, is
A.B = A₁B₁ + A₂B₂ + A₃B₃.

Proof

For this, let A and B be two non-zero vectors. Then, A, B and A – B vectors form a triangle. The length of the triangle is IAI, IBI and IA – BI.

Then, by the use of the cosine formula, we get

\begin{array}{l} {| A - B |}^{2} = {| A |}^{2} + {| B |}^{2} - 2 | A | | B | \cos θ \end{array}

Now,

\begin{array}{l} {| A - B |}^{2} = {| A |}^{2} + {| B |}^{2} - 2 A . B \dots . . (i) \end{array}

Since we know that,

\begin{array}{l} {| A - B |}^{2} = (A_{1} - B_{1})^{2} + (A_{2} - B_{2})^{2} + (A_{3} - B_{3})^{2} \end{array}

\begin{array}{l} = A_{1}^{2} - 2 A_{1} B_{1} + B_{1}^{2} + A_{2}^{2} - 2 A_{2} B_{2} + B_{2}^{2} + A_{3}^{2} - 2 A_{3} B_{3} + B_{3}^{2} \end{array}

\begin{array}{l} = (A_{1}^{2} + A_{2}^{2} + A_{3}^{2}) + (B_{1}^{2} + B_{2}^{2} + B_{3}^{2}) - 2 (A_{1} B_{1} + A_{2} B_{2} + A_{3} B_{3}) \end{array}

\begin{array}{l} = {| A |}^{2} + {| B |}^{2} - 2 (A_{1} B_{1} + A_{2} B_{2} + A_{3} B_{3}) \end{array}

Therefore, we have form equation (i),

\begin{array}{l} {| A |}^{2} + {| B |}^{2} - 2 (A_{1} B_{1} + A_{2} B_{2} + A_{3} B_{3}) = {| A - B |}^{2} = {| A |}^{2} + {| B |}^{2} - 2 A . B \end{array}

Then,

\begin{array}{l} A . B = A_{1} B_{1} + A_{2} B_{2} + A_{3} B_{3} \end{array}

Cross Product

Scalar Triple Product

Solved Examples on Vector Dot Product

Example 1: If

\begin{array}{l} \vec{a} = \hat{j} - \hat{k} a n d \vec{c} = \hat{i} - \hat{j} - \hat{k}, \end{array}

\begin{array}{l} then vector \vec{b} satisfying \vec{a} \times \vec{b} + \vec{c} = 0 and \vec{a} \cdot \vec{b} = 3 is \end{array}

Solution:

Unit vector perpendicular to the plane of vectors a and b –

\begin{array}{l} \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |} \\ Scalar Product of two vectors (dot product) - \\ \vec{a} \vec{b} = | a | | b | C o s θ \\ where θ is the angle between the vectors \vec{a} and \vec{b} \\ \vec{c} = \vec{b} \times \vec{a} => \vec{b} . \vec{c} = 0 \\ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . (\hat{i} - \hat{j} - \hat{k}) = 0 \\ b_{1} - b_{2} - b_{3} = 0 \\ \vec{a} . \vec{b} = 3 \\ b_{2} - b_{3} = 3 \\ b_{1} = b_{2} + b_{3} = 3 + 2 b_{3} \\ b_{3} = - 2 \\ \vec{b} = (3 + 2 b_{3}) \hat{i} + (3 + b_{3}) \hat{j} + b_{3} \hat{k} \\ So \vec{b} = - \hat{i} + \hat{j} - 2 \hat{k} \end{array}

Example 2: Find the angle between the vectors A and B where

\begin{array}{l} \vec{A} = 2 i + 3 j + 3 k, \vec{B} = i + 2 j - 3 k . \end{array}

Solution:

We know that

\begin{array}{l} \vec{A} . \vec{B} = | A | | B | c o s θ where | A | = \sqrt (22 + 32 + 32) = \sqrt 22 a n d | B | = \sqrt (12 + 22 + 32) = \sqrt 14 \\ c o s θ = (\vec{A} . \vec{B}) / (| A | | B |) \\ = ((2 i + 3 j + 3 k) (i + 2 j - 3 k)) / (\sqrt 22 \times \sqrt 14) \\ = (2 + 6 - 9) / (2 \sqrt 77) = (- 1) / (2 \sqrt 77) \\ θ = c o s - 1 ((- 1) / (2 \sqrt 77)) \end{array}

Example 3: Find the angle between two vectors

\begin{array}{l} \vec{a} and \vec{b} with magnitude 1 and 2 respectively and \vec{a} . \vec{b} = 1. \end{array}

Solution:

We have

\begin{array}{l} | \vec{a} | = 1 \end{array}

\begin{array}{l} | \vec{b} | = 2 \end{array}

\begin{array}{l} \vec{a} . \vec{b} = 1 \end{array}

Let θ be the angle between a and b.

Then

\begin{array}{l} c o s θ = \frac{\vec{a} . \vec{b}}{| \vec{a} | | \vec{b} |} \end{array}

⇒ cos θ = 1/(2×1) = ½

⇒ θ = π/3 .

Hence, the required angle is π/3.

Example 4.

\begin{array}{l} Find \vec{a} . \vec{b} when \vec{a} = 3 \hat{i} + 4 \hat{j} + \hat{k} and \vec{b} = 5 \hat{i} - \hat{j} + 2 \hat{k} . \end{array}

Solution:

We have

\begin{array}{l} \vec{a} = 3 \hat{i} + 4 \hat{j} + \hat{k} \end{array}

\begin{array}{l} \vec{b} = 5 \hat{i} - \hat{j} + 2 \hat{k} \end{array}

\begin{array}{l} ∴ \vec{a} . \vec{b} = 3 \times 5 + 4 \times (- 1) + 1 \times 2 \end{array}

= 15 – 4 + 2

= 13

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Frequently Asked Questions

Give the formula for the dot product of two vectors.

If A and B are two vectors, then the dot product is given by A.B = AB cos θ.

Is the dot product commutative?

Yes. The dot product of two vectors is commutative.

Can a dot product be equal to zero?

Yes. The dot product of perpendicular vectors is equal to zero.

What is the dot product of two parallel vectors?

The dot product of two parallel vectors is equal to the product of the magnitude of the two vectors.

Dot Product of Two Vectors

The Formula for Vectors Dot Product

Double Dot Product

Properties

Dot Product of Two Parallel Vectors

Dot Product of Opposite Vectors

Derivative of Dot Product

Geometric Interpretation of Dot Product

Proof

Related Articles

Solved Examples on Vector Dot Product

Video Lessons

Vector and 3D – Important Questions

Vector and 3D – Important Topics

Frequently Asked Questions

Give the formula for the dot product of two vectors.

Is the dot product commutative?

Can a dot product be equal to zero?

What is the dot product of two parallel vectors?

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