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 = ax+ay and b = bx+by, then the dot product or scalar product between them is defined as
a.b = axbx + ayby
The Formula for Vectors Dot Product
\(\begin{array}{l}\text{Let}\ \vec A\ \text{and}\ \vec{B}\ \text{be two non zero vectors.}\end{array} \)
Then, the scalar product is denoted by \(\begin{array}{l}\vec A . \vec B\end{array} \)
and is defined as the scalar; \(\begin{array}{l}|\vec A||\vec B| \cos \theta\end{array} \)
\(\begin{array}{l}\vec A . \vec B = |\vec A||\vec B| \cos \theta = AB \cos \theta\end{array} \)
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
\(\begin{array}{l}\text{Let}\ \vec A, \vec B\ \text{and}\ \vec C\ \text{be three non zero vectors.}\end{array} \)
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
\(\begin{array}{l}m \vec A . n \vec B = mn(\vec A . \vec B) = (mn \vec A) . \vec B = \vec A . (mn \vec B)\end{array} \)
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
\(\begin{array}{l}{\displaystyle \mathbf {a} \cdot (r\mathbf {b} +\mathbf {c} )=r(\mathbf {a} \cdot \mathbf {b} )+(\mathbf {a} \cdot \mathbf {c} ).}\end{array} \)
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,
\(\begin{array}{l}\vec A . \vec B = |\vec A||\vec B| \cos \theta\end{array} \)
= AB
Dot Product of Opposite Vectors
If the two vectors are opposite in direction, then
\(\begin{array}{l}\theta = \pi\end{array} \)
Here,
\(\begin{array}{l}\cos \pi = -1\end{array} \)
Now,
\(\begin{array}{l}\vec A . \vec B = |\vec A||\vec B| \cos \pi\end{array} \)
= – AB
Derivative of Dot Product
If we have A(x) = A1(x), A2(x), …., An(x) and B(x) = B1(x), B2(x), …., Bn(x) are two vectors, then the derivative of the dot product is given by
\(\begin{array}{l}\frac{d}{dx}[A(x) . B(x)] = A'(x).B(x) + A(x).B'(x)\end{array} \)
Proof
We can prove this with the help of the definition of dot product and product of real number property as
\(\begin{array}{l}\frac{d}{dx}[A(x) . B(x)] = \frac{d}{dx}\left [ \sum_{j = 1}^{n}A_{j}(x)B_{j}(x) \right ]\end{array} \)
\(\begin{array}{l}=\sum_{j = 1}^{n}\frac{d}{dx}\left [A_{j}(x)B_{j}(x) \right ]\end{array} \)
\(\begin{array}{l}=\sum_{j = 1}^{n} A’_{j}(x)B_{j}(x) + \sum_{j = 1}^{n} A_{j}(x)B’_{j}(x)\end{array} \)
\(\begin{array}{l}=A'(x) . B(x) + A(x) . B'(x)\end{array} \)
.
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
\(\begin{array}{l}A . B = |A| . |B| \cos \theta\end{array} \)
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
\(\begin{array}{l}\left | A-B \right |^{2}= \left | A \right |^{2}+ \left | B \right |^{2}-2\left | A \right |\left | B \right | \cos \theta\end{array} \)
Now,
\(\begin{array}{l}\left | A – B \right |^{2}=\left | A \right |^{2} + \left | B \right |^{2} – 2 A.B…..(i)\end{array} \)
Since we know that,
\(\begin{array}{l}\left | A – B \right |^{2} = (A_{1} – B_{1})^{2} + (A_{2} – B_{2})^{2} + (A_{3} – B_{3})^{2}\end{array} \)
\(\begin{array}{l}=A_{1}^{2} – 2A_{1}B_{1} + B_{1}^{2} + A_{2}^{2} – 2A_{2}B_{2} + B_{2}^{2} + A_{3}^{2} – 2A_{3}B_{3} + B_{3}^{2}\end{array} \)
\(\begin{array}{l}=(A_{1}^{2} + A_{2}^{2} + A_{3}^{2}) + (B_{1}^{2} + B_{2}^{2} + B_{3}^{2}) – 2(A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3})\end{array} \)
\(\begin{array}{l}=\left | A \right |^{2} + \left | B \right |^{2} – 2(A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3})\end{array} \)
Therefore, we have form equation (i),
\(\begin{array}{l}\left | A \right |^{2}+\left | B \right |^{2}-2(A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3})= \left | A-B \right |^{2}= \left | A \right |^{2}+\left | B \right |^{2}- 2A.B\end{array} \)
Then,
\(\begin{array}{l}A.B = A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3}\end{array} \)
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Solved Examples on Vector Dot Product
Example 1: If
\(\begin{array}{l}\vec{a}=\hat{j}-\hat{k}\; and\; \vec{c}=\hat{i}-\hat{j}-\hat{k},\end{array} \)
\(\begin{array}{l}\text{then vector}\ \vec{b}\ \text{satisfying}\ \vec{a}\times \vec{b}+ \vec{c}=0\; \text{and}\ \vec{a}\cdot \vec{b}=3\ \text{is}\end{array} \)
Solution:
Unit vector perpendicular to the plane of vectors a and b –
\(\begin{array}{l}\frac{\vec{a}\times \vec{b}}{\left | \vec{a}\times \vec{b} \right |} \\ \text{Scalar Product of two vectors (dot product)} -\\ \vec{a}\vec{b}=\left | a \right |\left | b \right |Cos\theta\\ \text{where}\ \theta\ \text{is the angle between the vectors}\ \vec{a}\ \text{and}\ \vec{b}\\ \vec{c}=\vec{b}\times \vec{a}=> \vec{b}\:.\vec{c}=0\\ (b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}).(\hat{i}-\hat{j}-\hat{k})=0\\ b_{1}-b_{2}-b_{3}=0\\ \vec{a}\:.\:\vec{b}=3\\ b_{2}-b_{3}=3\\ b_{1}=b_{2}+b_{3}=3+2b_{3}\\ b_{3}=\:-2\\ \vec{b}=(3+2b_{3})\hat{i}+(3+b_{3})\hat{j}+b_{3}\hat{k}\\ \text{So}\ \vec{b}=-\hat{i}+\hat{j}-2\hat{k}\\\end{array} \)
Example 2: Find the angle between the vectors A and B where
\(\begin{array}{l}\vec{A} = 2i + 3j + 3k, \vec{B} = i + 2j – 3k.\end{array} \)
Solution:
We know that
\(\begin{array}{l}\vec{A}.\vec{B} = |A||B| cosθ\ \text{where}\ |A| = √(22 +32 + 32) = √22 \,and\ |B| =√(12 + 22+ 32) = √14 \\ cosθ = (\vec{A}.\vec{B})/(|A||B|)\\ =((2i +3j +3k )(i +2j -3k ))/(√22×√14)\\ =(2+6-9)/(2√77)=(-1)/(2√77) \\ θ = cos-1((-1)/(2√77)) \\\end{array} \)
Example 3: Find the angle between two vectors \(\begin{array}{l}\vec{a}\ \text{and}\ \vec{b}\ \text{with magnitude 1 and 2 respectively and}\ \vec{a}.\vec{b}=1.\end{array} \)
Solution:
We have \(\begin{array}{l}\left | \vec{a} \right |=1\end{array} \)
\(\begin{array}{l}\left | \vec{b} \right |=2\end{array} \)
\(\begin{array}{l}\vec{a}.\vec{b}=1\end{array} \)
Let θ be the angle between a and b.
Then \(\begin{array}{l}cos\ \theta = \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\end{array} \)
⇒ cos θ = 1/(2×1) = ½
⇒ θ = π/3 .
Hence, the required angle is π/3.
Example 4. \(\begin{array}{l}\text{Find}\ \vec{a}.\vec{b}\ \text{when}\ \vec{a}=3\hat{i}+4\hat{j}+\hat{k}\ \text{and}\ \vec{b} = 5\hat{i}-\hat{j}+2\hat{k}.\end{array} \)
Solution:
We have \(\begin{array}{l}\vec{a} = 3\hat{i}+4\hat{j}+\hat{k}\end{array} \)
\(\begin{array}{l}\vec{b} = 5\hat{i}-\hat{j}+2\hat{k}\end{array} \)
.
\(\begin{array}{l}\therefore \vec{a}.\vec{b}= 3\times 5 + 4\times (-1) + 1\times 2\end{array} \)
= 15 – 4 + 2
= 13
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Vector and 3D – Important Questions
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Frequently Asked Questions
Q1
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.
Q2
Is the dot product commutative?
Yes. The dot product of two vectors is commutative.
Q3
Can a dot product be equal to zero?
Yes. The dot product of perpendicular vectors is equal to zero.
Q4
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.
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