Multiplication of a Vector

In mathematics and physics, vector multiplication is a key operation used to analyze motion, force, and direction. It involves multiplying a vector either by a scalar (a number) or another vector. The result depends on the type of multiplication: scalar multiplication stretches or shrinks the vector, while vector multiplication yields a new vector or a scalar based on the rule used.

1.0Multiplication of Vectors Definition

Multiplication of a vector refers to scaling a vector by a number (scalar) or combining it with another vector using dot product or cross product. The operation you use determines whether the result is a scalar or another vector.

2.0Multiplication of a Vector by a Scalar (Number)

When you perform the multiplication of a vector by a scalar, you are changing the magnitude of the vector but not its direction (unless the scalar is negative, which reverses the direction).

Multiplication of a Vector by a Number Formula

If $\vec{A}=a\hat{i}+b\hat{j}$, and k is a scalar, then:

$k\cdot \vec{A}=k(a\hat{i}+b\hat{j})=(ka)\hat{i}+(kb)\hat{j}$

This is also called scalar multiplication of a vector.

3.0How to Multiply Vector Components

To multiply a vector by a number:

1. Multiply each component of the vector by the scalar.
2. Keep the same direction (or reverse it if the scalar is negative).
3. The result is another vector.

Example:
Given $\vec{A}=2\hat{i}-3\hat{j}$ and scalar k = -2

$k\cdot \vec{A}=-2(2\hat{i}-3\hat{j})=-4\hat{i}+6\hat{j}$

This is multiplication of vector components by a scalar.

4.0What Happens When a Vector Is Multiplied by a Scalar?

• The magnitude of the vector changes.
• The direction remains the same if scalar is positive.
• The direction reverses if scalar is negative.
• The vector becomes a zero vector if multiplied by 0.

5.0Multiplication of Vectors by Another Vector

There are two types of vector-by-vector multiplication:

1. Dot Product (Scalar Product)

• Result: Scalar
• Formula:

$\vec{A}\cdot \vec{B}=|\vec{A}||\vec{B}|\cos (\theta )$

• Application: Work done, projection

2. Cross Product (Vector Product)

• Result: Vector
• Formula:

$\vec{A}\times \vec{B}=|\vec{A}||\vec{B}|\sin (\theta )\hat{n}$ 

where It is a unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.

• Application: Torque, rotational motion

6.0What Is the Rule for Vector Multiplication?

7.0Multiplication of Vectors Example

Example 1: Scalar Multiplication of $\vec{A}=3\hat{i}+4\hat{j},k=2$

Solution: 

$\vec{A}=2(3\hat{i}+4\hat{j})=6\hat{i}+8\hat{j}$

Example 2: Dot Product of $\vec{A}=⟨2,3⟩,\vec{B}=⟨4,-1⟩$ 

Solution: 

$\vec{A}\cdot \vec{B}=(2)(4)+(3)(-1)=8-3=5$

Example 3: Cross Product of $\vec{A}=\hat{i}+2\hat{j},\vec{B}=3\hat{i}+\hat{j}$

Solution: 

$\vec{A}\times \vec{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 0\\ 3& 1& 0\end{array}\right|=(0)\hat{i}-(0)\hat{j}+(1\cdot 1-2\cdot 3)\hat{k}=-5\hat{k}$

8.0JEE Advanced-Level Questions & Solutions on Multiplication of Vectors

Q1. Let $\vec{A}=\hat{i}+2\hat{j}+\hat{k}\vec{B}=2\hat{i}-\hat{j}+\hat{k}$, and $\vec{C}=a\hat{i}+b\hat{j}+c\hat{k}$. If the vectors $\vec{A},\vec{B},\vec{C}$ are coplanar, find the relation between a, b, c.

Solution: 

Use the scalar triple product condition:

$\vec{A}\cdot (\vec{B}\times \vec{C})=0$

Evaluate:

$\begin{array}{rl}& \vec{B}\otimes \vec{C}\otimes =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ a& b& c\end{array}\right|\\ & \vec{B}\times \vec{C}=((-1)(c)-(1)(b))\hat{i}-((2)(c)-(1)(a))\hat{j}+((2)(b)-(-1)(a))\hat{k}\\ & \vec{B}\times \vec{C}=(-c-b)\hat{i}-(2c-a)\hat{j}+(2b+a)\hat{k}\end{array}$

Now dot with $\vec{A}$:

=−5c + a + b(1)(-c - b) + 2(-2c + a) + 1(2b + a) 

= - c - b - 4c + 2a + 2b + a = -5c + a + b 

Set equal to 0:

a + b - 5c = 0 

Answer: ca + b = 5c

Q2. Let A and B be $\vec{A}=3\hat{i}+4\hat{j},\vec{B}=5\hat{i}$. Find the projection of $\vec{A}on\vec{B}$.

Solution:

$\text{ Projection of }\vec{A}\text{ on }\vec{B}=\frac{\vec{A}\cdot \vec{B}}{|\vec{B}B|}$

$\text{ Projection }=\frac{\vec{A}\cdot \vec{B}}{\overrightarrow{|B|}}=\frac{3(5)}{5}=3$

Answer: 3

Q3. Find angle between two vectors A and B. $\vec{A}=2\hat{i}+\hat{j}+2\hat{k},\vec{B}=\hat{i}-\hat{j}+\hat{k}$

Solution:

$\vec{A}\cdot \vec{B}=2(1)+1(-1)+2(1)=2-1+2=3$

$|\vec{A}|=\sqrt{4+1+4}=\sqrt{9}=3,|\vec{B}|=\sqrt{1+1+1}=\sqrt{3}$

Using formula:

$\cos \theta =\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}$

$\cos \theta =\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer: $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Q4. Find area of parallelogram using cross product of  $\vec{u}=3\hat{i}-\hat{j}+\hat{k},\vec{v}=\hat{i}+2\hat{j}-2\hat{k}.$

Solution:

$\text{ Area }=|\vec{u}\times \vec{v}|$

$\begin{array}{rl}& \vec{u}\times \vec{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ 1& 2& -2\end{array}\right|\\ & \Rightarrow \vec{u}\times \vec{v}=(-1)(-2)-(1)(2)\hat{i}-[3(-2)-(1)(1)]\hat{j}+[3(2)-(-1)(1)]\hat{k}\\ & \Rightarrow \vec{u}\times \vec{v}=(2-2)\hat{i}+(6-1)\hat{j}+(6+1)\hat{k}\\ & \Rightarrow |\vec{u}\times \vec{v}|=0\hat{i}+5\hat{j}+7\hat{k}\\ & \Rightarrow |\vec{u}\times \vec{v}|=\sqrt{0^2+25+49}=\sqrt{74}\end{array}$ 

Area: $\sqrt{74}$

Q5. Find shortest distance between:

$\begin{array}{rl}& L_1:\vec{r}=\hat{i}+2\hat{j}+\hat{k}+\lambda (\hat{i}+\hat{j}),\\ & L_2:\vec{r}=2\hat{i}+\hat{j}+2\hat{k}+\mu (2\hat{i}-\hat{j}+\hat{k})\end{array}$

Solution:

Let:

• $\overrightarrow{a_1}=\hat{i}+2\hat{j}+\hat{k}$
• $\overrightarrow{a_2}=2\hat{i}+\hat{j}+2\hat{k}$
• ${\vec{b}}_1=\hat{i}+\hat{j},{\vec{b}}_2=2\hat{i}-\hat{j}+\hat{k}$

Solution: 

$\begin{array}{rl}& \overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{j}+\hat{k}\\ & \overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 2& -1& 1\end{array}\right|\\ & \overrightarrow{b_1}\times \overrightarrow{b_2}=(1)(1)-(0)(-1)\hat{i}-(1)(1)-(0)(2)\hat{j}+(1)(-1)-(1)(2)\hat{k}\\ & \overrightarrow{b_1}\times \overrightarrow{b_2}=\hat{i}-\hat{j}-3\hat{k}\end{array}$

Using formula of shortest distance between two lines:

$d=\frac{|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\cdot \left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$

$\begin{array}{rl}& \begin{array}{rl}& d=\frac{|(\hat{i}-\hat{j}+\hat{k})\cdot (\hat{i}-\hat{j}-3\hat{k})|}{|\hat{i}-\hat{j}-3\hat{k}|}\\ & d=\frac{|1+1-3|}{\sqrt{1+1+9}}\end{array}\\ & d=\frac{|-1|}{\sqrt{11}}\\ & d=\frac{1}{\sqrt{11}}\end{array}$

Answer: $\frac{1}{\sqrt{11}}$

Q6. Find a unit vector perpendicular to both $\vec{A}=2\hat{i}+3\hat{j}-\hat{k},\vec{B}=\hat{i}-\hat{j}+4\hat{k}$ 

Solution:

Unit vector = $\frac{\vec{A}\times \vec{B}}{|\vec{A}\times \vec{B}|}$

$\begin{array}{rl}& \vec{A}\vec{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 1& -1& 4\end{array}\right|\\ & \vec{A}\times \vec{B}=((3)(4)-(-1)(-1))\hat{i}-((2)(4)-(-1)(1))\hat{j}+((2)(-1)-(3)(1))\hat{k}\\ & \vec{A}\times \vec{B}=(12-1)\hat{i}-(8+1)\hat{j}+(-2-3)\hat{k}\\ & \vec{A}\times \vec{B}=11\hat{i}-9\hat{j}-5\hat{k}\end{array}$

$\begin{array}{rl}& \Rightarrow \text{ Unit Vector }=\frac{11\hat{i}-9\hat{j}-5\hat{k}}{\sqrt{121+81+25}}\\ & =\frac{11\hat{i}-9\hat{j}-5\hat{k}}{\sqrt{227}}\end{array}$

Answer: $\frac{11\hat{i}-9\hat{j}-5\hat{k}}{\sqrt{227}}$

Q7. If $\vec{A}=\hat{i}+2\hat{j},\vec{B}=2\hat{i}-3\hat{j},findavector\vec{C}suchthat:\vec{A}\cdot \vec{C}=0,\vec{B}\cdot \vec{C}=0$

Solution: 

Let $\vec{C}=x\hat{i}+y\hat{j}+z\hat{k}$

Then,

$\begin{array}{rl}& \vec{A}\cdot \vec{C}\\ & \vec{B}\cdot \vec{C}=x+2y=0\Rightarrow x=-2y\\ & \vec{B}\cdot \vec{C}=2x-3y=0\Rightarrow 2(-2y)-3y=-4y-3y=-7y=0\Rightarrow y=0\Rightarrow x=0\end{array}$ 

So

$\vec{C}=z\hat{k}$

Answer: Any vector along .

Q8: What is the multiplication of vectors formula?

• $Scalar:k\cdot \vec{A}=k{A}_x\hat{i}+k{A}_y\hat{j}$
• $Dot:\vec{A}\cdot \vec{B}=|\vec{A}||\vec{B}|\cos \theta$
• $Cross:\vec{A}\times \vec{B}=|\vec{A}||\vec{B}|\sin \theta \hat{n}$

9.0Practice Questions on Vector Multiplication

1. Multiply $\vec{A}=(2,-1)byscalar-3$.
2. Find the dot product of $\vec{A}=(4,2)and\vec{B}=(-1,5)$.
3. Calculate $\vec{A}\times \vec{B}where\vec{A}=(0,2,1),\vec{B}=(3,1,4)$.
4. If $\vec{V}=5\hat{i}-\hat{j},find0\cdot \vec{V}$.
5. What happens to a vector when multiplied by -1?