Multiplication Of Vectors

Types of Vector

Let's start with the types of vectors.

Unit Vector

A vector having a magnitude of one unit is called a unit vector. It is represented by a cap/hat over the letter.

$\mathrm{Eg}-\hat{R}$ is called a unit vector of $\vec{R}$. Its direction is along the $\vec{R}$ and magnitude is unit.

Unit vector along $\vec{R}$ -
$
\hat{R}=\frac{\vec{R}}{|\vec{R}|}
$

Orthogonal Unit Vectors

It is defined as the unit vectors described under the three-dimensional coordinate system along the x, y, and z-axis. The three unit vectors are denoted by i, j and k respectively.

Any vector (Let us say $\vec{R}$ ) can be written as
$
\vec{R}=x \hat{i}+y \hat{j}+z \hat{k}
$

$\text { Where } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \text { are components of } \vec{R} \text { along } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \text { direction respectively. }$

$|\vec{R}|=\sqrt{x^2+y^2+z^2}$

Unit vector

$\hat{R}=\frac{x \hat{i}+y \hat{j}+z \hat{k}}{\sqrt{x^2+y^2+z^2}}$

1. If a vector is multiplied by any scalar

$\vec{Z}=n \cdot \vec{Y}$

(n=1,2,3..)

$\text { Vector } \times \text { Scalar }=\text { Vector }$

We get again a vector.

If a vector is multiplied by any real number (eg 2 or -2) then again, we get a vector quantity.

E.g.

• $\text { If } \vec{A}$ is multiplied by 2 then the direction of the resultant vector is the same as that of the given vector.

$\text { Vector }=2 \vec{A}$

• $\text { If } \vec{A}$ is multiplied by (-2), then the direction of the resultant is opposite to that of the given vector.

$\text { Vector }=-2 \vec{A}$

Scalar or Dot or Inner Product

• Scalar product of two vectors $\vec{A} \& \vec{B} \text { written as } \vec{A} \cdot \vec{B}$

• $\vec{A} \cdot \vec{B}$ is a scalar quantity given by the product of the magnitude of $\vec{A} \& \vec{B}$ and the cosine of a smaller angle between them.

$\vec{A} \cdot \vec{B}=A B \cdot \cos \Theta $

Figure showing the representation of scalar products of vectors.

Important results-

$\begin{aligned}
& \hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0 \\
& \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1 \\
& \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}
\end{aligned}$

Vector or Cross-Product

• Vector or cross product of two vectors $ \vec{A} \& \vec{B} \text { written as } \vec{A} \times \vec{B}$

• $A \times B$ is a single vector whose magnitude is equal to the product of the magnitude of $\vec{A} \& \vec{B}$ and the sine of the smaller angle $\theta$ between them.

• $\vec{A} \times \vec{B}=A B \sin \theta$

The figure shows the representation of the cross-product of vectors.

Important results

$ \begin{aligned}
& \quad \hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j} \\
& \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0} \\
& \vec{A} \times \vec{B}=-\vec{B} \times \vec{A}
\end{aligned}$

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Solved Example Based on Multiplication of Vector

Example 1: When we multiply a vector $\vec{A}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ with a number -2, we get

1) A scalar with a magnitude same as that of the vector

2) A vector with the same magnitude and opposite direction.

3) A vector with double magnitude and opposite direction.

4) A vector with double magnitude and the same direction

Solution:

$\text { If } \vec{A}$ is multiplied by (-2), then the direction of the resultant is opposite to that of the given vector.

$\begin{aligned}
& \text { Vector }=-2 \vec{A} \\
& \vec{B}=-2 \cdot \vec{A}=2(-\vec{A})
\end{aligned}$

$\therefore$ The direction of $\vec{B} \text { is opposite of } \vec{A}$ and

Magnitude of $ |\vec{B}|=2 \times \text { magnitude of }|\vec{A}| $.

Hence, the answer is the option (3).

Example 2: Let $\vec{A}=(\hat{i}+\hat{j})$ and, $\vec{B}=(2 \hat{i}-\hat{j})$ The magnitude of a coplanar vector $\vec{C}$ such that $\vec{A} \cdot \vec{C}=\vec{B} \cdot \vec{C}=\vec{A} \cdot \vec{B}$ is given by:

1) $\sqrt{(10 / 9)}$
2) $\sqrt{5 / 9}$
3) $\sqrt{(20 / 9)}$
4) $\sqrt{(9 / 12)}$

Solution:

$\begin{aligned}
& \text { Let } \vec{c}=x \hat{i}+y \hat{j} \\
& \begin{array}{l}
\vec{A} \cdot \vec{C}=x+y \\
\vec{B} \cdot \vec{C}=2 x-y \\
\vec{A} \cdot \vec{B}=2-1=1 \\
\Rightarrow x+y=1 \\
2 x-y=1
\end{array}
\end{aligned}$

$\begin{aligned}
& 3 x=2 \quad \text { or } \mathrm{x}=2 / 3 \\
& \mathrm{y}=1 / 3 \\
& \therefore \vec{c}=2 / 3 \hat{i}+1 / 3 \hat{j} \\
& \therefore|\vec{c}|=\sqrt{(4 / 9+1 / 9)}=\sqrt{(5 / 9)}
\end{aligned}$

Hence, the answer is the option (2).

$\text { Example 3: The angle between }(\hat{l}+\hat{j}) \text { and }(\hat{l}-\hat{j}) \text { is (in degrees) }$

1) 90

2) 60

3) 45

4) 30

Solution:

As we learned

Scalar, Dot or Inner Product -

The scalar product of two vectors $\vec{A} \& \vec{B}$ written as $\vec{A} \cdot \vec{B}$ is a scalar quantity given by the product of the magnitude of $\vec{A} \& \vec{B}$ and the cosine of the smaller angle between them.

$\vec{A} \cdot \vec{B}=A B \cdot \cos \Theta$

- wherein

showing a representation of scalar products of vectors.

$\begin{aligned}
& (\hat{l}+\hat{j}) \cdot(\hat{l}-\hat{j})=|\hat{l}+\hat{j}||\hat{l}-\hat{j}| \cdot \cos \theta \\
& U \operatorname{sing} \vec{A} \cdot \vec{B}=A B \cos \theta \\
& \Rightarrow \cos \theta=\frac{(\hat{l}+\hat{j}) \cdot(\hat{l}-\hat{j})}{|\hat{l}+\hat{j}||\hat{l}-\hat{j}|}=\frac{0}{\sqrt{2} \cdot \sqrt{2}}=0 \\
& \therefore \theta=90^{\circ}
\end{aligned}$

Hence, the answer is option (1).

Example 4: A particle is thrown with $10 \mathrm{~m} / \mathrm{s}$ at an angle of $60^{\circ}$ with horizontal the time at which its velocity is perpendicular to the initial velocity is ( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}$ )

1) $\sqrt{3} \mathrm{sec}$
2) $2 \sqrt{3} \mathrm{sec}$
3) $\frac{2}{\sqrt{3}} \mathrm{sec}$
4) $\frac{4}{\sqrt{3}} \mathrm{sec}$

Solution:

$\vec{A} \cdot \vec{B}=A B \cdot \cos \Theta$

Showing the representation of scalar products of vectors.

$
\begin{aligned}
& u=10 \cos 60^{\circ} \hat{i}+10 \sin 60^{\circ} \hat{j} \\
& =5(\hat{i}+\sqrt{3} \hat{j}) \\
& V_y=u_y-g t=5 \sqrt{3}-10 t(\text { here use } g=10) \\
& \vec{V}=5 \hat{i}+(5 \sqrt{3}-10 t) \hat{j}
\end{aligned}
$
since $\vec{V}$ is perpendicular to $\vec{u} h$ ence $\vec{u} \cdot \vec{V}=0$
$
\begin{aligned}
& \Rightarrow 5(\hat{i}+\sqrt{3} \hat{j}) \cdot 5(\hat{i}+(\sqrt{3}-2 t) \hat{j})=0 \\
& \Rightarrow 1+\sqrt{3}(\sqrt{3}-2 t)=0
\end{aligned}
$
or $1+3-2 \sqrt{3} t=0$
$
\Rightarrow t=\frac{2}{\sqrt{3}} \sec
$

Hence, the answer is option (3).

Example 5: If $\vec{a}, \vec{b}$ are unit vectors such that the angle between $\vec{a}$ and $\vec{b}$ is -

1) 0
2) $\frac{\pi}{2}$
3) $\pi$
4) Indeterminate

Solution:

$\vec{A} \times \vec{B}=A B \sin \theta$

The figure shows the representation of vectors or cross product of vectors.

shows the representation of vector or cross product of vectors

$\begin{aligned}
& 0=(\vec{a}+\vec{b}) \cdot(2 \vec{a}+3 \vec{b}) \times(3 \vec{a}-2 \vec{b})=0 \\
& 0=(\vec{a}+\vec{b}) \cdot(-4 \vec{a} \times \vec{b}-9 \vec{a} \times \vec{b})=-13 \cdot(\vec{a}+\vec{b}) \cdot(\vec{a} \times \vec{b})
\end{aligned}$

which is true for all values of $\vec{a} \text { and } \vec{b}$

Summary

Vector multiplication can be divided into two categories. A vector has both a magnitude and a direction, and as a result, the dot product of two vectors and the cross product of two vectors are the two methods of multiplying vectors. Because the resultant value is a scalar quantity, the dot product of two vectors is often referred to as the scalar product of two vectors. A vector has both a magnitude and a direction associated with it.