Chapter 7.13 - Mirror Image of a Vector

In the previous section, we saw the six terms in the cross product. In this section we will see a tabular form, which helps to obtain those six terms easily. Later in this section, we will also see reflected vectors and distributive law

1. Consider the table shown below:

• There is a 'header column on the left' + 3 additional columns
• There is a 'header row at top' + 3 additional rows
■ The header column is for writing the components of the first vector $\mathbf\small{\vec{a}}$ 
■ The header row is for writing the components of the second vector $\mathbf\small{\vec{b}}$.
2. The combinations are written inside the red box
• There are a total of 9 combinations possible
• But those coming diagonally are obtained from similar components. They are related to scalar product. They must not be used for vector product. Hence a '❌' mark is given to show that, those diagonal combinations are prohibited
• Such an elimination leaves 6 combinations
3. The signs of the combinations are also given
    ♦ An example: In the combination between row 1 and column 3, we have: $\mathbf\small{\hat{i}\times \hat{k}=-\hat{j}}$
• There are 3 positive and 3 negative combinations
• However, the final sign will depend also on the 'signs of the coefficients in the given vectors'

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Let us see an example to demonstrate the application of the table:

• Find the vector product of $\mathbf\small{\vec{a}=3\hat{i}-4\hat{j}+5\hat{k}}$ and $\mathbf\small{\vec{b}=-2\hat{i}+\hat{j}-3\hat{k}}$
Solution:
1. The table is shown below:

• We get: $\mathbf\small{\vec{a}\times\vec{b}=(-5+12)\hat{i}+(-10+9)\hat{j}+(-8+3)\hat{k}}$
$\mathbf\small{\Rightarrow \vec{a}\times\vec{b}=7\hat{i}-\hat{j}-5\hat{k}}$
2. Table for the reverse multiplication $\mathbf\small{\vec{b}\times\vec{a}}$ can also be formed:

• We get: $\mathbf\small{\vec{b}\times\vec{a}=(5-12)\hat{i}+(10-9)\hat{j}+(8-3)\hat{k}}$
$\mathbf\small{\Rightarrow \vec{a}\times\vec{b}=-7\hat{i}+\hat{j}+5\hat{k}}$
■ We see that: $\mathbf\small{(\vec{a}\times\vec{b})=-(\vec{b}\times\vec{a})}$

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Another method for obtaining the 6 terms:

• If we are familiar with determinants, we can use it for vector cross products
• Consider the formula given below:
$\mathbf\small{\vec{a}\times \vec{b}=\left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{matrix} \right| }$
• If we expand the determinant on the right side, we will get the 6 terms. The reader may try this method also

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Mirror image of a vector

1. Suppose that, we have a vector $\mathbf\small{\vec{a}=a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k}}$  
• The magnitudes of the components are a_x, a_y and a_z
• Fig.7.67(a) below shows how the 3 components help to define $\mathbf\small{\vec{a}}$ 

2. $\mathbf\small{\vec{a}}$ is shown in magenta color
• The red line indicates a_x
• The green line indicates a_y
• The blue line indicates a_z
3. Each of the above 3 lines are indispensable. For example, the green line has a definite length. But it is useless if we do not know where to place it. It's position is given by the red line
• Also remember that, the lengths of the red, green and blue lines are the coordinates (x,y,z) of the tip of the magenta vector $\mathbf\small{\vec{a}}$. That is: (x,y,z) = (a_x, a_y, a_z) 
4. Now we will see the 'mirror image' of $\mathbf\small{\vec{a}}$
• For that, we change the sign of the components
    ♦ a_x will become -a_x
    ♦ a_y will become -a_y
    ♦ a_z will become -a_z.
• These new components are shown in fig.7.67(b) above
• An example: -a_x is on the negative side of y-axis because, it has to be at a distance -a_y of from the x axis 
• In this way, each new component should be accurately positioned
5. When those positions are finalized, we get the tip of the 'mirror image' of $\mathbf\small{\vec{a}}$
• The coordinates of this new tip will obviously be (-x,-y,-z)
• Also we have: (x,y,z) = (-a_x, -a_y, -a_z)
■ We see the following 3 facts:
(i) The new vector falls on the same line as $\mathbf\small{\vec{a}}$
(ii) The new vector has the same magnitude as $\mathbf\small{\vec{a}}$
[∵ distance between O and (a_x, a_y, a_z) = distance between O and (-a_x, -a_y, -a_z)]
(iii) The new vector has the direction opposite to that of $\mathbf\small{\vec{a}}$
■ So we can write: The new vector, which is the mirror image, is $\mathbf\small{-\vec{a}}$
• Usually, we denote the mirror image of $\mathbf\small{\vec{a}}$ as $\mathbf\small{\vec{a}'}$
■ So we can write: $\mathbf\small{\vec{a}'=-\vec{a}}$

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In the above example, the tail end of $\mathbf\small{\vec{a}}$ is at the origin O. So calculations were easy. Now, in fig.7.68(a) below, the tail end is away from 'O'

Fig.7.68

We will write the steps:
1. The red, green and blue lines of the tail end are drawn. They have lengths x₁, y₁ and z₁ respectively. So the coordinates of the tail end are (x₁, y₁, z₁)
• The red, green and blue lines of the head are drawn. They have lengths x₂, y₂ and z₂ respectively. So the coordinates of the head are (x₂, y₂, z₂)
2. So we can write the components of $\mathbf\small{\vec{a}}$
    ♦ Magnitude of the x-component = a_x = (x₂ - x₁)
    ♦ Magnitude of the y-component = a_y = (y₂ - y₁)
    ♦ Magnitude of the z-component = a_z = (z₂ - z₁)
• These are shown as red, blue and green dashed lines in fig.7.68(b)
3. So how do we obtain the mirror image of $\mathbf\small{\vec{a}}$ ?
Ans: First, let us change the signs of both the coordinates 
• Thus:
    ♦ (x₁, y₁, z₁) will become (-x₁, -y₁, -z₁)
    ♦ (x₂, y₂, z₂) will become (-x₂, -y₂, -z₂)
4. We get a new tail end and a new head end. The vector drawn between these new head and tail is shown in fig.7.69 below:

Fig.7.69

• Now we can write the components of this new vector:
• Magnitude of the x-component of the new vector = [(-x₂ - (-x₁)] = [x₁ - x₂] 
• But we saw that:
    ♦ Magnitude of the x-component of the original vector = a_x = (x₂ - x₁)
    ♦ There is only a difference in sign
• That is: Magnitude of the x-component of the new vector = -a_x
Similarly we can write:
• Magnitude of the y-component of the new vector = [(-y₂ - (-y₁)] = [y₁ - y₂] = -a_y
• Magnitude of the z-component of the new vector = [(-z₂ - (-z₁)] = [z₁ - z₂] = -a_z
• They are shown in fig.7.70 below:

Fig.7.70

5. So it is clear that:
• To obtain the mirror image of any vector, all we need to do is, change the sign of each component.
• That is: Mirror image of $\mathbf\small{a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k}}$ is $\mathbf\small{-a_x\,\hat{i}-a_y\,\hat{j}-a_z\,\hat{k}}$
• But $\mathbf\small{-a_x\,\hat{i}-a_y\,\hat{j}-a_z\,\hat{k}=-(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})=-\vec{a}}$
• That means: Mirror image of $\mathbf\small{\vec{a}}$ = $\mathbf\small{\vec{a}'=-\vec{a}}$

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So we have seen two cases:

Case 1: The tail end of the vector is at O 
Case 2: The tail end of the vector is away from O
In both cases, we see that: $\mathbf\small{\vec{a}'=-\vec{a}}$

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Now we will see an interesting case:

• Given two vectors $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$
• We know how to find ($\mathbf\small{\vec{a}\times\vec{b}}$)
• What about ($\mathbf\small{\vec{a}'\times\vec{b}'}$)?
Solution:
1. We have: $\mathbf\small{\vec{a}'\times\vec{b}'=-\vec{a}\times-\vec{b}}$
• We know that:
    ♦ $\mathbf\small{\vec{a}}$ lies along the same line as $\mathbf\small{-\vec{a}}$
    ♦ $\mathbf\small{\vec{b}}$ lies along the same line as $\mathbf\small{-\vec{b}}$
This is shown in fig.7.71(a) below:

Fig.7.71

2. So all the four vectors can be brought together to a single point on the same plane as shown in fig.7.71(b) above
• We find that:
Angle between $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ = Angle between $\mathbf\small{-\vec{a}}$ and $\mathbf\small{-\vec{b}}$ 
(∵ they are opposite angles)
3. Once we get that angle between $\mathbf\small{-\vec{a}}$ and $\mathbf\small{-\vec{b}}$, we can find the cross product:
• $\mathbf\small{|(-\vec{a}\times-\vec{b})|=|-\vec{a}|\times |-\vec{b}|\times \sin \theta}$
$\mathbf\small{\Rightarrow |(-\vec{a}\times-\vec{b})|=|\vec{a}|\times |\vec{b}|\times \sin \theta}$
4. But $\mathbf\small{|\vec{a}|\times |\vec{b}|\times \sin \theta=|(\vec{a}\times \vec{b})|}$
• Thus we get: $\mathbf\small{|(-\vec{a}\times-\vec{b})|=|(\vec{a}\times \vec{b})|}$
5. But $\mathbf\small{|(-\vec{a}\times-\vec{b})|=|(\vec{a}'\times \vec{b}')|}$
• Thus we get: $\mathbf\small{|(\vec{a}'\times \vec{b}')|=|(\vec{a}\times \vec{b})|}$
■ That is., 
Magnitude of the cross product of two vectors = Magnitude of the cross product of their mirror images  
6. Now we want directions:
In fig.7.71(b), we see the following information:
(i) To find the direction of $\mathbf\small{(\vec{a}\times \vec{b})}$, we would turn the screw from $\mathbf\small{\vec{a}}$ to $\mathbf\small{\vec{b}}$  
(ii) This is the same direction in which we would turn the screw to find the direction of $\mathbf\small{(-\vec{a}\times -\vec{b})}$ 
(iii) So the screw will be moving in the same direction in both the cases
(iv) Thus we get: Direction of both the cross products are the same
7. Magnitudes and directions are the same. So we can write:
$\mathbf\small{(-\vec{a}\times -\vec{b})=(\vec{a}\times \vec{b})}$
Thus we get:
Eq.7.15:  $\mathbf\small{(\vec{a}'\times \vec{b}')=(\vec{a}\times \vec{b})}$

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Distributive property of vector cross product

■ The cross product obeys distributive law
Proof:
1. We have to prove that: $\mathbf\small{\vec{a}\times(\vec{b}+\vec{c})=(\vec{a}\times \vec{b})+(\vec{a}\times \vec{c})}$
2. Let:
$\mathbf\small{\vec{a}=a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k}}$
$\mathbf\small{\vec{b}=b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k}}$
$\mathbf\small{\vec{c}=c_x\,\hat{i}+c_y\,\hat{j}+c_z\,\hat{k}}$
Then we get: $\mathbf\small{(\vec{b}+\vec{c})=(b_x+c_x)\,\hat{i}+(b_y+c_y)\,\hat{j}+(b_z+c_z)\,\hat{k}}$
3. On the L.H.S of (1), we have: $\mathbf\small{\vec{a}\times(\vec{b}+\vec{c})}$
This will become: $\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})\times[(b_x+c_x)\,\hat{i}+(b_y+c_y)\,\hat{j}+(b_z+c_z)\,\hat{k}]}$
4. The cross product in (3) can be calculated using Table 1 below:

Table 1

• On the right side, the like terms are added together
• This completes our calculations on L.H.S of (1)
5. The R.H.S of (1) has two cross products
• The first one is $\mathbf\small{(\vec{a}\times \vec{b})}$
• This can be calculated using the Table 2 below:

Table.3

• On the right side, the like terms are added together
6. The second cross product in the R.H.S of (1) is $\mathbf\small{(\vec{a}\times \vec{c})}$
• This can be calculated using the Table 3 below:

Table 3

• On the right side, the like terms are added together
7. Now we examine the right sides of the 3 tables carefully. We find 3 facts:
(i) The $\mathbf\small{\hat{i}}$ component in Table 1 = ($\mathbf\small{\hat{i}}$ component in Table 2 + $\mathbf\small{\hat{i}}$ component in Table 3)
(ii) The $\mathbf\small{\hat{j}}$ component in Table 1 = ($\mathbf\small{\hat{j}}$ component in Table 2 + $\mathbf\small{\hat{j}}$ component in Table 3)
(iii) The $\mathbf\small{\hat{k}}$ component in Table 1 = ($\mathbf\small{\hat{k}}$ component in Table 2 + $\mathbf\small{\hat{k}}$ component in Table 3)
8. Thus we get:
L.H.S in (1) = R.H.S in (1)

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In the next section, we will see some solved examples related to cross products

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Chapter 7.12 - Magnitude of cross product

In the previous section, we saw the direction of the product vector. In this section we will see the magnitude. Later in this section, we will also see some properties of vector products

1. We denote the vector obtained as a result of the 'vector multiplication of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$' as: $\mathbf\small{\vec{c}}$
• So we can write: $\mathbf\small{\vec{a}\times \vec{b}=\vec{c}}$
2. The magnitude of this $\mathbf\small{\vec{c}}$ is given by:
Eq.7.13: $\mathbf\small{|\vec{c}|=|\vec{a}|\times |\vec{b}|\times \sin \theta}$
• Where θ is the angle between $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ 
3. Based on this we can write the reverse multiplication also:
If $\mathbf\small{\vec{b}\times \vec{a}=\vec{c}'}$, then:
$\mathbf\small{|\vec{c}'|=|\vec{b}|\times |\vec{a}|\times \sin \theta}$
4. The results in (2) and (3) are the same. So we can write:
$\mathbf\small{|(\vec{a}\times \vec{b})|=|(\vec{b}\times \vec{a})|}$
$\mathbf\small{\Rightarrow |\vec{c}|=|\vec{c}'|}$
5. We must be careful while selecting the angle between $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$
• There will be two angles between any two vectors  (Details here) 
• Even if the two tails coincide, there will be two angles:
    ♦ One is θ
    ♦ The other is (360-θ)  
• We must always choose the one which is less than 180^o

• This is shown in fig.7.62 below:

Fig.7.62

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A '×' sign is used to indicate the vector product. Because of that, the vector product is also know as cross product. We read $\mathbf\small{\vec{a}\times \vec{b}}$ as 'a cross b'

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Properties of Cross products

• In chapter 6, we saw that, dot products obey commutative law
That is: $\mathbf\small{\vec{a}\;.\vec{b}=\vec{b}\;.\vec{a}}$ 
• Can we say the same about cross products?
That is:
Is $\mathbf\small{\vec{a}\times \vec{b}}$ equal to $\mathbf\small{\vec{b}\times \vec{a}}$?  
• The answer is 'No'. Because, though they have the same magnitudes, the directions are opposite to each other. We saw this when we learned about the right hand screw rule in the previous section  
• So we can write:
$\mathbf\small{(\vec{a}\times \vec{b})\neq (\vec{b}\times \vec{a})}$
• However, we can write:
$\mathbf\small{(\vec{a}\times \vec{b})=-(\vec{b}\times \vec{a})}$

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Before moving on to more properties, we will see some elementary cross products:
1. Cross product of a vector with itself:
$\mathbf\small{(\vec{a}\times \vec{a})=\vec{0}}$
Where $\mathbf\small{\vec{0}}$ is a null vector
(A 'null vector' is a vector having zero magnitude)
Proof:
(i) We have: $\mathbf\small{|(\vec{a}\times \vec{a})|=|\vec{a}|\times |\vec{a}|\times \sin \theta}$
• But the angle between a vector and itself is zero
• Also we have: sin 0 = 0
(ii) Thus we get: $\mathbf\small{|(\vec{a}\times \vec{a})|=|\vec{a}|\times |\vec{a}|\times \sin 0 =|\vec{a}|\times |\vec{a}|\times 0 = 0}$
• If a vector has zero magnitude, it is a null vector. So we can write:
$\mathbf\small{(\vec{a}\times \vec{a})=\vec{0}}$
2. Based on the above result, we can write the following 3 results:
(i) $\mathbf\small{(\hat{i}\times \hat{i})=\vec{0}}$
(ii) $\mathbf\small{(\hat{j}\times \hat{j})=\vec{0}}$
(iii) $\mathbf\small{(\hat{k}\times \hat{k})=\vec{0}}$
3. Cross product of the unit vector $\mathbf\small{\hat{i}}$ with another perpendicular unit vector $\mathbf\small{\hat{j}}$:
$\mathbf\small{(\hat{i}\times \hat{j})=\hat{k}}$
Proof:
(i) We have: $\mathbf\small{|(\hat{i}\times \hat{j})|=|\hat{i}|\times |\hat{j}|\times \sin \theta}$
• But $\mathbf\small{\hat{i}}$ and $\mathbf\small{\hat{j}}$ are perpendicular to each other
• So the angle between them is 90
• Also we have: sin 90 = 1
• Thus we get: $\mathbf\small{|(\hat{i}\times \hat{j})|=1\times 1\times 1=1}$
• So we obtained the magnitude of $\mathbf\small{(\hat{i}\times \hat{j})}$
(ii) Next we want the direction:
• Let $\mathbf\small{\hat{i}}$ lie along the positive side of the x-axis 
• Let $\mathbf\small{\hat{j}}$ lie along the positive side of the y-axis
• Let their tails meet at 'O'. This is shown in fig,7.63(a) below:

Fig.7.63

(iii) Now we can find the required direction by applying the right hand screw rule
• For that, place a right handed screw at 'O'
• The screw must be perpendicular to the 'plane containing $\mathbf\small{\hat{i}}$ and $\mathbf\small{\hat{j}}$'
(iv) This plane is obviously, the xy-plane
• So the screw should be placed along the z-axis
• It's head should be at 'O'
• It's tip should be pointing towards the positive side of the z-axis
(v) Now, rotate the screw from $\mathbf\small{\hat{i}}$ towards $\mathbf\small{\hat{j}}$
• This rotation is indicated by the yellow curved arrow in fig.a 
• Due to this rotation, the screw will move towards the positive side of the z-axis
• Thus we get the direction of $\mathbf\small{(\hat{i}\times \hat{j})}$:
The direction is towards the positive side of the z-axis
• So we have two information:
    ♦ The vector $\mathbf\small{(\hat{i}\times \hat{j})}$ has a magnitude of '1'
    ♦ The vector $\mathbf\small{(\hat{i}\times \hat{j})}$ has a direction pointing towards the positive side of z-axis
• Obviously, such a vector is $\mathbf\small{\hat{k}}$
• So we can write: $\mathbf\small{(\hat{i}\times \hat{j})=\hat{k}}$
4. Cross product of the unit vector $\mathbf\small{\hat{j}}$ with another perpendicular unit vector $\mathbf\small{\hat{k}}$:
$\mathbf\small{(\hat{j}\times \hat{k})=\hat{i}}$
Proof:
(i) We have: $\mathbf\small{|(\hat{j}\times \hat{k})|=|\hat{j}|\times |\hat{k}|\times \sin \theta}$
• But $\mathbf\small{\hat{j}}$ and $\mathbf\small{\hat{k}}$ are perpendicular to each other
• So the angle between them is 90
• Also we have: sin 90 = 1
• Thus we get: $\mathbf\small{|(\hat{j}\times \hat{k})|=1\times 1\times 1=1}$
• So we obtained the magnitude of $\mathbf\small{(\hat{j}\times \hat{k})}$
(ii) Next we want the direction:
• Let $\mathbf\small{\hat{j}}$ lie along the positive side of the y-axis 
• Let $\mathbf\small{\hat{k}}$ lie along the positive side of the z-axis
• Let their tails meet at 'O'. This is shown in fig,7.63(b) above
(iii) Now we can find the required direction by applying the right hand screw rule
• For that, place a right handed screw at 'O'
• The screw must be perpendicular to the 'plane containing $\mathbf\small{\hat{j}}$ and $\mathbf\small{\hat{k}}$'
(iv) This plane is obviously, the yz-plane
• So the screw should be placed along the x-axis
• It's head should be at 'O'
• It's tip should be pointing towards the positive side of the x-axis
(v) Now, rotate the screw from $\mathbf\small{\hat{j}}$ towards $\mathbf\small{\hat{k}}$ 
• This rotation is indicated by the yellow curved arrow in fig.b
• Due to this rotation, the screw will move towards the positive side of the x-axis
• Thus we get the direction of $\mathbf\small{(\hat{i}\times \hat{j})}$:
The direction is towards the positive side of the x-axis
• So we have two information:
    ♦ The vector $\mathbf\small{(\hat{j}\times \hat{k})}$ has a magnitude of '1'
    ♦ The vector $\mathbf\small{(\hat{j}\times \hat{k})}$ has a direction pointing towards the positive side of x-axis
• Obviously, such a vector is $\mathbf\small{\hat{i}}$
• So we can write: $\mathbf\small{(\hat{j}\times \hat{k})=\hat{i}}$
5. By writing steps similar to the above (3) and (4), we will get: $\mathbf\small{(\hat{k}\times \hat{i})=\hat{j}}$
• The reader is advised to draw the diagram and write all the steps in his/her own note book
6. The results in (3), (4), (5) are obtained when we multiply the unit vectors in a cyclic order shown in fig.7.64(a) below:

Fig.7.64

• What if we multiply the unit vectors in a reverse cyclic order shown in fig.7.64(b)?
• We will get three products: $\mathbf\small{(\hat{i}\times \hat{k}),(\hat{k}\times \hat{j}),(\hat{j}\times \hat{i})}$
■ So all together, there are 6 possible products. They can be grouped into 2 sets:
(i) $\mathbf\small{(\hat{i}\times \hat{j}),(\hat{j}\times \hat{k}),(\hat{k}\times \hat{i})}$ 
(ii) $\mathbf\small{(\hat{i}\times \hat{k}),(\hat{k}\times \hat{j}),(\hat{j}\times \hat{i})}$ 
• These 6 are the only possible combinations
    ♦ The first set is obtained from fig.7.64(a)
    ♦ The second set is obtained from fig.7.64(b)
• We have already seen the result for each case in the first set
• Based on those results we can find the result for each case in the second set
• Let us see how this is done:
7. We want $\mathbf\small{(\hat{i}\times \hat{k})}$ 
(i) We have already seen this: $\mathbf\small{(\hat{k}\times \hat{i})=\hat{j}}$
(ii) We have seen at the beginning of this section that, for any two vectors $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$:
$\mathbf\small{(\vec{a}\times \vec{b})=-(\vec{b}\times \vec{a})}$
(iii) Applying this rule, we get:
$\mathbf\small{(\hat{i}\times \hat{k})=-(\hat{k}\times \hat{i})=-\hat{j}}$
8. We want $\mathbf\small{(\hat{k}\times \hat{j})}$ 
(i) We have already seen this: $\mathbf\small{(\hat{j}\times \hat{k})=\hat{i}}$
(ii) We know this rule:
$\mathbf\small{(\vec{a}\times \vec{b})=-(\vec{b}\times \vec{a})}$
(iii) Applying this rule, we get:
$\mathbf\small{(\hat{k}\times \hat{j})=-(\hat{j}\times \hat{k})=-\hat{i}}$
9. We want $\mathbf\small{(\hat{j}\times \hat{i})}$ 
(i) We have already seen this: $\mathbf\small{(\hat{i}\times \hat{j})=\hat{k}}$
(ii) We know this rule:
$\mathbf\small{(\vec{a}\times \vec{b})=-(\vec{b}\times \vec{a})}$
(iii) Applying this rule, we get:
$\mathbf\small{(\hat{j}\times \hat{i})=-(\hat{i}\times \hat{j})=-\hat{k}}$
■ So we see that, the reverse cyclic order in fig.7.64(b) gives all negative unit vectors

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• We have completed the discussion on the 'cross products of unit vectors'

• Now we will get back to the 'cross products of ordinary vectors':
1. Some times we get vectors in component form:
• $\mathbf\small{\vec{a}=a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k}}$
• $\mathbf\small{\vec{b}=b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k}}$
■ We want: $\mathbf\small{\vec{a}\times \vec{b}}$ 
That is., we want: $\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})\times (b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k})}$
2. Recall that, we have seen the corresponding dot product in chapter 6:
$\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k}).(b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k})=a_xb_x+a_yb_y+a_zb_z}$
(Details here)
• We see that, in the result, each component is obtained by the multiplication of similar components
• That is.,
    ♦ $\mathbf\small{a_xb_x}$ is obtained from the x-component of $\mathbf\small{\vec{a}}$ and x-component of $\mathbf\small{\vec{b}}$
    ♦ $\mathbf\small{a_yb_y}$ is obtained from the y-component of $\mathbf\small{\vec{a}}$ and y-component of $\mathbf\small{\vec{b}}$
    ♦ $\mathbf\small{a_zb_z}$ is obtained from the z-component of $\mathbf\small{\vec{a}}$ and z-component of $\mathbf\small{\vec{b}}$
• The above 3 terms of the scalar product, can be pictorially represented as in figs.7.65(i), (ii) and (iii) below:

Fig.7.65

• The yellow arrows shows how the components are formed. We see that like components are combined together to form the three terms in the resulting scalar product
3. But the vector product is not so simple. In the vector product, no like components are combined
• Only unlike components are combined. This can be pictorially represented as in figs.7.66(i), (ii) and (iii) below:

Fig.7.66

We see that:
(i) a_x does not combine with b_x. Instead, it combines with b_y and b_z
    ♦ Thus we get: a_xb_y and a_xb_z
(ii) a_y does not combine with b_y. Instead, it combines with b_x and b_z
    ♦ Thus we get: a_yb_x and a_yb_z
(iii) a_z does not combine with b_z. Instead, it combines with b_x and b_y
    ♦ Thus we get: a_zb_x and a_zb_y
4. Thus we get 6 terms. But remember that, this is a vector product. The result is a vector. So there will be directions also
(i) The first term in 3(i) is $\mathbf\small{a_xb_y}$
• It is formed by the multiplication between an $\mathbf\small{\hat{i}}$ component and a $\mathbf\small{\hat{j}}$ component
• We have: $\mathbf\small{(\hat{i}\times \hat{j})=\hat{k}}$
■ So the first term in 3(i) is $\mathbf\small{a_xb_y\,\hat{k}}$
• The second term in 3(i) is $\mathbf\small{a_xb_z}$
• It is formed by the multiplication between an $\mathbf\small{\hat{i}}$ component and a $\mathbf\small{\hat{k}}$ component
• We have: $\mathbf\small{(\hat{i}\times \hat{k})=-\hat{j}}$
■ So the second term in 3(i) is $\mathbf\small{-a_xb_z\,\hat{j}}$
(ii) The first term in 3(ii) is $\mathbf\small{a_yb_x}$
• It is formed by the multiplication between a $\mathbf\small{\hat{j}}$ component and an $\mathbf\small{\hat{i}}$ component
• We have: $\mathbf\small{(\hat{j}\times \hat{i})=-\hat{k}}$
■ So the first term in 3(ii) is $\mathbf\small{-a_yb_x\,\hat{k}}$
• The second term in 3(ii) is $\mathbf\small{a_yb_z}$
• It is formed by the multiplication between a $\mathbf\small{\hat{j}}$ component and a $\mathbf\small{\hat{k}}$ component
• We have: $\mathbf\small{(\hat{j}\times \hat{k})=\hat{i}}$
■ So the second term in 3(ii) is $\mathbf\small{a_yb_z\,\hat{i}}$
(iii) The first term in 3(iii) is $\mathbf\small{a_zb_x}$
• It is formed by the multiplication between a $\mathbf\small{\hat{k}}$ component and an $\mathbf\small{\hat{i}}$ component
• We have: $\mathbf\small{(\hat{k}\times \hat{i})=\hat{j}}$
■ So the first term in 3(iii) is $\mathbf\small{a_zb_x\,\hat{j}}$
• The second term in 3(iii) is $\mathbf\small{a_zb_y}$
• It is formed by the multiplication between a $\mathbf\small{\hat{k}}$ component and a $\mathbf\small{\hat{j}}$ component
• We have: $\mathbf\small{(\hat{k}\times \hat{j})=-\hat{i}}$
■ So the second term in 3(iii) is $\mathbf\small{-a_zb_y\,\hat{j}}$
5. Writing all the 6 terms together, we get:
$\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})\times (b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k})}$
= $\mathbf\small{a_xb_y\,\hat{k}-a_xb_z\,\hat{j}- a_yb_x\,\hat{k}+ a_yb_z\,\hat{i}+ a_zb_x\,\hat{j}- a_zb_y\,\hat{i}}$
• Combining like components, we get:
Eq.7.14:
$\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})\times (b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k})}$
= $\mathbf\small{(a_yb_z-a_zb_y)\hat{i}+ (a_zb_x-a_xb_z)\hat{j}+(a_xb_y-a_yb_x)\hat{k}}$

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In the next section, we will see a tabular form which is useful to obtain the 6 terms easily

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Chapter 4.5 - Addition and Subtraction of Vectors using Formula

In the previous section we saw an analytical method to find the resultant vector. In this section, we will see another analytical method which uses a formula.
The steps are given below:

1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.18(a) below:

Fig.4.18

2. Shift them so that, their tails coincide at a point O. This is shown in fig.(b) 
    ♦ Let the tip of $\small{\vec{A}}$ be P
    ♦ Let the tip of $\small{\vec{B}}$ be Q
• Draw a green line parallel to OP
• Draw a yellow line parallel to OQ
• The green and yellow lines meet at S
3. Place a vector between O and S. 
• We know that this vector between O and S is $\small{\vec{R}}$, which is the resultant of $\small{\vec{A}}$ and $\small{\vec{B}}$ (Details in a the previous section)
• The length of OS will give the magnitude of $\small{\vec{R}}$
4. So our first aim is to find the length OS
• For that, extend OP 
• Drop a perpendicular SN on to that extension
5. Now we have a new right triangle: ⊿ONS
• The base ON of this triangle = (OP + PN)
    ♦ Instead of OP, we can write $\left | \vec{A} \right |$

    ♦ Instead PN, we can write PS cos $\theta$

    ♦ But PS =  $\left | \vec{B} \right |$

    ♦ So instead of PN, we can write $\left | \vec{B} \right |$ cos $\theta$

• So the base ON = (OP+PN) = $\left | \vec{A} \right |$  +  $\left | \vec{B} \right |$ cos $\theta$

6. Now, altitude of⊿ONS = SN = PS sin $\theta$ =  $\left | \vec{B} \right |$ sin $\theta$

• Applying Pythagoras theorem, we get: 

OS² = ON² + SN²

$\small{\Rightarrow {OS}^2\, =\, \left ( \left | \vec{A} \right |\, +\, \left | \vec{B} \right |\, cos\, \theta  \right )^2\: +\: \left ( \left | \vec{B} \right |\, sin\, \theta  \right )^2}$
• The right side can be expanded and simplified:
$\small{\Rightarrow {OS}^2\, =\, \left ( \left | \vec{A} \right |^2\, +\,2\left | \vec{A} \right |\, \left | \vec{B} \right |\, cos\,\theta \, +\,   \left | \vec{B} \right |^2 \, {cos}^2\, \theta  \right )\: +\: \left ( \left | \vec{B} \right |^2\, {sin}^2\, \theta  \right )}$
• Group the last two terms together:
$\small{\Rightarrow {OS}^2\, =\, \left | \vec{A} \right |^2\, +\,2\left | \vec{A} \right |\, \left | \vec{B} \right |\, cos\,\theta \, +\,   \left ( \left | \vec{B} \right |^2 \, {cos}^2\, \theta  \: +\:  \left | \vec{B} \right |^2\, {sin}^2\, \theta  \right )}$
$\small{\Rightarrow\, {OS}^2\: =\: \left | \vec{A} \right |^2 \, +\, 2\left | \vec{A} \right | \left | \vec{B} \right |cos\, \theta \, +\, \left | \vec{B} \right |^2\, \left ( {cos}^2\, \theta \, +\, {sin}^2\, \theta  \right )}$
• But $\small{ \left ( {cos}^2\, \theta \, +\, {sin}^2\, \theta  \right ) = 1}$
• So we get:
$\small{{OS}^2\: =\: \left | \vec{A} \right |^2 \, +\, 2\left | \vec{A} \right | \left | \vec{B} \right |cos\, \theta \, +\, \left | \vec{B} \right |^2}$
Thus we can write:
Eq.4.4:
OS = $\small{\left | \vec{R} \right |=\left ( \left | \vec{A}+\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 + 2\left | \vec{A} \right |\left | \vec{B} \right |cos \theta}}$
■ Thus we get the magnitude of $\small{\vec{R}}$
■ Eq.4.4 is known as the Law of cosines
7. Next we want the direction of $\small{\vec{R}}$
• For that, we consider 𝜟ONS
• We have: $tan\,  \alpha \, =\, \frac{SN}{ON}$ = $\frac{SN}{OP\, +\, PN}$
• We have already obtained the expressions for SN, OP and PN 
• Thus we get:
Eq.4.5:
$tan\,  \alpha \, =\, \frac{\left | \vec{B} \right |\, sin\, \theta }{\left | \vec{A} \right |\, +\,\left | \vec{B} \right |\, cos\, \theta }$
• From this expression, we get 'α', which is the angle made by $\small{\vec{R}}$ with one of the vectors $\small{\vec{A}}$
■ We may not be always able to denote the vectors as $\small{\vec{A}}$ and $\small{\vec{B}}$. So remember that, α in Eq.4.5 is the angle between $\small{\vec{R}}$ and the first term (the term without the cosine ratio) in the denominator 
8. Another method to find 'α':
We have two right triangles: ⊿ONS and ⊿PNS
• Consider ⊿ONS. We have:
SN = OS sin α = |$\small{\vec{R}}$| sin α
• Consider ⊿PNS. We have:
SN = SP sin θ = |$\small{\vec{B}}$| sin θ.
• Equating the two, we get: |$\small{\vec{R}}$| sin α = |$\small{\vec{B}}$| sin θ. 
So we can write: 
Eq.4.6:
$\large{\frac{\left | \vec{R} \right |}{sin\, \theta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• So, after calculating the magnitude of $\small{\vec{R}}$, we can use Eq.4.6 to calculate 'α'
9. Now we want the angle 'β', which $\small{\vec{R}}$ makes with the other vector $\small{\vec{B}}$
• For that, we drop a perpendicular from P onto OS. This is shown in fig.4.18(c)
• Now we have two new right triangles: ⊿OPM and ⊿SPM
■ Note that, in ⊿SPM, the angle at S is the same $\small{\beta}$ that we are trying to find. This is because, ∠QOS and ∠OSP are alternate angles 
10. Consider ⊿OPM. We have:
PM = OP sin α = |$\small{\vec{A}}$| sin α
• Consider ⊿SPM. We have:
PM = SP sin β = |$\small{\vec{B}}$| sin β.
• Equating the two, we get: |$\small{\vec{A}}$| sin α = |$\small{\vec{B}}$| sin β.
$\large{\Rightarrow \, \frac{\left | \vec{A} \right |}{sin\, \beta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• But Eq.4.6 above gives another expression for $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• So we can equate the three items. We get:
Eq.4.7:
$\large{\frac{\left | \vec{R} \right |}{sin\, \theta}}$ = $\large{\frac{\left | \vec{A} \right |}{sin\, \beta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
■ Eq.4.7 gives the relation between the three vectors and the angles. It is known as the Law of sines.
• We can see a pattern:
(i) When $\small{\left | \vec{A} \right |}$ is in the numerator
    ♦ Angle 'β', which is related to $\small{\vec{B}}$ is in the denominator
(ii) When $\small{\left | \vec{B} \right |}$ is in the numerator
    ♦ Angle 'α', which is related to $\small{\vec{A}}$ is in the denominator
(iii) When $\small{\left | \vec{R} \right |}$ is in the numerator
    ♦ Angle 'θ', which is related to both $\small{\vec{A}}$ and $\small{\vec{B}}$ is in the denominator
• If we select the first two ratios from Eq.4.7, we will be using $\small{\left | \vec{R} \right |}$, $\small{\left | \vec{A} \right |}$ and θ to calculate β
• If we select the last two ratios from Eq.4.7, we will be using $\small{\left | \vec{A} \right |}$, $\small{\left | \vec{B} \right |}$ and α to calculate β
11. Special cases:
(i) Maximum magnitude for resultant
• Consider Eq.4.4 that we saw above. It gives the magnitude of the resultant. We want to know the 'condition for obtaining the maximum magnitude' 
• From our math classes, we know that, the range of cosine function varies from -1 to +1 both inclusive
• In most cases, the cosine value that we will be using in Eq.4.4 will be a fraction
• We know that, when a quantity is multiplied by a fraction, it's value will decrease
• The only 'non-fraction values' that cos θ can take are: -1 and +1
• So, maximum value of $\small{\left | \vec{R} \right |}$ will be obtained when cos θ = 1
• For cos θ to be 1, θ should be equal to zero
• That is., for maximum magnitude, the two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ should have the same direction  
(ii) Minimum magnitude for resultant
• We want to know the 'condition for obtaining the least possible magnitude'
• When we add two vectors, we may get a resultant which has a negative direction
• But even if the direction is negative, the resultant will have a magnitude.
• So the least possible magnitude is zero. 
• That is., for the least possible magnitude, the two vectors should have the same magnitude but opposite directions
• Resultant of such a pair will have a zero magnitude
• Resultant of all other pairs will have a certain magnitude

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Now we will see a solved example.

We saw Solved example 4.2 in the previous section. We will do the same problem by the new method. The link to the new file is given below:

Solved example 4.3
We find that, the results obtained for $\small{\left | \vec{R} \right |}$, α, β, and $\small{\theta_R}$ are the same in both methods
• One more solved example is given below:
Solved example 4.4

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Now we will see the analytical method for vector subtraction
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.19(a) below:

Fig.4.19

• We want to find ($\small{\vec{A}}$ - $\small{\vec{B}}$)
• That is., we want to find [$\small{\vec{A}}$ + (-$\small{\vec{B}}$)]
2. Draw a vector with the same magnitude of $\small{\vec{B}}$, but opposite in direction.
• This new vector is $\small{\vec{-B}}$. It is also shown in fig.a
3. Shift $\small{\vec{-B}}$ until it's tail coincide with the tail of $\small{\vec{A}}$. This is shown in fig.b
Clearly:
IF angle between $\small{\vec{A}}$ and $\small{\vec{B}}$ is θ,
THEN angle between $\small{\vec{A}}$ and $\small{\vec{-B}}$ is (180-θ).
4. Draw green line parallel to $\small{\vec{A}}$
• Draw yellow line parallel to $\small{\vec{B}}$  
• Thus we get the parallelogram
• Based on the parallelogram, we can derive expressions for the following quantities:
$\small{\left | \vec{R} \right |}$, α, β, and $\small{\theta_R}$
(i) Magnitude of resultant:

$\small{\left | \vec{R} \right |=\left ( \left | \vec{A}-\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 + 2\left | \vec{A} \right |\left | \vec{B} \right |cos(180-\theta)}}$
• But cos (180-θ) = -cos θ. So we get:
Eq.4.4(a):
$\small{\left | \vec{R} \right |=\left ( \left | \vec{A}-\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 - 2\left | \vec{A} \right |\left | \vec{B} \right |cos \theta}}$
(ii) Direction of resultant:
$\tan\alpha =\frac{\left | \vec{B} \right |\sin(180-\theta )}{\left | \vec{A} \right |+\left | \vec{B} \right |\cos(180-\theta )}$
• But sin (180-θ) = sin θ and 
• cos (180-θ) = -cos θ. So we get:
Eq.4.5(a):
$\tan\alpha =\frac{\left | \vec{B} \right |\sin \theta}{\left | \vec{A} \right |-\left | \vec{B} \right |\cos \theta}$
■ In the same way, the other equations can also be written. Readers are advised to write all steps in their own notebooks and derive Eqs.4.4(a), 4.5(a), 4.6(a) and 4.7(a) independently.

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■ A  note about 'angle between two vectors':

1. Consider fig.4.19(c) above.
• Two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ are placed in such a way that, the head of $\small{\vec{B}}$ coincide with the tail of $\small{\vec{A}}$
2. In this position, the angle between the two vectors is $\theta_1$.
• But this is not the correct way for finding the angle between two vectors
3. Both the tails must coincide at the same point. This is shown in fig.d
• In this position, the angle between the two vectors is $\theta_2$.
■ So we can write:
• The angle $\theta$ between the two vectors is $\theta_2$.

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In the next section, we will see vectors applied to motion in a plane.

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