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'
• 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: $\mathbf\small{(\vec{a}\times\vec{b})=-(\vec{b}\times\vec{a})}$
• 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
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 ax, ay and az
• 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 ax
• The green line indicates ay
• The blue line indicates az
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) = (ax, ay, az)
4. Now we will see the 'mirror image' of $\mathbf\small{\vec{a}}$
• For that, we change the sign of the components
♦ ax will become -ax
♦ ay will become -ay
♦ az will become -az.
• These new components are shown in fig.7.67(b) above
• An example: -ax is on the negative side of y-axis because, it has to be at a distance -ay 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) = (-ax, -ay, -az)
■ 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 (ax, ay, az) = distance between O and (-ax, -ay, -az)]
(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: $\mathbf\small{\vec{a}'=-\vec{a}}$
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'
Let us see an example to demonstrate the application of the table:
Solution:
1. The table is shown below:
$\mathbf\small{\Rightarrow \vec{a}\times\vec{b}=7\hat{i}-\hat{j}-5\hat{k}}$
2. Table for the reverse multiplication $\mathbf\small{\vec{b}\times\vec{a}}$ can also be formed:
$\mathbf\small{\Rightarrow \vec{a}\times\vec{b}=-7\hat{i}+\hat{j}+5\hat{k}}$
■ We see that: $\mathbf\small{(\vec{a}\times\vec{b})=-(\vec{b}\times\vec{a})}$
Another method for obtaining the 6 terms:
• 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
Mirror image of a vector
• The magnitudes of the components are ax, ay and az
• 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 ax
• The green line indicates ay
• The blue line indicates az
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) = (ax, ay, az)
4. Now we will see the 'mirror image' of $\mathbf\small{\vec{a}}$
• For that, we change the sign of the components
♦ ax will become -ax
♦ ay will become -ay
♦ az will become -az.
• These new components are shown in fig.7.67(b) above
• An example: -ax is on the negative side of y-axis because, it has to be at a distance -ay 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) = (-ax, -ay, -az)
■ 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 (ax, ay, az) = distance between O and (-ax, -ay, -az)]
(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: $\mathbf\small{\vec{a}'=-\vec{a}}$
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'
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 = ax = (x2 - x1)
♦ Magnitude of the y-component = ay = (y2 - y1)
♦ Magnitude of the z-component = az = (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:
• 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 = ax = (x2 - x1)
♦ There is only a difference in sign
• That is: Magnitude of the x-component of the new vector = -ax
Similarly we can write:
• Magnitude of the y-component of the new vector = [(-y2 - (-y1)] = [y1 - y2] = -ay
• Magnitude of the z-component of the new vector = [(-z2 - (-z1)] = [z1 - z2] = -az
• They are shown in fig.7.70 below:
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 $\mathbf\small{\vec{a}}$ = $\mathbf\small{\vec{a}'=-\vec{a}}$
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: $\mathbf\small{\vec{a}'=-\vec{a}}$
• 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:
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: $\mathbf\small{(\vec{a}'\times \vec{b}')=(\vec{a}\times \vec{b})}$
■ 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:
• 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:
• 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:
• 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)
 |
| Fig.7.68 |
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 = ax = (x2 - x1)
♦ Magnitude of the y-component = ay = (y2 - y1)
♦ Magnitude of the z-component = az = (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 |
• 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 = ax = (x2 - x1)
♦ There is only a difference in sign
• That is: Magnitude of the x-component of the new vector = -ax
Similarly we can write:
• Magnitude of the y-component of the new vector = [(-y2 - (-y1)] = [y1 - y2] = -ay
• Magnitude of the z-component of the new vector = [(-z2 - (-z1)] = [z1 - z2] = -az
• They are shown in fig.7.70 below:
 |
| Fig.7.70 |
• 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 $\mathbf\small{\vec{a}}$ = $\mathbf\small{\vec{a}'=-\vec{a}}$
So we have seen two cases:
Case 2: The tail end of the vector is away from O
In both cases, we see that: $\mathbf\small{\vec{a}'=-\vec{a}}$
Now we will see an interesting case:
• 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 |
• 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: $\mathbf\small{(\vec{a}'\times \vec{b}')=(\vec{a}\times \vec{b})}$
Distributive property of vector cross product
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 |
• 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 |
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)
In the next section, we will see some solved examples related to cross products
No comments:
Post a Comment