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 180o
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:
$\mathbf\small{(\vec{a}\times \vec{b})=-(\vec{b}\times \vec{a})}$
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 180o
• This is shown in fig.7.62 below:
 |
| Fig.7.62 |
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'
Properties of Cross products
• In chapter 6, we saw that, dot products obey commutative lawThat 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:
$\mathbf\small{(\vec{a}\times \vec{b})=-(\vec{b}\times \vec{a})}$
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:
(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:
• 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
• 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:
• 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:
We see that:
(i) ax does not combine with bx. Instead, it combines with by and bz
♦ Thus we get: axby and axbz
(ii) ay does not combine with by. Instead, it combines with bx and bz
♦ Thus we get: aybx and aybz
(iii) az does not combine with bz. Instead, it combines with bx and by
♦ Thus we get: azbx and azby
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:
$\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})\times (b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k})}$
= $\mathbf\small{(a_yb_z-a_zb_y)\hat{i}+ (a_zb_x-a_xb_z)\hat{j}+(a_xb_y-a_yb_x)\hat{k}}$
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 |
• 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 |
• 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
• We have completed the discussion on the 'cross products of unit 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:
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| Fig.7.65 |
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:
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| Fig.7.66 |
(i) ax does not combine with bx. Instead, it combines with by and bz
♦ Thus we get: axby and axbz
(ii) ay does not combine with by. Instead, it combines with bx and bz
♦ Thus we get: aybx and aybz
(iii) az does not combine with bz. Instead, it combines with bx and by
♦ Thus we get: azbx and azby
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:
$\mathbf\small{(a_x\,\hat{i}+a_y\,\hat{j}+a_z\,\hat{k})\times (b_x\,\hat{i}+b_y\,\hat{j}+b_z\,\hat{k})}$
= $\mathbf\small{(a_yb_z-a_zb_y)\hat{i}+ (a_zb_x-a_xb_z)\hat{j}+(a_xb_y-a_yb_x)\hat{k}}$
In the next section, we will see a tabular form which is useful to obtain the 6 terms easily
