In the previous section, we saw the details about cross products. In this section we will see some solved examples
Solved example 7.11
Show that the area of the triangle contained in between $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ is one half the magnitude of ($\mathbf\small{\vec{a}\times \vec{b}}$)
Solution:
1. To find the cross product of two vectors, we bring their tail ends to a single point
• This point can be considered as one of the 3 vertices of a triangle
• Then the two vectors will become two sides of that triangle. This situation is shown in fig.7.72(a) below:
• The angle between the two vectors is denoted as θ
• The third side is shown as a dashed line
2. In fig.b, a perpendicular (shown in red color) is dropped from the apex onto the base
• The length of this perpendicular, is obviously $\mathbf\small{|\vec{a}|\times \sin \theta}$
• This 'perpendicular' is the altitude of the triangle
3. Now, area of any triangle is given by: $\mathbf\small{\frac{1}{2}\times \text{base}\times \text{altitude}}$
• Thus we get: Area of our present triangle = $\mathbf\small{\frac{1}{2}\times |\vec{b}|\times |\vec{a}|\times \sin \theta}$
• Rearranging this, we get: Area of the triangle = $\mathbf\small{\frac{1}{2}\times (|\vec{a}|\times |\vec{b}|\times \sin \theta)}$
4. But $\mathbf\small{(|\vec{a}|\times |\vec{b}|\times \sin \theta)}$ is $\mathbf\small{|(\vec{a}\times \vec{b})|}$
• Thus we get: Area of the triangle = $\mathbf\small{\frac{1}{2}\times |(\vec{a}\times \vec{b})|}$
Solved example 7.12
Show that, the area of a parallelogram whose adjacent sides are $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ is equal to the magnitude of ($\mathbf\small{\vec{a}\times \vec{b}}$)
• This point can be considered as one of the 4 vertices of a parallelogram
• Then the two vectors will become two adjacent sides of that parallelogram. This situation is shown in fig.7.73(a) below:
• The angle between the two vectors is denoted as θ
• The third and fourth sides are shown as a dashed lines
2. In fig.b, a perpendicular (shown in red color) is dropped from the top left vertex onto the base
• The length of this perpendicular, is obviously $\mathbf\small{|\vec{a}|\times \sin \theta}$
• This 'perpendicular' is the height of the parellelogram
3. Now, area of any parallelogram is given by: base × height
• Thus we get: Area of our present parallelogram = $\mathbf\small{|\vec{b}|\times |\vec{a}|\times \sin \theta}$
• Rearranging this, we get: Area of the parallelogram = $\mathbf\small{(|\vec{a}|\times |\vec{b}|\times \sin \theta)}$
4. But $\mathbf\small{(|\vec{a}|\times |\vec{b}|\times \sin \theta)}$ is |($\mathbf\small{\vec{a}\times \vec{b}}$)|
• Thus we get: Area of the parallelogram = |$\mathbf\small{(\vec{a}\times \vec{b})}$|
Solved example 7.13
Show that $\mathbf\small{\vec{a}.(\vec{b}\times \vec{c})}$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors $\mathbf\small{\vec{a}}$, $\mathbf\small{\vec{b}}$ and $\mathbf\small{\vec{c}}$
Solution:
Note:
• If we see a '×' sign between two vectors, it indicates a 'cross product
• If the '×' sign is not in between two vectors, it indicates an ordinary multiplication between two scalars
1. In fig.7.74(a) below, a rectangular prism rests on a plane surface
• It has 6 faces:
♦ Front, back
♦ Left, right
♦ Top, bottom
• All the 6 faces are rectangles
2. It rests on it's bottom face
• Keeping this bottom face fixed, we slide the top face horizontally towards the right as indicated by the red arrow in fig.b
• Now the front and back faces are no longer rectangles. They are parallelograms
• The newly formed front and back parallelograms are identical
■ The object has now become a parallelepiped
3. We can further slide the top face horizontally towards the front or rear. Then the sides also will become parallelograms
• In general, a parallelepiped is a 6 sided solid, which has identical parallelograms on opposite sides
• For our present problem, we will consider the parallelepiped in fig.b
• It is shown again in fig.7.75(a) below:
4. We see that, the vectors $\mathbf\small{\vec{a}}$, $\mathbf\small{\vec{b}}$ and $\mathbf\small{\vec{c}}$ form the adjacent edges of the parallelepiped
• Vectors have both magnitude and direction. So, if three vectors form adjacent edges at a vertex, that parallelepiped can be completely defined
5. Based on the previous solved example 7.12, we can write:
• Area of the base of the parallelepiped in fig.7.75(a) = $\mathbf\small{|(\vec{b}\times \vec{c})|}$
• If we multiply this area by the height of the parallelepiped, we will get the volume
6. So our next aim is to find the height
• In fig.7.75(b), the vector $\mathbf\small{(\vec{b}\times \vec{c})}$ is shown in green color
• That green vector will be perpendicular to the base of the parallelepiped
• So, if we can find the projection of $\mathbf\small{\vec{a}}$ on to the green vector, we get the height
• That is., we want to find the projection of $\mathbf\small{\vec{a}}$ onto $\mathbf\small{(\vec{b}\times \vec{c})}$
(This is clear from the black dashed line. This dashed line is drawn in alignment with the top edge of the parallelepiped)
7. If the angle between $\mathbf\small{\vec{a}}$ and $\mathbf\small{(\vec{b}\times \vec{c})}$ is Φ, then we get:
• The required projection = height of the parallelepiped = $\mathbf\small{|\vec{a}|\times \cos \Phi}$
• Note that, this result is a scalar quantity
8. So volume of the parallelepiped = area × height = $\mathbf\small{|(\vec{b}\times \vec{c})|\times |\vec{a}|\times\cos \Phi}$
• But '$\mathbf\small{|(\vec{b}\times \vec{c})|\times |\vec{a}|\times\cos \Phi}$' is the scalar product of two vectors: $\mathbf\small{(\vec{b}\times \vec{c})}$ and $\mathbf\small{\vec{a}}$
♦ Where Φ is the angle between those two vectors
(Details here)
9. Thus we can write:
• Volume of the parallelepiped = magnitude of [$\mathbf\small{(\vec{b}\times \vec{c}).\vec{a}}$]
• Since scalar products obeys commutative law, we can write:
$\mathbf\small{(\vec{b}\times \vec{c}).\vec{a}=\vec{a}.(\vec{b}\times \vec{c})}$
Thus we get:
• Volume of the parallelepiped = magnitude of $\mathbf\small{[\vec{a}.(\vec{b}\times \vec{c})]}$
10. Why do we specifically write: 'magnitude of'?
Ans: volume is a scalar quantity. So it's magnitude is what we want
• If instead of $\mathbf\small{(\vec{b}\times \vec{c})}$, we take $\mathbf\small{(\vec{c}\times \vec{b})}$, the final result will have an opposite sign
• But the magnitudes will be the same
Solved example 7.14
Find a unit vector which is perpendicular to both the vectors given below:
$\mathbf\small{\vec{a}=2\hat{i}+3\hat{j}+\hat{k}}$
$\mathbf\small{\vec{b}=\hat{i}-\hat{j}+\hat{k}}$
Solution:
1. The cross product of any two vectors will be perpendicular to both of them
• So we first find the cross product $\mathbf\small{(\vec{a}\times\vec{b})}$ of the given vectors
• The table is shown below:
2. We get: $\mathbf\small{(\vec{a}\times\vec{b})=(1+3)\hat{i}+(1-2)\hat{j}+(-3+-2)\hat{k}=4\hat{i}-\hat{j}-5\hat{k}}$
• So magnitude of $\mathbf\small{(\vec{a}\times\vec{b})}$ = $\mathbf\small{|(\vec{a}\times\vec{b})|}$ = $\mathbf\small{\sqrt{(4)^2+(-1)^2+(-5)^2}=\sqrt{42}}$
• Thus we get:
Unit vector perpendicular to both $\mathbf\small{\vec{a}\;\text{and}\;\vec{b}}$ = $\mathbf\small{\frac{1}{\sqrt{42}}(4\hat{i}-\hat{j}-5\hat{k})}$
Solved example 7.11
Show that the area of the triangle contained in between $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ is one half the magnitude of ($\mathbf\small{\vec{a}\times \vec{b}}$)
Solution:
1. To find the cross product of two vectors, we bring their tail ends to a single point
• This point can be considered as one of the 3 vertices of a triangle
• Then the two vectors will become two sides of that triangle. This situation is shown in fig.7.72(a) below:
 |
| Fig.7.72 |
• The third side is shown as a dashed line
2. In fig.b, a perpendicular (shown in red color) is dropped from the apex onto the base
• The length of this perpendicular, is obviously $\mathbf\small{|\vec{a}|\times \sin \theta}$
• This 'perpendicular' is the altitude of the triangle
3. Now, area of any triangle is given by: $\mathbf\small{\frac{1}{2}\times \text{base}\times \text{altitude}}$
• Thus we get: Area of our present triangle = $\mathbf\small{\frac{1}{2}\times |\vec{b}|\times |\vec{a}|\times \sin \theta}$
• Rearranging this, we get: Area of the triangle = $\mathbf\small{\frac{1}{2}\times (|\vec{a}|\times |\vec{b}|\times \sin \theta)}$
4. But $\mathbf\small{(|\vec{a}|\times |\vec{b}|\times \sin \theta)}$ is $\mathbf\small{|(\vec{a}\times \vec{b})|}$
• Thus we get: Area of the triangle = $\mathbf\small{\frac{1}{2}\times |(\vec{a}\times \vec{b})|}$
Solved example 7.12
Show that, the area of a parallelogram whose adjacent sides are $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ is equal to the magnitude of ($\mathbf\small{\vec{a}\times \vec{b}}$)
Solution:
To find the cross product of two vectors, we bring their tail ends to a single point• This point can be considered as one of the 4 vertices of a parallelogram
• Then the two vectors will become two adjacent sides of that parallelogram. This situation is shown in fig.7.73(a) below:
 |
| Fig.7.73 |
• The third and fourth sides are shown as a dashed lines
2. In fig.b, a perpendicular (shown in red color) is dropped from the top left vertex onto the base
• The length of this perpendicular, is obviously $\mathbf\small{|\vec{a}|\times \sin \theta}$
• This 'perpendicular' is the height of the parellelogram
3. Now, area of any parallelogram is given by: base × height
• Thus we get: Area of our present parallelogram = $\mathbf\small{|\vec{b}|\times |\vec{a}|\times \sin \theta}$
• Rearranging this, we get: Area of the parallelogram = $\mathbf\small{(|\vec{a}|\times |\vec{b}|\times \sin \theta)}$
4. But $\mathbf\small{(|\vec{a}|\times |\vec{b}|\times \sin \theta)}$ is |($\mathbf\small{\vec{a}\times \vec{b}}$)|
• Thus we get: Area of the parallelogram = |$\mathbf\small{(\vec{a}\times \vec{b})}$|
Solved example 7.13
Show that $\mathbf\small{\vec{a}.(\vec{b}\times \vec{c})}$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors $\mathbf\small{\vec{a}}$, $\mathbf\small{\vec{b}}$ and $\mathbf\small{\vec{c}}$
Solution:
Note:
• If we see a '×' sign between two vectors, it indicates a 'cross product
• If the '×' sign is not in between two vectors, it indicates an ordinary multiplication between two scalars
1. In fig.7.74(a) below, a rectangular prism rests on a plane surface
 |
| Fig.7.74 |
♦ Front, back
♦ Left, right
♦ Top, bottom
• All the 6 faces are rectangles
2. It rests on it's bottom face
• Keeping this bottom face fixed, we slide the top face horizontally towards the right as indicated by the red arrow in fig.b
• Now the front and back faces are no longer rectangles. They are parallelograms
• The newly formed front and back parallelograms are identical
■ The object has now become a parallelepiped
3. We can further slide the top face horizontally towards the front or rear. Then the sides also will become parallelograms
• In general, a parallelepiped is a 6 sided solid, which has identical parallelograms on opposite sides
• For our present problem, we will consider the parallelepiped in fig.b
• It is shown again in fig.7.75(a) below:
 |
| Fig.7.75 |
• Vectors have both magnitude and direction. So, if three vectors form adjacent edges at a vertex, that parallelepiped can be completely defined
5. Based on the previous solved example 7.12, we can write:
• Area of the base of the parallelepiped in fig.7.75(a) = $\mathbf\small{|(\vec{b}\times \vec{c})|}$
• If we multiply this area by the height of the parallelepiped, we will get the volume
6. So our next aim is to find the height
• In fig.7.75(b), the vector $\mathbf\small{(\vec{b}\times \vec{c})}$ is shown in green color
• That green vector will be perpendicular to the base of the parallelepiped
• So, if we can find the projection of $\mathbf\small{\vec{a}}$ on to the green vector, we get the height
• That is., we want to find the projection of $\mathbf\small{\vec{a}}$ onto $\mathbf\small{(\vec{b}\times \vec{c})}$
(This is clear from the black dashed line. This dashed line is drawn in alignment with the top edge of the parallelepiped)
7. If the angle between $\mathbf\small{\vec{a}}$ and $\mathbf\small{(\vec{b}\times \vec{c})}$ is Φ, then we get:
• The required projection = height of the parallelepiped = $\mathbf\small{|\vec{a}|\times \cos \Phi}$
• Note that, this result is a scalar quantity
8. So volume of the parallelepiped = area × height = $\mathbf\small{|(\vec{b}\times \vec{c})|\times |\vec{a}|\times\cos \Phi}$
• But '$\mathbf\small{|(\vec{b}\times \vec{c})|\times |\vec{a}|\times\cos \Phi}$' is the scalar product of two vectors: $\mathbf\small{(\vec{b}\times \vec{c})}$ and $\mathbf\small{\vec{a}}$
♦ Where Φ is the angle between those two vectors
(Details here)
9. Thus we can write:
• Volume of the parallelepiped = magnitude of [$\mathbf\small{(\vec{b}\times \vec{c}).\vec{a}}$]
• Since scalar products obeys commutative law, we can write:
$\mathbf\small{(\vec{b}\times \vec{c}).\vec{a}=\vec{a}.(\vec{b}\times \vec{c})}$
Thus we get:
• Volume of the parallelepiped = magnitude of $\mathbf\small{[\vec{a}.(\vec{b}\times \vec{c})]}$
10. Why do we specifically write: 'magnitude of'?
Ans: volume is a scalar quantity. So it's magnitude is what we want
• If instead of $\mathbf\small{(\vec{b}\times \vec{c})}$, we take $\mathbf\small{(\vec{c}\times \vec{b})}$, the final result will have an opposite sign
• But the magnitudes will be the same
Solved example 7.14
Find a unit vector which is perpendicular to both the vectors given below:
$\mathbf\small{\vec{a}=2\hat{i}+3\hat{j}+\hat{k}}$
$\mathbf\small{\vec{b}=\hat{i}-\hat{j}+\hat{k}}$
Solution:
1. The cross product of any two vectors will be perpendicular to both of them
• So we first find the cross product $\mathbf\small{(\vec{a}\times\vec{b})}$ of the given vectors
• The table is shown below:
• So magnitude of $\mathbf\small{(\vec{a}\times\vec{b})}$ = $\mathbf\small{|(\vec{a}\times\vec{b})|}$ = $\mathbf\small{\sqrt{(4)^2+(-1)^2+(-5)^2}=\sqrt{42}}$
• Thus we get:
Unit vector perpendicular to both $\mathbf\small{\vec{a}\;\text{and}\;\vec{b}}$ = $\mathbf\small{\frac{1}{\sqrt{42}}(4\hat{i}-\hat{j}-5\hat{k})}$
In the next section, we will see angular velocity
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