Chapter 6.2 - Solved examples on Scalar product

In the previous section, we saw the properties of scalar product of vectors. In this section, we will continue that discussion and see some properties related to unit vectors. After that we will see some solved examples also.

1. What is $\mathbf\small{\hat{i}.\hat{i}}$?
Answer:
• This is a dot product of two vectors $\mathbf\small{\hat{i}\;\; \text{and}\;\;\hat{i}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is 1
(ii) The magnitude of the second vector
• In this case, it is 1
(iii) Cosine of the angle θ between the two vectors
• In this case, both the vectors lie along the x axis. That means they are parallel
• The angle θ between them will be 0
• So cos θ = cos 0 = 1
■ Thus we get: $\mathbf\small{\hat{i}.\hat{i}=1 \times 1 \times \cos 0 =1 \times 1 \times 1 = 1}$
2. What is $\mathbf\small{\hat{j}.\hat{j}}$?
Answer:
• This is a dot product of two vectors $\mathbf\small{\hat{j}\;\; \text{and}\;\;\hat{j}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is 1
(ii) The magnitude of the second vector
• In this case, it is 1
(iii) Cosine of the angle θ between the two vectors
• In this case, both the vectors lie along the y axis. That means they are parallel
• The angle θ between them will be 0
• So cos θ = cos 0 = 1
■ Thus we get: $\mathbf\small{\hat{j}.\hat{j}=1 \times 1 \times \cos 0 =1 \times 1 \times 1 = 1}$
3. Using the same procedure, we will get: $\mathbf\small{\hat{k}.\hat{k}=1}$
■ We will write them as a result for easy reference
Eq.6.13:
$\mathbf\small{\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1}$

4. What is $\mathbf\small{\hat{i}.\hat{j}}$?

Answer:
• This is a dot product of two vectors $\mathbf\small{\hat{i}\;\; \text{and}\;\;\hat{j}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is 1
(ii) The magnitude of the second vector
• In this case, it is 1
(iii) Cosine of the angle θ between the two vectors
• In this case, the first vector lies along the x axis and the second vector lies along the y axis. That means they are perpendicular
• The angle θ between them will be 90^o
• So cos θ = cos 90 = 0
(This is obvious because, we cannot project a vector onto another perpendicular vector)
■ Thus we get: $\mathbf\small{\hat{i}.\hat{i}=1 \times 1 \times \cos 90 =1 \times 1 \times 0 = 0}$
5. What is $\mathbf\small{\hat{i}.\hat{k}}$?
Answer:
• This is a dot product of two vectors $\mathbf\small{\hat{i}\;\; \text{and}\;\;\hat{k}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is 1
(ii) The magnitude of the second vector
• In this case, it is 1
(iii) Cosine of the angle θ between the two vectors
• In this case, the first vector lies along the x axis and the second vector lies along the z axis. That means they are perpendicular
• The angle θ between them will be 90^o
• So cos θ = cos 90 = 0
(This is obvious because, we cannot project a vector onto another perpendicular vector)
■ Thus we get: $\mathbf\small{\hat{i}.\hat{k}=1 \times 1 \times \cos 90 =1 \times 1 \times 0 = 0}$
3. Using the same procedure, we will get: $\mathbf\small{\hat{j}.\hat{k}=0}$
■ We will write them as a result for easy reference
Eq.6.14:
$\mathbf\small{\hat{i}.\hat{j}=\hat{i}.\hat{k}=\hat{j}.\hat{k}=0}$

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Let us write all the 14 results together:

Eq.6.1: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
• Where θ is the angle between $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$
Eq.6.2: $\mathbf\small{\vec{A}.\vec{A}=|\vec{A}|^2}$
Eq.6.3: $\mathbf\small{\vec{A}.\vec{B}=\vec{B}.\vec{A}}$
Eq.6.4: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
Eq.6.5:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}}$
$\mathbf\small{\text{And}\;\;\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=A_xB_x+A_yB_y}$
Eq.6.6:
$\mathbf\small{\text{If}\;\;\vec{A}\;\;\text{and}\;\;\vec{B}\;\;\text{are parallel}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=|\vec{A}|\times |\vec{B}|}$
Eq.6.7:
$\mathbf\small{\text{If}\;\;\vec{A}\;\;\text{and}\;\;\vec{B}\;\;\text{are perpendicular}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=0}$
Eq.6.8:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k}}$
$\mathbf\small{\text{And}\;\;\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}+B_z\,\hat{k}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=A_xB_x+A_yB_y+A_zB_z}$
Eq.6.9:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{A}=A_x^2+A_y^2}$
Eq.6.10:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{A}=A_x^2+A_y^2+A_z^2}$
Eq.6.11:
$\mathbf\small{\text{If}\;\;\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}}$
$\mathbf\small{\text{Then}\;\;\lambda\vec{B}=(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$
Eq.6.12:
$\mathbf\small{\vec{A}.(\lambda \vec{B})=\lambda(\vec{A}.\vec{B})}$
Eq.6.13:
$\mathbf\small{\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1}$
Eq.6.14:
$\mathbf\small{\hat{i}.\hat{j}=\hat{i}.\hat{k}=\hat{j}.\hat{k}=0}$

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Now we will see some solved examples

Solved example 6.1
Find the angle between force $\mathbf\small{\vec{F}=3\hat{i}+4\hat{j}-5\hat{k}}$ unit and displacement $\mathbf\small{\vec{d}=5\hat{i}+4\hat{j}+3\hat{k}}$ unit. Also find the projection of $\mathbf\small{\vec{F}}$ on $\mathbf\small{\vec{d}}$.
Solution:
Part (a):
1. Using Eq.6.8, we get: 
$\mathbf\small{\vec{F}.\vec{d}=F_xd_x+F_yd_y+F_zd_z}$ = 3×5 + 4×4 + -5×3 = 16
2. Using Eq.4.2, we have: $\mathbf\small{|\vec{F}|=\sqrt{F_x^2+F_y^2+F_z^2}=\sqrt{3^2+4^2+(-5)^2}=\sqrt{50}}$ 
3. We have: $\mathbf\small{|\vec{d}|=\sqrt{d_x^2+d_y^2+d_z^2}=\sqrt{5^2+4^2+3^2}=\sqrt{50}}$ 
4. Using Eq.6.1, we get: $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}| |\vec{d}| \cos \theta}$
= $\mathbf\small{\sqrt{50} \times \sqrt{50} \times\cos \theta=50 \cos \theta}$
5. Equating (1) and (4), we get: 16 = 50 cos θ
Thus we get: $\mathbf\small{\theta=\cos^{-1}\frac{16}{50}=\cos^{-1}0.32}$
Part (b):
1. Projection of $\mathbf\small{\vec{F}}$ on $\mathbf\small{\vec{d}}$ is $\mathbf\small{|\vec{F}|\cos \theta}$  
2. In (2) above, we calculated $\mathbf\small{|\vec{F}|}$
In (5) above, we calculated $\mathbf\small{\cos \theta}$
3. So $\mathbf\small{|\vec{F}|\cos \theta=\sqrt{50} \times \frac{16}{50}=\frac{16}{\sqrt{50}}}$ 

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Note:
• We know that if $\mathbf\small{\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}}$, then $\mathbf\small{|\vec{A}|=A_x^2+A_y^2+A_z^2}$
• Now, after learning dot product, we get a new method to find $\mathbf\small{|\vec{A}|}$
• We will write the steps:
1. Using Eq.6.10 we have: $\mathbf\small{\vec{A}.\vec{A}=A_x^2+A_y^2+A_z^2}$
2. Using Eq.6.2 we have: $\mathbf\small{\vec{A}.\vec{A}=|\vec{A}|^2}$
3. Equating (1) and (20, we get: $\mathbf\small{|\vec{A}|^2=A_x^2+A_y^2+A_z^2}$
Thus we get: $\mathbf\small{|\vec{A}|=\sqrt{A_x^2+A_y^2+A_z^2}=\frac{\vec{A}.\vec{A}}{|\vec{A}|}}$

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Solved example 6.2
Prove that the vectors $\mathbf\small{\vec{A}=2\hat{i}-3\hat{j}+\hat{k}}$ and $\mathbf\small{\vec{B}=\hat{i}+\hat{j}+\hat{k}}$ are mutually perpendicular
Solution:
1. Using Eq.6.8, we get: $\mathbf\small{\vec{A}.\vec{B}=A_xB_x+A_yB_y+A_zB_z}$ = 2×1 - 3×1 + 1×1 = 0
2. We have $\mathbf\small{|\vec{A}|=\sqrt{A_x^2+A_y^2+A_z^2}=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}}$
3. We have $\mathbf\small{|\vec{B}|=\sqrt{B_x^2+B_y^2+B_z^2}=\sqrt{1^2+1^2+1^2}=\sqrt{3}}$
4. Using Eq.6.3 we get: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta=\sqrt{14} \times \sqrt{3} \times \cos \theta=\sqrt{42}\,\cos \theta}$
5. Equating (1) and (4) we get: $\mathbf\small{0=\sqrt{42}\,\cos \theta}$
$\mathbf\small{\Rightarrow \cos \theta=0\;\;\Rightarrow \theta = 90^o}$

Solved example 6.3
Find the component of ($\mathbf\small{\vec{A}+\vec{B}}$) along (i) x axis (ii) $\mathbf\small{\vec{C}}$
Given that: $\mathbf\small{\vec{A}=\hat{i}-2\hat{j},\;\;\vec{B}=2\hat{i}+3\hat{k},\;\;\vec{C}=\hat{i}+\hat{j}}$
Solution:
Part (a):
1. Let $\mathbf\small{\vec{R}=\vec{A}+\vec{B}}$
Then we get: $\mathbf\small{\vec{R}=(\hat{i}-2\hat{j})+(2\hat{i}+3\hat{k})=3\hat{i}-2\hat{j}+3\hat{k}}$
2. Component of $\mathbf\small{\vec{R}}$ along the x axis is $\mathbf\small{|\vec{R}|\cos \theta}$
• Where θ is the angle between $\mathbf\small{\vec{R}}$ and x axis
• So our next aim is to find θ.
3. The vector representation of x axis is $\mathbf\small{\hat{i}}$
• So dot product of $\mathbf\small{\vec{R}}$ and x axis is: $\mathbf\small{\vec{R}.\hat{i}}$
$\mathbf\small{\vec{R}.\hat{i}=(3\hat{i}-2\hat{j}+3\hat{k}).(\hat{i})=3}$
4. Also we have: $\mathbf\small{\vec{R}.\hat{i}=|\vec{R}||\hat{i}|\cos \theta}$
• $\mathbf\small{|\vec{R}|=\sqrt{3^2+(-2)^2+3^2}=\sqrt{22}}$
• $\mathbf\small{|\hat{i}|=1}$
• Thus we get: $\mathbf\small{\vec{R}.\hat{i}=|\vec{R}||\hat{i}|\cos \theta=\sqrt{22}\cos \theta}$
5. Equating (3) and (4) we get: $\mathbf\small{3=\sqrt{22}\cos \theta}$
• Thus we get: $\mathbf\small{\cos \theta=\frac{3}{\sqrt{22}}}$  
6. From (2), we get:
The required component = $\mathbf\small{\sqrt{22} \times \frac{3}{\sqrt{22}}=3}$
Part (b):
1. We have to find the component of $\mathbf\small{\vec{R}}$ along $\mathbf\small{\vec{C}}$
• The required component is $\mathbf\small{|\vec{R}|\cos \theta_1}$
• Where θ₁ is the angle between $\mathbf\small{\vec{R}}$ and $\mathbf\small{\vec{C}}$
• So our next aim is to find θ₁
2. We have: $\mathbf\small{\vec{R}.\vec{C}=3 \times 1 - 2 \times 1 = 1}$
3. Also we have: $\mathbf\small{\vec{R}.\vec{C}=|\vec{R}||\vec{C}|\cos \theta_1=\sqrt{22}\times \sqrt{2}\times \cos \theta_1}$
4. Equating (8) and (9) we get: $\mathbf\small{1=\sqrt{22}\times \sqrt{2}\times \cos \theta_1}$
$\mathbf\small{\Rightarrow \cos \theta_1=\frac{1}{\sqrt{22}\times \sqrt{2}}}$
5. Thus the required component = $\mathbf\small{|\vec{R}|\cos \theta_1=\sqrt{22}\times\frac{1}{\sqrt{22}\times \sqrt{2}}=\frac{1}{\sqrt{2}}}$

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In the next section, we will see work done by a force

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