Rectangular components of a vector MCQ Quiz - Objective Question with Answer for Rectangular components of a vector - Download Free PDF

CONCEPT:

The dot product of two vectors

• The dot product or scalar product is the sum of the products of the corresponding entries of the two sequences of numbers.
• Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them.
• Let the two vectors are A and B then dot Product of two vectors:

⇒  A⋅B = |A||B| cos θ 

Where |A| = Magnitude of vectors A and |B| = Magnitude of vectors B and θ is the angle between A and B.

The above equation can be rewritten as:

$$\begin{gathered} \Rightarrow \overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|cos\theta \\ \Rightarrow cos\theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \end{gathered}$$

• The dot product of unit vectors

$$\begin{gathered} \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1 \\ \hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0 \end{gathered}$$

​CALCULATION:

Given- The two vectors are $\overrightarrow{A}=4\hat{i}+3\hat{j}-4\hat{k}$ and $\overrightarrow{B}=4\hat{i}+3\hat{j}+6\hat{k}$.

• The dot product of given vectors:

$$\begin{gathered} \Rightarrow \overrightarrow{A}.\overrightarrow{B}=(4\hat{i}+3\hat{j}-4\hat{k}).(4\hat{i}+3\hat{j}+6\hat{k}) \\ \Rightarrow \overrightarrow{A}.\overrightarrow{B}=(4\times 4)(\hat{i}.\hat{i})+(3\times 3)(\hat{j}.\hat{j})+(-4\times 6)(\hat{k}.\hat{k}) \\ \Rightarrow \overrightarrow{A}.\overrightarrow{B}=16+9-24=1 \end{gathered}$$

• The magnitude of vector A:

$|\overrightarrow{A}|=\sqrt{4^2+3^2+4^2}=\sqrt{41}$

• The magnitude of vector B:

$|\overrightarrow{B}|=\sqrt{4^2+3^2+6^2}=\sqrt{61}$

• The angle between vectors A and B:

$$\begin{gathered} \Rightarrow \theta =co{s}^{-1}\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \\ \Rightarrow \theta =co{s}^{-1}\frac{1}{\sqrt{41}\times \sqrt{61}} \\ \Rightarrow \theta ={88.8}^o\approx {89}^{\circ } \end{gathered}$$

Therefore, option 4 is correct.

Important Points

• Cross product of unit vectors

$$\begin{gathered} \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0 \\ \hat{i}\times \hat{j}=\hat{k} \\ \hat{j}\times \hat{k}=\hat{i} \\ \hat{k}\times \hat{i}=\hat{j} \end{gathered}$$​

CONCEPT:​

• Resolution of vectors into components: We have a vector (F) where the magnitude of the vector is F and the angle with horizontal is θ.

The vector has two components: 1. Vertical component and 2. Horizontal component

Vertical component (Fy) = F Sinθ

Horizontal component (Fx) = F Cosθ 

CALCULATION:

• Resolving tension T₁ and T₂ into two rectangular components, we have

T₁ cos 30° acts horizontally and T1 sin 30° vertically upwards

T2 cos 60° acts horizontally and T2 sin 60° vertically upwards

• As the system is in equilibrium, so

⇒ T1 cos 30° = T2 cos 60°

$\Rightarrow \frac{\surd 3T_1}{2}=\frac{T_2}{2}$

⇒ T₂ = √3T₁ 

And,

⇒ T1 sin30° + T2 sin60° = W

$\Rightarrow \frac{T_1}{2}+\frac{\sqrt{3}{T}_2}{2}=W$

$\Rightarrow \frac{T_1}{2}+\frac{3T_1}{2}=W$          [∵ T2 = √3T1]

$\Rightarrow 2T_1=W$

$\Rightarrow T_1=\frac{W}{2}$ and $T_2=\frac{\sqrt{3}W}{2}$

Concept:

If two forces of equal magnitude (F) act at an angle of β to each other, then the resultant of these forces is given by

$\mathbf{R}=\sqrt{{\mathbf{F}}^2+{\mathbf{F}}^2+2\times \mathbf{F}\times \mathbf{F}\times \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\beta }}$

$R=\sqrt{2{\mathrm{F}}^2+(2{\mathrm{F}}^2\times \mathrm{c}\mathrm{o}\mathrm{s}\beta )}$

$R=\sqrt{2F^2(1+\cos \beta )}$       

 $\therefore \mathbf{R}=2\mathbf{F}\mathbf{c}\mathbf{o}\mathbf{s}\left(\frac{\mathbf{\beta }}{2}\right)$      $(\because 1+\mathrm{c}\mathrm{o}\mathrm{s}\beta =2{\cos }^2\left(\frac{\beta }{2}\right))$

CONCEPT:

The dot product of two vectors

• The dot product or scalar product is the sum of the products of the corresponding entries of the two sequences of numbers.
• Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them.
• Let the two vectors are A and B then dot Product of two vectors:

⇒  A⋅B = |A||B| cos θ 

Where |A| = Magnitude of vectors A and |B| = Magnitude of vectors B and θ is the angle between A and B.

The above equation can be rewritten as:

$$\begin{gathered} \Rightarrow \overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|cos\theta \\ \Rightarrow cos\theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \end{gathered}$$

• The dot product of unit vectors

$$\begin{gathered} \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1 \\ \hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0 \end{gathered}$$

​CALCULATION:

Given- The two vectors are $\overrightarrow{A}=4\hat{i}+3\hat{j}-4\hat{k}$ and $\overrightarrow{B}=4\hat{i}+3\hat{j}+6\hat{k}$.

• The dot product of given vectors:

$$\begin{gathered} \Rightarrow \overrightarrow{A}.\overrightarrow{B}=(4\hat{i}+3\hat{j}-4\hat{k}).(4\hat{i}+3\hat{j}+6\hat{k}) \\ \Rightarrow \overrightarrow{A}.\overrightarrow{B}=(4\times 4)(\hat{i}.\hat{i})+(3\times 3)(\hat{j}.\hat{j})+(-4\times 6)(\hat{k}.\hat{k}) \\ \Rightarrow \overrightarrow{A}.\overrightarrow{B}=16+9-24=1 \end{gathered}$$

• The magnitude of vector A:

$|\overrightarrow{A}|=\sqrt{4^2+3^2+4^2}=\sqrt{41}$

• The magnitude of vector B:

$|\overrightarrow{B}|=\sqrt{4^2+3^2+6^2}=\sqrt{61}$

• The angle between vectors A and B:

$$\begin{gathered} \Rightarrow \theta =co{s}^{-1}\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \\ \Rightarrow \theta =co{s}^{-1}\frac{1}{\sqrt{41}\times \sqrt{61}} \\ \Rightarrow \theta ={88.8}^o\approx {89}^{\circ } \end{gathered}$$

Therefore, option 4 is correct.

Important Points

• Cross product of unit vectors

$$\begin{gathered} \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0 \\ \hat{i}\times \hat{j}=\hat{k} \\ \hat{j}\times \hat{k}=\hat{i} \\ \hat{k}\times \hat{i}=\hat{j} \end{gathered}$$​

Concept:

If two forces of equal magnitude (F) act at an angle of β to each other, then the resultant of these forces is given by

$\mathbf{R}=\sqrt{{\mathbf{F}}^2+{\mathbf{F}}^2+2\times \mathbf{F}\times \mathbf{F}\times \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\beta }}$

$R=\sqrt{2{\mathrm{F}}^2+(2{\mathrm{F}}^2\times \mathrm{c}\mathrm{o}\mathrm{s}\beta )}$

$R=\sqrt{2F^2(1+\cos \beta )}$       

 $\therefore \mathbf{R}=2\mathbf{F}\mathbf{c}\mathbf{o}\mathbf{s}\left(\frac{\mathbf{\beta }}{2}\right)$      $(\because 1+\mathrm{c}\mathrm{o}\mathrm{s}\beta =2{\cos }^2\left(\frac{\beta }{2}\right))$

CONCEPT: 

The Resultant of two vectors is given by:

R2 = (A2 + B2 + 2AB cosθ) 

where R is the resultant vector, A and B are the two vectors, and θ is the angle between two vectors.

• The angle between resultant and one vector: If α is the angle between resultant and one of the vectors, then

$tan\alpha =\frac{Bsin\theta }{A+Bcos\theta }$

where α is the angle between resultant and vector A, and θ is the angle between both vectors.

CALCULATION:

Given that R = 10 kg wt; θ = 120°; α = 90°

$tan\alpha =\frac{Bsin\theta }{A+Bcos\theta }$

$tan90=\frac{F_{2}sin\theta }{F_1+F_{2}cos\theta }$

F₁ + F₂ cosθ = 0

F1 + F2 cos120 = 0

F1 = F2 (1/2) .............(i)

Resultant R2 = F₁2 + F₂2 + 2F₁F₂ cosθ 

10² = F12 + 4F12 + 2F1(2F₁) cos 120

100 = 5F₁² - 2F₁²

F₁ = $\frac{10}{\sqrt{3}}kgwt$

So the correct answer is option 4.

CONCEPT:​

• Resolution of vectors into components: We have a vector (F) where the magnitude of the vector is F and the angle with horizontal is θ.

The vector has two components: 1. Vertical component and 2. Horizontal component

Vertical component (Fy) = F Sinθ

Horizontal component (Fx) = F Cosθ 

CALCULATION:

• Resolving tension T₁ and T₂ into two rectangular components, we have

T₁ cos 30° acts horizontally and T1 sin 30° vertically upwards

T2 cos 60° acts horizontally and T2 sin 60° vertically upwards

• As the system is in equilibrium, so

⇒ T1 cos 30° = T2 cos 60°

$\Rightarrow \frac{\surd 3T_1}{2}=\frac{T_2}{2}$

⇒ T₂ = √3T₁ 

And,

⇒ T1 sin30° + T2 sin60° = W

$\Rightarrow \frac{T_1}{2}+\frac{\sqrt{3}{T}_2}{2}=W$

$\Rightarrow \frac{T_1}{2}+\frac{3T_1}{2}=W$          [∵ T2 = √3T1]

$\Rightarrow 2T_1=W$

$\Rightarrow T_1=\frac{W}{2}$ and $T_2=\frac{\sqrt{3}W}{2}$

CONCEPT:

The dot product of two vectors

• The dot product or scalar product is the sum of the products of the corresponding entries of the two sequences of numbers.
• Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them.
• Let the two vectors are A and B then dot Product of two vectors:

⇒  A⋅B = |A||B| cos θ 

Where |A| = Magnitude of vectors A and |B| = Magnitude of vectors B and θ is the angle between A and B.

The above equation can be rewritten as:

$$\begin{gathered} \Rightarrow \overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|cos\theta \\ \Rightarrow cos\theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \end{gathered}$$

• The dot product of unit vectors

$$\begin{gathered} \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1 \\ \hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0 \end{gathered}$$

​CALCULATION:

Given- The two vectors are $\overrightarrow{A}=4\hat{i}+3\hat{j}-4\hat{k}$ and $\overrightarrow{B}=4\hat{i}+3\hat{j}+6\hat{k}$.

• The dot product of given vectors:

$$\begin{gathered} \Rightarrow \overrightarrow{A}.\overrightarrow{B}=(4\hat{i}+3\hat{j}-4\hat{k}).(4\hat{i}+3\hat{j}+6\hat{k}) \\ \Rightarrow \overrightarrow{A}.\overrightarrow{B}=(4\times 4)(\hat{i}.\hat{i})+(3\times 3)(\hat{j}.\hat{j})+(-4\times 6)(\hat{k}.\hat{k}) \\ \Rightarrow \overrightarrow{A}.\overrightarrow{B}=16+9-24=1 \end{gathered}$$

• The magnitude of vector A:

$|\overrightarrow{A}|=\sqrt{4^2+3^2+4^2}=\sqrt{41}$

• The magnitude of vector B:

$|\overrightarrow{B}|=\sqrt{4^2+3^2+6^2}=\sqrt{61}$

• The angle between vectors A and B:

$$\begin{gathered} \Rightarrow \theta =co{s}^{-1}\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \\ \Rightarrow \theta =co{s}^{-1}\frac{1}{\sqrt{41}\times \sqrt{61}} \\ \Rightarrow \theta ={88.8}^o\approx {89}^{\circ } \end{gathered}$$

Therefore, option 4 is correct.

Important Points

• Cross product of unit vectors

$$\begin{gathered} \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0 \\ \hat{i}\times \hat{j}=\hat{k} \\ \hat{j}\times \hat{k}=\hat{i} \\ \hat{k}\times \hat{i}=\hat{j} \end{gathered}$$​

Concept:

If two forces of equal magnitude (F) act at an angle of β to each other, then the resultant of these forces is given by

$\mathbf{R}=\sqrt{{\mathbf{F}}^2+{\mathbf{F}}^2+2\times \mathbf{F}\times \mathbf{F}\times \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\beta }}$

$R=\sqrt{2{\mathrm{F}}^2+(2{\mathrm{F}}^2\times \mathrm{c}\mathrm{o}\mathrm{s}\beta )}$

$R=\sqrt{2F^2(1+\cos \beta )}$       

 $\therefore \mathbf{R}=2\mathbf{F}\mathbf{c}\mathbf{o}\mathbf{s}\left(\frac{\mathbf{\beta }}{2}\right)$      $(\because 1+\mathrm{c}\mathrm{o}\mathrm{s}\beta =2{\cos }^2\left(\frac{\beta }{2}\right))$

CONCEPT: 

The Resultant of two vectors is given by:

R2 = (A2 + B2 + 2AB cosθ) 

where R is the resultant vector, A and B are the two vectors, and θ is the angle between two vectors.

• The angle between resultant and one vector: If α is the angle between resultant and one of the vectors, then

$tan\alpha =\frac{Bsin\theta }{A+Bcos\theta }$

where α is the angle between resultant and vector A, and θ is the angle between both vectors.

CALCULATION:

Given that R = 10 kg wt; θ = 120°; α = 90°

$tan\alpha =\frac{Bsin\theta }{A+Bcos\theta }$

$tan90=\frac{F_{2}sin\theta }{F_1+F_{2}cos\theta }$

F₁ + F₂ cosθ = 0

F1 + F2 cos120 = 0

F1 = F2 (1/2) .............(i)

Resultant R2 = F₁2 + F₂2 + 2F₁F₂ cosθ 

10² = F12 + 4F12 + 2F1(2F₁) cos 120

100 = 5F₁² - 2F₁²

F₁ = $\frac{10}{\sqrt{3}}kgwt$

So the correct answer is option 4.

CONCEPT:​

• Resolution of vectors into components: We have a vector (F) where the magnitude of the vector is F and the angle with horizontal is θ.

The vector has two components: 1. Vertical component and 2. Horizontal component

Vertical component (Fy) = F Sinθ

Horizontal component (Fx) = F Cosθ 

CALCULATION:

• Resolving tension T₁ and T₂ into two rectangular components, we have

T₁ cos 30° acts horizontally and T1 sin 30° vertically upwards

T2 cos 60° acts horizontally and T2 sin 60° vertically upwards

• As the system is in equilibrium, so

⇒ T1 cos 30° = T2 cos 60°

$\Rightarrow \frac{\surd 3T_1}{2}=\frac{T_2}{2}$

⇒ T₂ = √3T₁ 

And,

⇒ T1 sin30° + T2 sin60° = W

$\Rightarrow \frac{T_1}{2}+\frac{\sqrt{3}{T}_2}{2}=W$

$\Rightarrow \frac{T_1}{2}+\frac{3T_1}{2}=W$          [∵ T2 = √3T1]

$\Rightarrow 2T_1=W$

$\Rightarrow T_1=\frac{W}{2}$ and $T_2=\frac{\sqrt{3}W}{2}$