Informative line

### Taking Components Of Vectors

Learn subtracting & adding vectors using components. Practice Addition & Subtraction of Two Vectors by Taking Component and resolution of forces. Find components of vector along x-axis and y-axis.

# Components of a Vector

• Let, there be a point A(3,4) in x-y plane. Practically, there are infinite ways to reach the point A from origin (O). But, here, we will discuss only one path i.e., from O to B and then B to A.

Vectorially

$$\vec {OB}+\vec {BA}=\vec {OA}$$

These two vectors $$\vec {OB}$$ and $$\vec {BA}$$ are known as the components of $$\vec {OA}$$.

Since $$\vec {OB}$$ is along x-axis, therefore it is called x-component and $$\vec {BA}$$ is along y-axis therefore it is called y-component.

$$\vec {OA}=OB\,\hat i+BA\,\hat j$$

$$\vec {OA}=3\,\hat i+4\hat j$$

## Rectangular components of a Vector

• Components of a vector taken along two mutually perpendicular axes (generally the x and y axis of a Cartesian co-ordinate system), are known as rectangular components.
• Consider a vector $$\vec {OA}$$ making an angle $$\theta_x$$ with x-axis and $$\theta_y$$ with y-axis.

In $$\Delta\; OAB$$

$$cos\,\theta_x=\dfrac {OB}{OA}$$

$$OB=OA\;cos\,\theta_x$$

In $$\Delta\; OAC$$

$$cos\,\theta_y=\dfrac {OC}{OA}$$

$$OC=OA\;cos\,\theta_y$$

$$OB$$ is x-component of $$OA$$

$$OC$$ is y-component of $$OA$$

## To find component along x-axis and y-axis

Let

$$\theta_x=$$ Angle between the vector and x-axis.

$$\theta_y=$$ Angle between the vector and y-axis.

Magnitude of Vector = Length of vector

 Find $$\theta_x$$ and $$\theta_y$$ $$x-$$Component = (Magnitude of vector) × $$cos\;\theta_x$$ $$y-$$Component = (Magnitude of vector ) × $$cos\;\theta_y$$

Note: $$cos\;(90-\theta)=sin\,\theta$$

#### Calculate $$x$$ and $$y$$ components of  $$\vec{OA}$$ . Given $$|\vec {OA}|=10$$

A $$5\sqrt 3, \;5$$

B $$5, \;2$$

C $$2, \;5\sqrt 3$$

D $$10, \;8$$

×

Magnitude of vector $$=|\vec {OA}|=10$$

$$\theta_{x}=30°$$$$\theta_{y}=60°$$

$$x-$$component

=$$|\vec {OA}|×cos\,30°$$

$$=10×\dfrac {\sqrt 3}{2}$$

$$=5 {\sqrt 3}$$

$$y-$$component

=$$|\vec {OA}|×cos\,60°$$

$$=10×\dfrac {1}{2}$$

$$=5$$

### Calculate $$x$$ and $$y$$ components of  $$\vec{OA}$$ . Given $$|\vec {OA}|=10$$

A

$$5\sqrt 3, \;5$$

.

B

$$5, \;2$$

C

$$2, \;5\sqrt 3$$

D

$$10, \;8$$

Option A is Correct

# Rectangular Components of Vector Along Inclined

• Component of vector along an inclined plane can be taken by considering two mutually perpendicular axes (Generally the $$x$$ and $$y$$ axis of the Cartesian coordinate) along the inclined plane, as shown in figure.
• Consider the vector $$\vec {OA}$$ making an angle $$\theta_x$$ with the $$x$$-axis of an inclined plane and $$\theta_y$$ with y-axis of an inclined plane.

X-component of OA along inclined axis

In  $$\triangle\; OAB$$,

$$cos\;\theta_x=\dfrac {OB}{OA}$$

$$OB=OA\,cos\;\theta_x$$

Y-component of OA along inclined axis

In  $$\triangle\; OAC$$,

$$cos\;\theta_y=\dfrac {OC}{OA}$$

$$OC=OA\,cos\;\theta_y$$

• Find component along x-axis and y-axis of an inclined plane.
•            $$\theta_x=$$Angle between the vector and inclined x-axis

$$\theta_y=$$Angle between the vector and inclined y-axis

• Magnitude of vector is same as length of vector.

### Steps to calculate components of vector along inclined

Step-1: Find $$\theta_x$$ and $$\theta_y$$

Step-2: x-component of vector along inclined = Magnitude of Vector × $$cos\;\theta_x$$

Step-3: y-component of vector along inclined = Magnitude of Vector × $$cos\;\theta_y$$

Note: $$cos(90°-\theta)=sin\;\theta$$

#### Calculate the $$x$$ and $$y$$ component of $$\vec {OA}$$ along inclined axis. Given $$|\vec {OA}|=10$$ and $$\theta_x=30°$$

A $$OA_x=10,\;\;OA_y=5$$

B $$OA_x=5\sqrt 3,\;\;OA_y=5$$

C $$OA_x=5,\;\;OA_y=10$$

D $$OA_x=5,\;\;OA_y=5$$

×

Magnitude of vector  $$=|\vec {OA}|=10$$

$$\theta_x=30°$$

$$\theta_y=60°$$

X-component along inclined $$x$$-axis

$$x$$ component $$=|\vec {OA}|×cos\,\theta_x$$

$$=10×cos\,30°$$

$$=10×\dfrac {\sqrt 3}{2}$$

$$=5\;\sqrt 3$$

$$Y$$-component along inclined $$y$$-axis

$$y$$ component $$=|\vec {OA}|×cos\,\theta_y$$

$$=10×cos\,60°$$

$$=10×\dfrac {1}{2}$$

$$=5$$

### Calculate the $$x$$ and $$y$$ component of $$\vec {OA}$$ along inclined axis. Given $$|\vec {OA}|=10$$ and $$\theta_x=30°$$

A

$$OA_x=10,\;\;OA_y=5$$

.

B

$$OA_x=5\sqrt 3,\;\;OA_y=5$$

C

$$OA_x=5,\;\;OA_y=10$$

D

$$OA_x=5,\;\;OA_y=5$$

Option B is Correct

# Applications of Vectors

• Consider a vector $$\vec{OA}$$ making an angle $$\alpha _x$$ with $$x-$$ axis and vector $$\vec{OB}$$ making an angle $$\beta _x$$ with $$x-$$ axis.
• Vector addition of both vectors can be done by adding the component of both vectors along $$x-$$ axis and $$y-$$ axis.

### Components of $$\vec{OA}$$

Component of $$\vec{OA}$$ along $$x-$$ axis

In $$\Delta \,OAB$$

$$cos\,\alpha_x=\dfrac{OB}{OA}$$

$$OB=OA\,cos\,\alpha_x$$

Component of $$\vec{OA}$$ along $$y-$$ axis

In $$\Delta \,OAC$$

$$cos\,\alpha_y=\dfrac{OC}{OA}$$

$$OC=OA\,cos\,\alpha_y$$

### Components of $$\vec{OB}$$

Component of $$\vec{OB}$$ along $$x-$$ axis

In $$\Delta \,OPB$$

$$cos\,\beta_x=\dfrac{OP}{OB}$$

$$OP=OB\,cos\,\beta_x$$

Component of $$\vec{OB}$$ along $$y-$$ axis

In $$\Delta \,OQB$$

$$cos\,\beta_y=\dfrac{OQ}{OB}$$

$$OQ=OB\,cos\,\beta_y$$

### Resultant component along $$x-$$ axis

• Resultant component along $$x-$$ axis can be achieved by summing the components of both $$\vec{OA}$$ and $$\vec{OB}$$ along the $$x-$$ axis.

$$R_x=OA\;cos\,\alpha_x+OB\,cos\,\beta_x$$

### Resultant component along $$y-$$ axis

• Resultant component along $$y-$$ axis can be achieved by summing the components of both $$\vec{OA}$$ and $$\vec{OB}$$ along the $$y-$$ axis.

$$R_y=OA\;cos\,\alpha_y+OB\,cos\,\beta_y$$

#### On a certain day, rain was falling vertically with a speed of $$5\,m/sec.$$ A wind starts blowing after some time with a speed of $$10\sqrt3\,m/s$$ towards south. In which direction should a girl hold her umbrella to avoid the rain?

A $$tan^{-1}\,\left(2\sqrt3\right)$$ with the vertical

B $$tan^{-1}\,\sqrt2$$ with the vertical

C $$20°$$ with the vertical

D $$60°$$ with the vertical

×

Let $$\vec v_F$$ is the resultant velocity

velocity of rain $$(\vec v_r)=5\,m/s$$ vertically downward

velocity of wind $$(\vec v_w)=10\sqrt3\;m/s$$ towards south

Resultant velocity of rain

$$\vec V_F=\vec V_w+\vec V_r$$

$$=(-10\sqrt3)\,\hat i-5\,\hat j$$

$$tan\,\alpha=\dfrac{-10\sqrt3}{-5}$$

$$tan\,\alpha =2\sqrt3$$

$$\alpha=tan^{-1}\,\left(2\sqrt3\right)$$ with vertical

Option (A) is correct.

### On a certain day, rain was falling vertically with a speed of $$5\,m/sec.$$ A wind starts blowing after some time with a speed of $$10\sqrt3\,m/s$$ towards south. In which direction should a girl hold her umbrella to avoid the rain?

A

$$tan^{-1}\,\left(2\sqrt3\right)$$ with the vertical

.

B

$$tan^{-1}\,\sqrt2$$ with the vertical

C

$$20°$$ with the vertical

D

$$60°$$ with the vertical

Option A is Correct

# Components of Force when it is oblique to x or y axis

• Consider a force vector, as shown in figure.

• To calculate force along $$x$$ and $$y$$–axis, selection of axis is done as shown in figure.
• Due to the impact of this force, this force is resolved in $$x$$ and $$y$$–axis.
• Suppose the force vector $$\vec{F}$$ makes angle $$\theta$$ with positive $$x$$–axis.

The components of force can be resolved as follows.

If the force vector $$\vec{F}$$ makes angle $$\theta$$ with positive $$y$$–axis, then the components of $$\vec{F}$$ will be as follows:

#### The system shown in figure is in equilibrium. Calculate tension in the string, if $$\theta = 30 °$$ and mass of block is $$m = 10 \; kg$$. Given:- $$g =10\;m/s^2$$

A $$100 \,N$$

B $$150\, N$$

C $$160\, N$$

D $$200\, N$$

×

Mass of block $$m = 10 \; kg$$

Acceleration due to gravity, $$g = 10 \; m/s^2$$

Tension in string = T

Resolving tension into its components:

component along $$y$$ axis is $$T$$ $$sin\; 30°$$

Total force acting along y axis is $$2T\;sin\; 30°$$

Resolving in $$x$$ direction

$$\sum\; F_x = 0$$

$$T \; cos \; 30° = T \; cos \; 30°$$

So, block is in equilibrium in $$x$$ direction.

The system is in equilibrium, thus, the force acting on opposite sides of block must be equal in  y direction

Thus,

$$T \; sin\; 30 ° + T \; sin\; 30 ° = mg$$

$$2T \; sin \;30 ° = 10 \times10$$

$$T \left(\dfrac{1}{2}\right) = \dfrac{100}{2}$$

$$T = 100\; N$$

### The system shown in figure is in equilibrium. Calculate tension in the string, if $$\theta = 30 °$$ and mass of block is $$m = 10 \; kg$$. Given:- $$g =10\;m/s^2$$

A

$$100 \,N$$

.

B

$$150\, N$$

C

$$160\, N$$

D

$$200\, N$$

Option A is Correct

# Addition of Two Vectors by Taking Component

• Consider a vector $$\vec{OA}$$ making an angle $$\alpha _x$$ with $$x-$$ axis and vector $$\vec{OB}$$ making an angle $$\beta _x$$ with $$x-$$ axis.
• Vector addition of both vectors can be done by adding the component of both vectors along $$x-$$ axis and $$y-$$ axis.

### Components of $$\vec{OA}$$

• Component of $$\vec{OA}$$ along $$x-$$ axis

In $$\Delta \,OAB$$

$$cos\,\alpha_x=\dfrac{OB}{OA}$$

$$OB=OA\,cos\,\alpha_x$$

• Component of $$\vec{OA}$$ along $$y-$$ axis

In $$\Delta \,OAC$$

$$cos\,\alpha_y=\dfrac{OC}{OA}$$

$$OC=OA\,cos\,\alpha_y$$

### Components of $$\vec{OB}$$

• Component of $$\vec{OB}$$ along $$x-$$ axis

In $$\Delta \,OPB$$

$$cos\,\beta_x=\dfrac{OP}{OB}$$

$$OP=OB\,cos\,\beta_x$$

• Component of $$\vec{OB}$$ along $$y-$$ axis

In $$\Delta \,OQB$$

$$cos\,\beta_y=\dfrac{OQ}{OB}$$

$$OQ=OB\,cos\,\beta_y$$

### Resultant component along $$x-$$ axis

• Resultant component along $$x-$$ axis can be achieved by summing the components of both $$\vec{OA}$$ and $$\vec{OB}$$ along the $$x-$$ axis.

$$R_x=OA\;cos\,\alpha_x+OB\,cos\,\beta_x$$

### Resultant component along $$y-$$ axis

• Resultant component along $$y-$$ axis can be achieved by summing the components of both $$\vec{OA}$$ and $$\vec{OB}$$ along the $$y-$$ axis.

$$R_y=OA\;cos\,\alpha_y+OB\,cos\,\beta_y$$

#### Find the components of the resultant vector obtained by addition of two vectors shown in figure. Given $$|\vec{OA}|=10,\;|\vec{OB}|=20,\;\alpha_x=30°\;and\;\beta_x=60°$$

A $$5(\sqrt{3}+2),\;5(1+2\sqrt{3})$$

B $$5(\sqrt{3}+2),\;(5+2\sqrt{3})$$

C $$(5\sqrt{3}+2),\;(5+2\sqrt{3})$$

D $$(5\sqrt{3}+2),\;(5+10\sqrt{3})$$

×

Components of $$\vec{OA}$$

$$\alpha_x=30°$$

$$\alpha_y=90°-30°$$

$$=60°$$

$$x-$$ component of $$\vec{OA}=OA\;cos\,\alpha_x$$

$$=10\,cos\,30°$$

$$=10×\dfrac{\sqrt3}{2}$$

$$=5\sqrt3$$

$$y-$$ component of $$\vec{OA}=OA\;cos\,\alpha_y$$

$$=10×cos\,60°$$

$$=10×\dfrac{1}{2}$$

$$=5$$

Components of $$\vec{OB}$$

$$\beta_x=60°$$

$$\beta_y=90°-60°$$

$$=30°$$

$$x-$$ component of $$\vec{OB}=OB\;cos\,\beta_x$$

$$=20\,cos\,60°$$

$$=20×\dfrac{1}{2}$$

$$=10$$

$$y-$$ component of $$\vec{OB}=OB\;cos\,\beta_y$$

$$=20×cos\,30°$$

$$=20×\dfrac{\sqrt3}{2}$$

$$=10\sqrt3$$

Resultant component along $$x-$$ axis

$$R_x=x-\text{component of}\;\vec{OA}\;+x-\text{component of}\;\vec{OB}$$

$$R_x=5\sqrt3+10$$

$$=5(\sqrt3+2)$$

Resultant component along $$y-$$ axis

$$R_y=y-\text{component of}\;\vec{OA}\;+y-\text{component of}\;\vec{OB}$$

$$R_y=5+10\sqrt3$$

$$R_y=5(1+2\sqrt3)$$

### Find the components of the resultant vector obtained by addition of two vectors shown in figure. Given $$|\vec{OA}|=10,\;|\vec{OB}|=20,\;\alpha_x=30°\;and\;\beta_x=60°$$

A

$$5(\sqrt{3}+2),\;5(1+2\sqrt{3})$$

.

B

$$5(\sqrt{3}+2),\;(5+2\sqrt{3})$$

C

$$(5\sqrt{3}+2),\;(5+2\sqrt{3})$$

D

$$(5\sqrt{3}+2),\;(5+10\sqrt{3})$$

Option A is Correct

# Subtraction of Two Vectors by Taking Component

• Consider a vector $$\vec{OA}$$ making an angle $$\alpha _x$$ with $$x-$$ axis and vector $$\vec{OB}$$ making an angle $$\beta _x$$ with $$x-$$ axis.
• Vector subtraction of both vectors can be done by subtracting the components of both vectors along $$x-$$ axis and $$y-$$ axis.

### Components of $$\vec{OA}$$

Component of $$\vec{OA}$$ along $$x-$$ axis

In $$\Delta \,OAB$$

$$cos\,\alpha_x=\dfrac{OB}{OA}$$

$$OB=OA\,cos\,\alpha_x$$

Component of $$\vec{OA}$$ along $$y-$$ axis

In $$\Delta \,OAC$$

$$cos\,\alpha_y=\dfrac{OC}{OA}$$

$$OC=OA\,cos\,\alpha_y$$

### Components of $$\vec{OB}$$

Component of $$\vec{OB}$$ along $$x-$$ axis

In $$\Delta \,OPB$$

$$cos\,\beta_x=\dfrac{OP}{OB}$$

$$OP=OB\,cos\,\beta_x$$

Component of $$\vec{OB}$$ along $$y-$$ axis

In $$\Delta \,OQB$$

$$cos\,\beta_y=\dfrac{OQ}{OB}$$

$$OQ=OB\,cos\,\beta_y$$

### Resultant component along $$x-$$ axis

• Resultant component along $$x-$$ axis can be achieved by subtracting the components of both $$\vec{OA}$$ and $$\vec{OB}$$ along the $$x-$$ axis.

$$R_x=OA\;cos\,\alpha_x-OB\,cos\,\beta_x$$

### Resultant component along $$y-$$ axis

• Resultant component along $$y-$$ axis can be achieved by subtracting the components of both $$\vec{OA}$$ and $$\vec{OB}$$ along the $$y-$$ axis.

$$R_y=OA\;cos\,\alpha_y-OB\,cos\,\beta_y$$

#### Find the component of the resultant vector obtained by the subtraction of two vectors shown in figure. Given $$|\vec{OA}|=10,\;|\vec{OB}|=15,\;\alpha_x=37°\;and\;\beta_x=53°$$ $$cos37°=4/5\,,\,\,cos53°=3/5$$

A $$-1,\;5$$

B $$1,\,6$$

C $$-1,\;-6$$

D $$-1,\,-5$$

×

Components of $$\vec{OA}$$

$$x-$$ component of $$\vec{OA}=OA\;cos\,\alpha_x$$

$$=10\,cos\,37°$$

$$=10×\dfrac{4}{5}$$

$$=8$$

$$y-$$ component of $$\vec{OA}=OA\;cos\,\alpha_y$$

$$=10×cos\,(90-37°)$$

$$=10\,cos\,(53°)$$

$$=10×\dfrac{3}{5}$$

$$=6$$

Components of $$\vec{OB}$$

$$x-$$ component of $$\vec{OB}=OB\;cos\,\beta_x$$

$$=15\,cos\,(53°)$$

$$=15×\dfrac{3}{5}$$

$$=9$$

$$y-$$ component of $$\vec{OB}=OB\;cos\,\beta_y$$

$$=15×cos\,(90°-53°)$$

$$=15\,cos\,37°$$

$$=15×\dfrac{4}{5}$$

$$=12$$

Resultant component along $$x-$$ axis

$$R_x=(x-\text{component of}\;\vec{OA})\;-\;(x-\text{component of}\;\vec{OB})$$

$$R_x=8-9$$

$$R_x=-1$$

Resultant component along $$y-$$ axis

$$R_y=(y-\text{component of}\;\vec{OA})\;-\;(y-\text{component of}\;\vec{OB})$$

$$R_y=6-12$$

$$R_y=-6$$

### Find the component of the resultant vector obtained by the subtraction of two vectors shown in figure. Given $$|\vec{OA}|=10,\;|\vec{OB}|=15,\;\alpha_x=37°\;and\;\beta_x=53°$$ $$cos37°=4/5\,,\,\,cos53°=3/5$$

A

$$-1,\;5$$

.

B

$$1,\,6$$

C

$$-1,\;-6$$

D

$$-1,\,-5$$

Option C is Correct