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: \(cos\;(90-\theta)=sin\,\theta\) 

Illustration Questions

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 \(=|\vec {OA}|=10\)

image

\(\theta_{x}=30°\)\(\theta_{y}=60°\)

image

\(x-\)component

=\(|\vec {OA}|×cos\,30°\)

\(=10×\dfrac {\sqrt 3}{2}\)

\(=5 {\sqrt 3}\)

image

\(y-\)component

=\(|\vec {OA}|×cos\,60°\)

\(=10×\dfrac {1}{2}\)

\(=5\)

image

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

image
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: \(cos(90°-\theta)=sin\;\theta\)

 

Illustration Questions

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  \(=|\vec {OA}|=10\)

image

\(\theta_x=30°\)

\(\theta_y=60°\)

image

X-component along inclined \(x\)-axis

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

\(=10×cos\,30°\)

\(=10×\dfrac {\sqrt 3}{2}\)

\(=5\;\sqrt 3\)

image

\(Y\)-component along inclined \(y\)-axis

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

\(=10×cos\,60°\)

\(=10×\dfrac {1}{2}\)

\(=5\)

image

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

image
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

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.

\(R_y=OA\;cos\,\alpha_y+OB\,cos\,\beta_y\)

Illustration Questions

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 \(\vec{OA}=OA\;cos\,\alpha_y\)

\(=10×cos\,60°\)

\(=10×\dfrac{1}{2}\)

\(=5\)

image

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 \(\vec{OB}=OB\;cos\,\beta_y\)

\(=20×cos\,30°\)

\(=20×\dfrac{\sqrt3}{2}\)

\(=10\sqrt3\)

image

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 \(|\vec{OA}|=10,\;|\vec{OB}|=20,\;\alpha_x=30°\;and\;\beta_x=60°\)

image
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.

\(R_y=OA\;cos\,\alpha_y-OB\,cos\,\beta_y\)

Illustration Questions

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 \(\vec{OA}=OA\;cos\,\alpha_y\)

\(=10×cos\,(90-37°)\)

\(=10\,cos\,(53°)\)

\(=10×\dfrac{3}{5}\)

\(=6\)

image

Components of \(\vec{OB}\)

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

\(=15\,cos\,(53°)\)

\(=15×\dfrac{3}{5}\)

\(=9\)

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

\(=15×cos\,(90°-53°)\)

\(=15\,cos\,37°\)

\(=15×\dfrac{4}{5}\)

\(=12\)

image

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 \(|\vec{OA}|=10,\;|\vec{OB}|=15,\;\alpha_x=37°\;and\;\beta_x=53°\) \(cos37°=4/5\,,\,\,cos53°=3/5\)

image
A

\(-1,\;5\)

.

B

\(1,\,6\)

C

\(-1,\;-6\)

D

\(-1,\,-5\)

Option C 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.

\(R_y=OA\;cos\,\alpha_y+OB\,cos\,\beta_y\)

Illustration Questions

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

image

Resultant velocity of rain

\(\vec V_F=\vec V_w+\vec V_r\)

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

 

image

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

\(tan\,\alpha =2\sqrt3\)

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

 

image

Option (A) is correct.

image

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: 

Illustration Questions

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 \(2T\;sin\; 30°\)

image

Resolving in \(x\) direction 

\(\sum\; F_x = 0\)

\(T \; cos \; 30° = T \; cos \; 30°\)

So, block is in equilibrium in \(x\) direction. 

image

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

image

Thus,

\(T \; sin\; 30 ° + T \; sin\; 30 ° = mg\)

\(2T \; sin \;30 ° = 10 \times10\)

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

\(T = 100\; N\)

image

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:- \(g =10\;m/s^2\) 

image
A

\(100 \,N\)

.

B

\(150\, N\)

C

\(160\, N\)

D

\(200\, N\)

Option A is Correct

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