Question

Explain the resolution of vectors in two dimensions.

Answer 1

Resolution of vector in two Dimensions. The resolution of a vector into two mutually perpendicular vectors is called the rectangular resolution of vector in a plane or two dimensions.

Consider that a vector $\overrightarrow{OP}=\overrightarrow{A}$ has to be resolved into two component vector along the direction of two mutually perpendicular directions of X-axis and Y-axis. Let $\hat{i}$ and $\hat{j}$ be the unit vectors along X-axis and Y-axis respectively figure 2.19.

From point P, drop PM and PN perpendicular to X-axis and Y-axis respectively. From the parallelogram law of vector addition, it follows that

$\overrightarrow{OP}=\overrightarrow{OM}+\overrightarrow{MP}$

If $OM=A_x$ and $ON=MP=A_y,$ then

$\overrightarrow{OB}=A_x\hat{i}$ and $\widehat{ON}=\overrightarrow{MP}=A_y\hat{j}$

Therefore, the above, equation becomes

$\overrightarrow{OP}=A_x\hat{i}+A_y\hat{j}$

or $\overrightarrow{A}=A_x\hat{i}+A_y\hat{y}$ .......(i)

The equation (i) describes vector, $\overrightarrow{A}$ into the component vector $A_x\hat{i}$ and $A_y\hat{i}.$ In practice $A_x\hat{i}$ and $A_y\hat{j}$ are called respectively x-component and y-component of vectors $\overrightarrow{A}.$ Further $A_x$ and $A_y$ are called magnitude of the two component vectors.

If A is the magnitude of the vector $\overrightarrow{A}$ and $\theta$ is its inclination with X-axis, then from the right angled triangle OMP,

$cos\theta =\frac{OM}{OP}=\frac{A_x}{A}$

or $A_x=Acos\theta$ .......(ii)

Also $sin\theta =\frac{MP}{OA}=\frac{A_y}{A}$

$A_y=Asin\theta$ .......(iii)

Adding the squares of $A_x$ and $A_y$, we get

$A_x^2+A_y^2=(Acos\theta {)}^2+(Asin\theta {)}^2$

or, $A_x^2+A_y^2=A^{2}co{s}^2\theta +A^{2}si{n}^2\theta$

$\Rightarrow A_x^2+A_y^2=A^2(co{s}^2\theta +si{n}^2\theta )$

or, $A_x^2+A_y^2=A^2(\because si{n}^2\theta +co{s}^2\theta =1)$

or, $A^2=A_x^2+A_y^2$

$\Rightarrow A_x^2+A_y^2=A^2(co{s}^2\theta +si{n}^2\theta )$

or, $A_x^2+A_y^2=A^2(\because si{n}^2\theta +co{s}^{\theta }=1)$

or, $A^2=A_x^2+A_y^2$

$\Rightarrow A=\sqrt{A_x^2+A_y^2}$ ........(iv)

Dividing $A_y$ and $A_x,$ we get

$\frac{A_y}{A_x}=\frac{Acos\theta }{Acos\theta }$

$\Rightarrow \frac{A_y}{A_x}=tan\theta$

or, $tan\theta =\frac{A_y}{A_x}$

or, $\theta =ta{n}^{-1}\left(\frac{A_y}{A_x}\right)$........(v)