Question

i cap and j cap are unit vectors along Xand Y axis resectively . what is the magnitude and direction of the vectors i cap +j cap, and i cap - j cap ? what are the components of a vector A= 2icap + 3j cap along the directions of i cap + j cap and i cap - j cap?

Answer 1

Magnitude and direction of $\hat{i}$$+\hat{j}$ :

If $\overrightarrow{B}$$=\hat{i}$$+\hat{j}$$=(B_x\hat{i}$$+B_y\widehat{j)}$
Then, $B_x=1and{B}_y=1$
So magnitude of $B=\sqrt{B_x^2+B_y^2}=\sqrt{1^2+1^2}=\sqrt{2}$
If $\theta$ is the angle which $\overrightarrow{B}$ makes with the x-axis, then
$$\begin{gathered} \tan \theta =\frac{B_x}{B_y}=\frac{1}{1}=1=\tan {45}^o \\ or\theta =-{45}^o \end{gathered}$$

Magnitude and direction of $\hat{i}$$-\hat{j}$ :

If $\overrightarrow{B}$$=\hat{i}$$-\hat{j}$$=(B_x\hat{i}$$-B_y\widehat{j)}$
Then, $B_x=1and{B}_y=-1$
So magnitude of $B=\sqrt{B_x^2+B_y^2}=\sqrt{1^2+(-1{)}^2}=\sqrt{2}$
If $\theta$ is the angle which $\overrightarrow{B}$ makes with the x-axis, then
$$\begin{gathered} \tan \theta =\frac{B_x}{By}=\frac{-1}{1}=-1=\tan {45}^o \\ or\theta =-{45}^o \end{gathered}$$

Components of vector $A=2\hat{i}$$+3\hat{j}$ along the directions of $\hat{i}$$+\hat{j}$ and $\hat{i}$$-\hat{j}$:

Here, $A_x=2and{A}_y=3$
So magnitude of $A=\sqrt{A_x^2+A_y^2}=\sqrt{2^2+3^2}=\sqrt{13}$
If $\alpha$ is the angle which $\overrightarrow{A}$ makes with the x-axis, then
$$\begin{gathered} \tan \alpha =\frac{A_x}{A_y}=\frac{3}{2}=1.5=\tan {56}^{o}18\text{'} \\ or\alpha ={56}^{o}18\text{'} \end{gathered}$$
Angle between $2\hat{i}$$+3\hat{j}$ and $\hat{i}$$+\hat{j}$ is, $\beta ={56}^{o}18\text{'}-{45}^o={11}^{o}18\text{'}$
Component of vector $\overrightarrow{A}$ along the direction of $\hat{i}$$+\hat{j}$ making an angle $\beta$ is
$$\begin{gathered} (B\cos \beta )\hat{B}=(\sqrt{13}\cos {11}^{o}18\text{'})\frac{(\hat{i}+\hat{j})}{\sqrt{2}} \\ =\sqrt{13}\times 0.9806\left(\frac{(\hat{i}+\hat{j})}{\sqrt{2}}\right)=2.5(\hat{i}+\hat{j}) \end{gathered}$$
If $\beta \text{'}$ is the angle which $\overrightarrow{A}$ makes with the direction of $\hat{i}$$-\hat{j}$ then $\beta \text{'}$$={56}^{o}18\text{'}+{45}^o={101}^{o}18\text{'}$
Component of vector $\overrightarrow{A}$ along the direction of ​ $\hat{i}$$-\hat{j}$ is
$$\begin{gathered} (B\cos \beta \text{'})\hat{B}=(\sqrt{13}\cos {101}^{o}18\text{'})\frac{(\hat{i}-\hat{j})}{\sqrt{2}} \\ =\sqrt{13}(-\cos {78}^{o}32\text{'})\left(\frac{(\hat{i}+\hat{j})}{\sqrt{2}}\right) \\ =-\sqrt{3}\times 0.1983\times \left(\frac{(\hat{i}+\hat{j})}{\sqrt{2}}\right)=-0.5(\hat{i}-\hat{j}) \end{gathered}$$