$\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along $x$ – and $y$-axis respectively.
Question.

$\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along $x$ – and $y$-axis respectively. What is the magnitude and direction of the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$, and $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ ? What are the components of a vector $\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ along the directions of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ ? [You may use graphical method]

solution:

Consider a vector $\bar{P}$, given as:

$\vec{P}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

On comparing the components on both sides, we get:

$P_{x}=P_{y}=1$

$\overrightarrow{|P|}=\sqrt{P_{x}^{2}+P_{y}^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$ $\ldots(i)$

Hence, the magnitude of the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ is $\sqrt{2}$.

Let $\theta$ be the angle made by the vector $\vec{P}$, with the $x$-axis, as shown in the following figure.

[You may use graphical method]

$\therefore \tan \theta=\left(\frac{P_{y}}{P_{x}}\right)$

$\theta=\tan ^{-1}\left(\frac{1}{1}\right)=45^{\circ}$ $\ldots(i i)$

Hence, the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ makes an angle of $45^{\circ}$ with the $x$-axis.

Let $\vec{Q}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$

$Q_{x} \hat{\mathbf{i}}-Q_{y} \hat{\mathbf{j}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$

$Q_{x}=Q_{y}=1$

$|\vec{Q}|=\sqrt{Q_{x}^{2}+Q_{y}^{2}}=\sqrt{2}$ $\ldots$ (iii)

Hence, the magnitude of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ is $\sqrt{2}$.

Let $\theta$ be the angle made by the vector $\vec{Q}$, with the $x$ – axis, as shown in the following figure.

[You may use graphical method]02

$\therefore \tan \theta=\left(\frac{Q_{y}}{Q_{x}}\right)$

$\theta=-\tan ^{-1}\left(-\frac{1}{1}\right)=-45^{\circ}$ $\ldots(i v)$

Hence, the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ makes an angle of $-45^{\circ}$ with the $x$-axis.

It is given that:

$\vec{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$

$A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$

On comparing the coefficients of $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$, we have:

$A_{x}=2$ and $A_{y}=3$

$|\vec{A}|=\sqrt{2^{2}+3^{2}}=\sqrt{13}$

Let $\vec{A}$, make an angle $\theta$ with the $x$-axis, as shown in the following figure.

 [You may use graphical method]03

$\therefore \tan \theta=\left(\frac{A_{y}}{A_{x}}\right)$

$\theta=\tan ^{-1}\left(\frac{3}{2}\right)$

$=\tan ^{-1}(1.5)=56.31^{\circ}$

Angle between the vectors $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})$ and $(\hat{\mathbf{i}}+\hat{\mathbf{j}}), \theta=56.31-45=11.31^{\circ}$

Component of vector $\vec{A}$, along the direction of $\vec{P}$, making an angle $\theta^{\prime}$

$=(A \cos \theta) \hat{P}=(A \cos 11.31) \frac{(\hat{i}+\hat{j})}{\sqrt{2}}$

$=\sqrt{13} \times \frac{0.9806}{\sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}})$

$=2.5(\hat{\mathbf{i}}+\hat{\mathbf{j}})$

$=\frac{25}{10} \times \sqrt{2}$

$=\frac{5}{\sqrt{2}}$ $\ldots(v)$

Let $\theta^{\prime}$ be the angle between the vectors $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})$ and $(\hat{\mathbf{i}}-\hat{\mathbf{j}})$.

$\theta^{\prime \prime}=45+56.31=101.31^{\circ}$

Component of vector $\vec{A}$, along the direction of $\vec{Q}$, making an angle $\theta$

$=\left(A \cos \theta^{\prime \prime}\right) \vec{Q}=\left(A \cos \theta^{\prime \prime}\right) \frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}$

$=\sqrt{13} \cos \left(901.31^{\circ}\right) \frac{(\hat{\mathbf{i}}-\hat{\mathbf{j}})}{\sqrt{2}}$

$=-\sqrt{\frac{13}{2}} \sin 11.30^{\circ}(\hat{\mathbf{i}}-\hat{\mathbf{j}})$

$=-2.550 \times 0.1961(\hat{\mathbf{i}}-\hat{\mathbf{j}})$

$=-0.5(\hat{\mathbf{i}}-\hat{\mathbf{j}})$

$=-\frac{5}{10} \times \sqrt{2}$

$=-\frac{1}{\sqrt{2}}$ $\ldots$ (vi)
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