and direction of the vectors $\hat{ i }+\hat{ j }$, and $\hat{ i }-\hat{ j }$ ? What are the components of a vector $A =2 \hat{ i }+3 \hat{ j }$ along the directions of $\hat{ i }+\hat{ j }$ and $\hat{ i }-\hat{ j } ?$

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

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

$P_{x} \hat{ i }+P_{y} \hat{ j }=\hat{ i }+\hat{ j }$

On comparing the components on both sides, we get:

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

$\overrightarrow{1 P 1}=\sqrt{P_{x}^{2}+P_{y}^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$

Hence, the magnitude of the vector $\vec{\imath}+\vec{\jmath}$ is $\sqrt{2}$

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

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

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

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

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

$Q, \hat{ i }-Q, \hat{ j }=\hat{ i }-\hat{ j }$

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

$|\vec{Q}|=\sqrt{Q_{r}^{2}+Q_{y}^{2}}=\sqrt{2}$

Hence, the magnitude of the vector $\vec{\imath}-\vec{\jmath}$ is $\sqrt{2}$

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

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

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

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

It is given that:

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

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

On comparing the coefficients of $\vec{\imath}$ and $\vec{\jmath},$ we have:

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{5}{\sqrt{2}}$

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

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

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

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

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

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

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

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

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

$=-\frac{1}{\sqrt{2}}$