Question

i and j are the unit vectors along x and y- axis respectively.what is the magnitude and direction of the vectors I+ j and i - j ? what are the components of Vector A vector= 2i + 3j along directions of I + j a d i- j ??​

Answer 1

Aηswer:

consider a vector $\displaystyle\sf \vec{P}$ given as:

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

$\displaystyle\sf P_x \hat{i} + P_y \hat{j} = \hat{i}+\hat{j}$

on comparing the components on both sides, we get :

$\displaystyle\sf P_x = P_y = 1$

$\displaystyle\sf |\vec{P}| = \sqrt{P^2_x + P^2_y} = \sqrt{1^2+1^2} = \sqrt{2}\:\;\;\dots(i)$

Hence, the magnitude of vector $\displaystyle\sf \hat{i}+\hat{j}$ is √2.

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Let θ be the angle made by the vector $\displaystyle\sf \vec{P}$, with the x axis (see attachment {i}).

$\displaystyle\sf$

$\displaystyle\sf \therefore \tan\theta = \left(\dfrac{P_y}{P_x}\right)$

$\displaystyle\sf \theta = \tan^{-1}\left(\dfrac{1}{1}\right)$

$\displaystyle\sf = 45^{\circ}\;\;\;\dots(ii)$

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

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

$\displaystyle\sf Q_x \hat{i} - Q_y \hat{j} = \hat{i} - \hat{j}$

$\displaystyle\sf Q_x = Q_y = 1$

$\displaystyle\sf |\vec{Q}| = \sqrt{Q^2_x + Q^2_y} = \sqrt{2}\;\;\;\dots(iii)$

Hence the magnitude of vector $\displaystyle\sf \hat{i}+\hat{j}$ is √2.

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Let θ be the angle made by the vector $\displaystyle\sf \vec{Q}$, with the x-axis, (see attachment {ii}).

$\displaystyle\sf$

$\displaystyle\sf \therefore \tan\theta = \left(\dfrac{Q_x}{Q_y}\right)$

$\displaystyle\sf \theta = -\tan^{-1}\left(-\dfrac{1}{1}\right)$

$\displaystyle\sf = 45^{\circ}\;\;\;\dots(iv)$

Hence, the vector $\displaystyle\sf \hat{i}+\hat{j}$ makes an angle of 45.

$\displaystyle\sf$

It is given that :

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

$\displaystyle\sf A_x \hat{i} + A_y \hat{j} = 2\hat{i} + 3\hat{j}$

On comparing the coefficients of $\displaystyle\sf \hat{i}\:\text{and}\:\hat{j}$,

A_x = 2 and A_y = 2

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

$\displaystyle\sf = \sqrt{13}$

$\displaystyle\sf$

Let $\displaystyle\sf \vec{A}_x$ make an angle of θ with the x-axis. (see attachment {iii}).

$\displaystyle\sf \therefore \tan\theta = \left(\dfrac{A_y}{A_x}\right)$

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

$\displaystyle\sf = \tan^{-1}(1.5)$

$\displaystyle\sf = 56.31^{\circ}$

$\displaystyle\sf$

Angle between the vectors $\displaystyle\sf (2\hat{i} + 3\hat{j})$ and $\displaystyle\sf (\hat{i}+\hat{j})$,

θ = 56.31 – 45

= 11.31°

$\displaystyle\sf$

Components of vector $\displaystyle\sf \vec{A}$, along the direction of $\displaystyle\sf \vec{P}$, making an angle θ:

$\displaystyle\sf = (A\cos\theta)\vec{P}$

$\displaystyle\sf = (A\cos 11.31)\dfrac{(\hat{i}+\hat{j})}{\hat{2}}$

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

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

$\displaystyle\sf = \dfrac{25}{12}\times\sqrt{2}$

$\displaystyle\sf = \dfrac{5}{\sqrt{2}}\:\;\;\dots(v)$

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Let θ be the angle between the vectors $\displaystyle\sf (2\hat{i}+3\hat{j})$ and $\displaystyle\sf (\hat{i}-\hat{j})$

$\displaystyle\sf \theta = 45 + 56.31$

$\displaystyle\sf = 101.31^{\circ}$

component of vector $\displaystyle\sf \hat{A}$, along the direction of $\displaystyle\sf \vec{Q}$, making an angle θ,

$\displaystyle\sf = (A\cos\theta) \vec{Q} = (A\cos\theta)\dfrac{\hat{i}-\hat{j})}{\sqrt{2}}$

$\displaystyle\sf = \sqrt{13}\cos(901.31^{\circ}) \dfrac{\hat{i}-\hat{j})}{\sqrt{2}}$

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

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

$\displaystyle\sf = -\dfrac{5}{10}\times\sqrt{2}$

$\displaystyle\sf = -\dfrac{1}{\sqrt{2}}\:\;\;\dots(iv)$

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