Resolution of a Vector into Rectangular Components Questions in English

(D) The component of a vector $\vec{F}$ making an angle $\theta$ with a specific axis is given by $F \cos \theta$ along that axis.
Here,the force $F = 5\, N$ makes an angle $\theta = 60^\circ$ with the vertical axis.
Therefore,the vertical component of the force is $F_v = F \cos 60^\circ$.
Substituting the values,we get $F_v = 5 \times \frac{1}{2} = 2.5\, N$.
Thus,the vertical component is $2.5\, N$.

(B) The vector is given by $\vec{A} = 2\hat{i} + 3\hat{j}$.
Here,the $x$-component is $A_x = 2$ and the $y$-component is $A_y = 3$.
The angle $\theta$ between the vector $\vec{A}$ and the $y$-axis is defined by the relation $\tan \theta = \frac{A_x}{A_y}$.
Substituting the values,we get $\tan \theta = \frac{2}{3}$.
Therefore,the angle $\theta = \tan^{-1}(2/3)$.

(C) Given vector $\vec{R} = 3\hat{i} + \hat{j} + 2\hat{k}$.
To find the length of the vector in the $XY$ plane,we consider only the $x$ and $y$ components of the vector.
The $x$-component is $R_x = 3$ and the $y$-component is $R_y = 1$.
The length in the $XY$ plane is given by the formula $\sqrt{R_x^2 + R_y^2}$.
Substituting the values: $\sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.

(B) Given vector $\vec{A} = \hat{i} + \hat{j}$.
Comparing this with $\vec{A} = A_x \hat{i} + A_y \hat{j}$,we get $A_x = 1$ and $A_y = 1$.
The angle $\theta$ made by the vector with the $x$-axis is given by $\tan \theta = \frac{A_y}{A_x}$.
Substituting the values,$\tan \theta = \frac{1}{1} = 1$.
Since $\tan 45^\circ = 1$,the angle $\theta = 45^\circ$.

(C) The process of splitting a single vector into two or more vectors such that their vector sum is equal to the original vector is known as the resolution of a vector.
Any vector $\vec{A}$ can be expressed as the sum of any number of vectors $\vec{A_1}, \vec{A_2}, \dots, \vec{A_n}$ such that $\vec{A} = \vec{A_1} + \vec{A_2} + \dots + \vec{A_n}$.
These component vectors do not necessarily have to be perpendicular or parallel to each other.
Therefore,any vector can be replaced by two or three arbitrary vectors that have the original vector as their resultant.

(A) Given: $v_y = 20$ and $v_x = 10$.
The velocity vector is given by $\vec{v} = v_x \hat{i} + v_y \hat{j} = 10 \hat{i} + 20 \hat{j}$.
The direction of motion $\theta$ with the horizontal ($X$-axis) is given by the formula:
$\tan \theta = \frac{v_y}{v_x}$
Substituting the given values:
$\tan \theta = \frac{20}{10} = 2$
Therefore,the angle $\theta$ is:
$\theta = \tan^{-1}(2)$.

(B) Given,$F_x = 5 \, N$ and $F_y = 5 \, N$.
The magnitude of the resultant force $F$ is given by:
$|\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, N$.
The direction $\theta$ with the $X$-axis is given by:
$\tan \theta = \frac{F_y}{F_x} = \frac{5}{5} = 1$.
Since $\tan \theta = 1$,we have $\theta = \pi/4$ radians (or $45^\circ$).
Thus,the magnitude is $5\sqrt{2} \, N$ and the direction is $\pi/4$.

(A) Given that the sum of the three vectors is zero,i.e.,$\vec{OA} + \vec{OB} + \vec{OC} = 0$.
From the figure,$\vec{OA}$ is acting downwards with a magnitude of $10 \, N$.
To balance the vertical component of $\vec{OC}$,the vertical component of $\vec{OC}$ must be equal to the magnitude of $\vec{OA}$.
Let $\theta = 45^\circ$ be the angle $\vec{OC}$ makes with the horizontal axis $OD$.
The vertical component of $\vec{OC}$ is $OC \sin(45^\circ) = OA = 10 \, N$.
Therefore,$OC = \frac{10}{\sin(45^\circ)} = \frac{10}{1/\sqrt{2}} = 10\sqrt{2} \, N$.
The horizontal component of $\vec{OC}$ is $OC \cos(45^\circ) = (10\sqrt{2}) \times (1/\sqrt{2}) = 10 \, N$.
Since the sum of vectors is zero,the horizontal component of $\vec{OC}$ must be balanced by $\vec{OB}$.
Thus,$OB = 10 \, N$.
So,the magnitudes are $OB = 10 \, N$ and $OC = 10\sqrt{2} \, N$.

(A) vector in a two-dimensional plane can be resolved into two mutually perpendicular components,typically along the $x$ and $y$ axes.
While a vector can be resolved into any number of components in different directions,the standard resolution in a plane involves exactly $2$ rectangular components.
Therefore,the maximum number of rectangular components in a plane is $2$.

(C) Two vectors are perpendicular if their dot product is zero.
Let $\vec{B} = B_x \hat{i} + B_y \hat{j}$.
For $\vec{A} \cdot \vec{B} = 0$,we have $(A \cos \theta)(B_x) + (A \sin \theta)(B_y) = 0$.
This implies $B_x \cos \theta = -B_y \sin \theta$,or $\frac{B_x}{B_y} = -\frac{\sin \theta}{\cos \theta}$.
Thus,we can choose $B_x = B \sin \theta$ and $B_y = -B \cos \theta$.
Therefore,$\vec{B} = \hat{i} B \sin \theta - \hat{j} B \cos \theta$.

(A) Given: Horizontal component $F_x = 40 \ N$,Angle $\theta = 60^\circ$.
The horizontal component is given by $F_x = F \cos \theta$.
Substituting the values: $40 = F \cos 60^\circ = F \times (1/2)$.
Therefore,the magnitude of the force $F = 40 \times 2 = 80 \ N$.
The vertical component is given by $F_y = F \sin \theta$.
Substituting the values: $F_y = 80 \sin 60^\circ = 80 \times (\sqrt{3}/2) = 40\sqrt{3} \ N$.
Using $\sqrt{3} \approx 1.732$,we get $F_y = 40 \times 1.732 = 69.28 \ N$.

(B) Given vector $\vec{A} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.
The unit vector along the $Z$-axis is $\hat{k}$,so $\vec{B} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.
The angle $\theta$ between two vectors is given by the dot product formula: $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.
First,calculate the dot product: $\vec{A} \cdot \vec{B} = (1)(0) + (1)(0) + (\sqrt{2})(1) = \sqrt{2}$.
Next,calculate the magnitude of $\vec{A}$: $|\vec{A}| = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.
The magnitude of $\vec{B}$ is $|\vec{B}| = \sqrt{0^2 + 0^2 + 1^2} = 1$.
Substitute these values into the formula: $\cos \theta = \frac{\sqrt{2}}{2 \times 1} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{\sqrt{2}}) = 45^{\circ}$.

(B) Let the vectors be $\vec{A}, \vec{B}, \vec{C}, \dots$ with $x$-components $A_x, B_x, C_x, \dots$ and magnitudes $A, B, C, \dots$.
The resultant vector is $\vec{R} = \vec{A} + \vec{B} + \vec{C} + \dots$.
The $x$-component of the resultant is $R_x = A_x + B_x + C_x + \dots$.
Thus,statement $(a)$ is correct.
Since the $x$-component of any vector is always less than or equal to its magnitude $(A_x \le A)$,the sum of $x$-components is $R_x = \sum A_x \le \sum A$.
This means $R_x$ is always less than or equal to the sum of the magnitudes of the vectors.
Therefore,$R_x$ can be less than the sum of magnitudes (statement $(b)$ is correct).
However,$R_x$ cannot be greater than the sum of magnitudes,so statement $(c)$ is incorrect.
$R_x$ is not necessarily equal to the sum of magnitudes (statement $(d)$ is incorrect).
Thus,statements $(a)$ and $(b)$ are correct.

(A) The vectors are given by:
$\vec{A} = 50(\cos 45^{\circ} \hat{i} + \sin 45^{\circ} \hat{j}) = 50(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$
$\vec{B} = 50(\cos 135^{\circ} \hat{i} + \sin 135^{\circ} \hat{j}) = 50(-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$
$\vec{C} = 50(\cos 315^{\circ} \hat{i} + \sin 315^{\circ} \hat{j}) = 50(\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j})$
Summing the vectors:
$\vec{R} = \vec{A} + \vec{B} + \vec{C} = 50(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) \hat{i} + 50(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{j}$
$\vec{R} = 50(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$
The magnitude of the resultant vector is:
$|\vec{R}| = \sqrt{(50/\sqrt{2})^2 + (50/\sqrt{2})^2} = \sqrt{1250 + 1250} = \sqrt{2500} = 50 \text{ units}$.

(B) vector in three-dimensional space can be resolved into components along the three mutually perpendicular axes: the $x$-axis,$y$-axis,and $z$-axis.
These components are known as rectangular components.
Since our physical space is three-dimensional,a vector can have at most $3$ rectangular components.

(D) vector can be resolved into an infinite number of components in different directions.
While we typically resolve a vector into $3$ mutually perpendicular (orthogonal) components in a $3D$ Cartesian coordinate system,mathematically,a vector can be decomposed into any number of components along arbitrary directions.

(D) Let the vector be $\vec{r}$ and its magnitude be $r$.
Given that the component along the $Y$-axis is $r_y = 10$ units.
The angle made with the $X$-axis is $\theta = 30^{\circ}$.
The component along the $Y$-axis is given by $r_y = r \sin(\theta)$.
Substituting the values: $10 = r \sin(30^{\circ})$.
Since $\sin(30^{\circ}) = 0.5$,we have $10 = r \times 0.5$.
Therefore,$r = \frac{10}{0.5} = 20$ units.

(D) Let $\vec{a} = 3\hat{i} + 4\hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.
The component of $\vec{a}$ along $\vec{b}$ is given by the formula: $\vec{a}_{\text{parallel}} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (3)(1) + (4)(1) = 3 + 4 = 7$.
Next,calculate the square of the magnitude of $\vec{b}$: $|\vec{b}|^2 = (1)^2 + (1)^2 = 1 + 1 = 2$.
Substitute these values into the formula: $\vec{a}_{\text{parallel}} = \left( \frac{7}{2} \right) (\hat{i} + \hat{j})$.
Thus,the component is $\frac{7}{2}(\hat{i} + \hat{j})$.

(D) The projection of a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ on the $xy$-plane is given by the vector $\vec{r}_{xy} = x\hat{i} + y\hat{j}$.
Given $\vec{r} = 3\hat{i} + 1\hat{j} + 2\hat{k}$,the projection on the $xy$-plane is $\vec{r}_{xy} = 3\hat{i} + 1\hat{j}$.
The magnitude of this projection is $|\vec{r}_{xy}| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.

(D) The vertical component of a force $\vec{F}$ making an angle $\theta$ with the vertical axis is given by $F_y = F \cos \theta$.
Given,$F = 5 \, N$ and $\theta = 60^\circ$.
Therefore,the vertical component is $F_y = 5 \cos 60^\circ$.
Since $\cos 60^\circ = 0.5$,we have $F_y = 5 \times 0.5 = 2.5 \, N$.

(A) Let the angles made by the vector $\overrightarrow{A}$ with the $x$,$y$,and $z$ axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given that the vector makes equal angles with the axes,we have $\alpha = \beta = \gamma$.
The direction cosines of the vector satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting $\alpha = \beta = \gamma$,we get $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \frac{1}{\sqrt{3}}$.
The components of the vector $\overrightarrow{A}$ are given by $A_x = A \cos \alpha$,$A_y = A \cos \beta$,and $A_z = A \cos \gamma$.
Therefore,$A_x = A_y = A_z = A \left( \frac{1}{\sqrt{3}} \right) = \frac{A}{\sqrt{3}}$.

(B) Let $\theta$ be the angle that the particle's path makes with the $x$-axis.
From the given coordinates $(x, y) = (\sqrt{3}, 3)$,the slope of the line is given by $\tan \theta = \frac{y}{x}$.
Substituting the values,we get $\tan \theta = \frac{3}{\sqrt{3}}$.
Simplifying this,$\tan \theta = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^o$.

(B) Let $AC = b$ and $AB = c$. The forces are $F_1 = 1/b$ acting along $AC$ and $F_2 = 1/c$ acting along $AB$.
Since $\angle A = 90^{\circ}$,the resultant force $F_{\text{net}}$ is given by:
$F_{\text{net}} = \sqrt{F_1^2 + F_2^2} = \sqrt{\frac{1}{b^2} + \frac{1}{c^2}} = \sqrt{\frac{b^2 + c^2}{b^2 c^2}}$
In $\Delta ABC$,$BC^2 = b^2 + c^2$,so $F_{\text{net}} = \frac{\sqrt{BC^2}}{bc} = \frac{BC}{bc}$.
The area of $\Delta ABC$ can be expressed as $\text{Area} = \frac{1}{2} \times b \times c = \frac{1}{2} \times BC \times AD$,where $AD$ is the altitude to the hypotenuse $BC$.
Therefore,$bc = BC \times AD$,which implies $\frac{BC}{bc} = \frac{1}{AD}$.
Thus,$F_{\text{net}} = \frac{1}{AD}$.

(B) Let the angle between the forces $P$ and $Q$ be $\theta$.
The resultant $R$ is given by the vector sum $\vec{R} = \vec{P} + \vec{Q}$.
The resolved part of $R$ in the direction of $P$ is given by $P + Q \cos \theta$.
According to the problem,this resolved part is equal to $Q$.
Therefore,$P + Q \cos \theta = Q$.
Rearranging the terms,we get $Q \cos \theta = Q - P$.
Thus,$\cos \theta = \frac{Q - P}{Q} = 1 - \frac{P}{Q}$.
Using the trigonometric identity $\cos \theta = 1 - 2 \sin^2 \left(\frac{\theta}{2}\right)$,we have:
$1 - 2 \sin^2 \left(\frac{\theta}{2}\right) = 1 - \frac{P}{Q}$.
Subtracting $1$ from both sides,we get $-2 \sin^2 \left(\frac{\theta}{2}\right) = -\frac{P}{Q}$.
Therefore,$\sin^2 \left(\frac{\theta}{2}\right) = \frac{P}{2Q}$.
Taking the square root,we get $\sin \left(\frac{\theta}{2}\right) = \left(\frac{P}{2Q}\right)^{1/2}$.
Finally,the angle $\theta = 2 \sin^{-1} \left(\frac{P}{2Q}\right)^{1/2}$.

(B) The vector is given by $\vec{A} = 3\hat{i} - 4\hat{j} + 10\hat{k}$.
The component of the vector in the $x-y$ plane is the resultant of its $x$ and $y$ components,given by $\sqrt{A_x^2 + A_y^2}$.
Substituting the values,we get $\sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The component of the vector along the $z$-axis is $A_z = 10$.
The ratio of the magnitude of the component in the $x-y$ plane to the component along the $z$-axis is $\frac{5}{10} = 0.5$.

(C) The angle $\gamma$ that a vector $P = P_x \hat{i} + P_y \hat{j} + P_z \hat{k}$ makes with the $z$-axis is given by $\cos \gamma = \frac{P_z}{|P|}$.
Given $P = 6 \hat{i} + 4 \sqrt{2} \hat{j} + 4 \sqrt{2} \hat{k}$.
The magnitude of the vector $P$ is $|P| = \sqrt{(6)^2 + (4 \sqrt{2})^2 + (4 \sqrt{2})^2}$.
$|P| = \sqrt{36 + (16 \times 2) + (16 \times 2)} = \sqrt{36 + 32 + 32} = \sqrt{100} = 10$.
The $z$-component is $P_z = 4 \sqrt{2}$.
Therefore,$\cos \gamma = \frac{4 \sqrt{2}}{10} = \frac{2 \sqrt{2}}{5}$.
Thus,$\gamma = \cos^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$.

(D) The force $F = 5 \ N$ makes an angle $\theta = 60^{\circ}$ with the vertical direction.
The vertical component of a vector $F$ making an angle $\theta$ with the vertical axis is given by $F_V = F \cos \theta$.
Substituting the given values:
$F_V = 5 \times \cos 60^{\circ}$
Since $\cos 60^{\circ} = 0.5$,we get:
$F_V = 5 \times 0.5 = 2.5 \ N$.
Therefore,the vertical component is $2.5 \ N$.

(A) The man walks a distance $d = 20\,m$ at an angle $\theta = 60^{\circ}$ north of east.
To find the distance travelled towards the east,we calculate the horizontal component of the displacement vector.
The horizontal component is given by $d_x = d \cos(\theta)$.
Substituting the given values: $d_x = 20 \cos(60^{\circ})$.
Since $\cos(60^{\circ}) = 0.5$,we get $d_x = 20 \times 0.5 = 10\,m$.
Therefore,the man has travelled $10\,m$ towards the east.

(A) The magnitude of the vector $\vec{A} = 3\hat{i} + 6\hat{j} + 2\hat{k}$ is given by $|\vec{A}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
Let $\alpha, \beta, \text{ and } \gamma$ be the angles that the vector makes with the $x, y, \text{ and } z$ axes respectively.
The direction cosines are given by $\cos \alpha = \frac{A_x}{|\vec{A}|}, \cos \beta = \frac{A_y}{|\vec{A}|}, \text{ and } \cos \gamma = \frac{A_z}{|\vec{A}|}$.
Substituting the values: $\cos \alpha = \frac{3}{7}, \cos \beta = \frac{6}{7}, \text{ and } \cos \gamma = \frac{2}{7}$.
Therefore,the angles are $\alpha = \cos^{-1}(\frac{3}{7}), \beta = \cos^{-1}(\frac{6}{7}), \text{ and } \gamma = \cos^{-1}(\frac{2}{7})$.

(C) From the figure,the force $\vec{F}$ makes an angle of $45^{\circ}$ with the positive $x$-axis in the fourth quadrant.
Therefore,the angle with the positive $x$-axis is $\theta = -45^{\circ}$.
The components of the force are:
$F_x = F \cos(-45^{\circ}) = 100 \times \frac{1}{\sqrt{2}} = 50\sqrt{2}$
$F_y = F \sin(-45^{\circ}) = 100 \times \left(-\frac{1}{\sqrt{2}}\right) = -50\sqrt{2}$
Thus,the force in unit vector notation is:
$\vec{F} = F_x \hat{i} + F_y \hat{j} = 50\sqrt{2} \hat{i} - 50\sqrt{2} \hat{j} = 50\sqrt{2}(\hat{i} - \hat{j})$

(N/A) Let $\vec{P} = \hat{i} + \hat{j}$. The magnitude is $|\vec{P}| = \sqrt{1^2 + 1^2} = \sqrt{2}$. The direction is $\theta = \tan^{-1}(1/1) = 45^{\circ}$ with the $x$-axis.
Let $\vec{Q} = \hat{i} - \hat{j}$. The magnitude is $|\vec{Q}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. The direction is $\theta = \tan^{-1}(-1/1) = -45^{\circ}$ with the $x$-axis.
To find the component of $\vec{A} = 2\hat{i} + 3\hat{j}$ along a vector $\vec{V}$,we use the formula $\text{Component} = \left( \frac{\vec{A} \cdot \vec{V}}{|\vec{V}|} \right) \hat{V}$.
For $\vec{V}_1 = \hat{i} + \hat{j}$,$|\vec{V}_1| = \sqrt{2}$. The unit vector is $\hat{u}_1 = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
Component $= (\vec{A} \cdot \hat{u}_1) \hat{u}_1 = \left( \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} \right) \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \left( \frac{2+3}{2} \right) (\hat{i} + \hat{j}) = 2.5(\hat{i} + \hat{j})$.
For $\vec{V}_2 = \hat{i} - \hat{j}$,$|\vec{V}_2| = \sqrt{2}$. The unit vector is $\hat{u}_2 = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.
Component $= (\vec{A} \cdot \hat{u}_2) \hat{u}_2 = \left( \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j})}{\sqrt{2}} \right) \frac{\hat{i} - \hat{j}}{\sqrt{2}} = \left( \frac{2-3}{2} \right) (\hat{i} - \hat{j}) = -0.5(\hat{i} - \hat{j})$.

(N/A) In figure $(a)$,$\vec{A}$ and $\vec{B}$ are coplanar and non-parallel vectors.
We want to resolve vector $\vec{R}$ into components along $\vec{A}$ and $\vec{B}$.
Suppose $\vec{OQ}$ represents $\vec{R}$.
In figure $(b)$,draw a line through $O$ parallel to $\vec{A}$ and another line through $Q$ parallel to $\vec{B}$. These two lines intersect at point $P$.
According to the triangle law of vector addition:
$\vec{OQ} = \vec{OP} + \vec{PQ}$
Since $\vec{OP} \parallel \vec{A}$,we can write $\vec{OP} = \lambda \vec{A}$.
Since $\vec{PQ} \parallel \vec{B}$,we can write $\vec{PQ} = \mu \vec{B}$.
(Here,$\lambda$ and $\mu$ are scalar constants).
Therefore,$\vec{R} = \lambda \vec{A} + \mu \vec{B}$.
This means $\vec{R}$ is expressed as the sum of its components in the directions of $\vec{A}$ and $\vec{B}$.

(N/A) Consider an $O$-$XY$ two-dimensional Cartesian coordinate system as shown in the figure.
Let the position vector of point $P$ be $\vec{A}$.
By drawing a projection from $P$ to the $X$-axis,we obtain $OM$,where $\vec{OM} = \vec{A}_x = A_x \hat{i}$ is the $X$-component of $\vec{A}$.
By drawing a projection from $P$ to the $Y$-axis,we obtain $ON$,where $\vec{ON} = \vec{A}_y = A_y \hat{j}$ is the $Y$-component of $\vec{A}$,where $A_x$ and $A_y$ are real numbers.
From the figure,by the parallelogram law of vector addition:
$\vec{A} = \vec{A}_x + \vec{A}_y$
$\vec{A} = A_x \hat{i} + A_y \hat{j}$
Suppose $\vec{A}$ makes an angle $\theta$ with the $X$-axis.
In the right-angled triangle $\Delta OMP$:
$\cos \theta = \frac{A_x}{A} \implies A_x = A \cos \theta$
$\sin \theta = \frac{A_y}{A} \implies A_y = A \sin \theta$
From these equations,it is evident that the components can be positive,negative,or zero depending on the angle $\theta$.
Vectors can be represented in a plane in two ways:
$(i)$ By their magnitude and direction.
$(ii)$ By their components ($X$ and $Y$ components).

(B) The resolution of a vector is the process of splitting a single vector into two or more components along specific directions (usually perpendicular axes like $x$ and $y$).
This technique is primarily required to simplify the addition and subtraction of vectors.
By resolving vectors into their rectangular components,we can add or subtract them algebraically by simply summing or subtracting their respective $x$ and $y$ components,which is much easier than using the parallelogram law or triangle law for every operation.

(A) Let the total distance covered be $d = 500 \,m$ at an angle $\theta = 60^o$ with the horizontal runway.
The horizontal component of the distance is given by $d_x = d \cos(\theta)$.
$d_x = 500 \cos(60^o) = 500 \times (1/2) = 250 \,m$.
The vertical component of the distance is given by $d_y = d \sin(\theta)$.
$d_y = 500 \sin(60^o) = 500 \times (\sqrt{3}/2) = 250 \sqrt{3} \,m$.
Therefore,the horizontal distance is $250 \,m$ and the vertical distance is $250 \sqrt{3} \,m$.

(N/A) No. Let a vector $\vec{A}$ have rectangular components $A_x$ and $A_y$ in a two-dimensional plane.
These components are given by:
$A_x = A \cos \theta$
$A_y = A \sin \theta$
where $A$ is the magnitude of the vector and $\theta$ is the angle it makes with the $x$-axis.
Since the maximum value of both $\sin \theta$ and $\cos \theta$ is $1$,the maximum value of the components $A_x$ and $A_y$ can only be equal to $A$ (when $\theta = 0^\circ$ or $90^\circ$).
Mathematically,since $|\cos \theta| \le 1$ and $|\sin \theta| \le 1$,it follows that $|A_x| \le |A|$ and $|A_y| \le |A|$.
Therefore,the magnitude of a rectangular component can never exceed the magnitude of the original vector.

(N/A) Let $\alpha, \beta,$ and $\gamma$ be the angles between the vector $\overrightarrow{A}$ and the $x, y,$ and $z$-axes,respectively.
The components of the vector $\overrightarrow{A}$ along the $x, y,$ and $z$-axes are given by:
$A_{x} = A \cos \alpha$
$A_{y} = A \cos \beta$
$A_{z} = A \cos \gamma$
In general,the vector $\overrightarrow{A}$ can be expressed in terms of its components as:
$\overrightarrow{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$
The magnitude of the vector $\overrightarrow{A}$ is given by:
$|\overrightarrow{A}| = A = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$
Similarly,a position vector $\vec{r}$ can be expressed as:
$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$
where $x, y,$ and $z$ are the components of $\vec{r}$ along the $x, y,$ and $z$-axes,respectively.
The magnitude of the position vector $\vec{r}$ is:
$|\vec{r}| = \sqrt{x^{2} + y^{2} + z^{2}}$

(A) Let the force $\overrightarrow{P}$ be applied at point $A$ in the downward vertical direction. We need to resolve this force along the directions of the rods $AB$ and $AC$.
Let $\overrightarrow{F_{AB}}$ and $\overrightarrow{F_{AC}}$ be the components of force $\overrightarrow{P}$ along $AB$ and $AC$ respectively.
According to the parallelogram law of vector addition,$\overrightarrow{P} = \overrightarrow{F_{AB}} + \overrightarrow{F_{AC}}$.
Using the Lami's theorem or the sine rule in the triangle formed by the force vectors:
The angle between the vertical force $\overrightarrow{P}$ and rod $AB$ is $35^{\circ}$ (as the angle between the vertical and the wall is $90^{\circ}$,and the angle between $AB$ and the wall is $70^{\circ}$,so the angle between $AB$ and the vertical is $90^{\circ} - 70^{\circ} = 20^{\circ}$ is incorrect based on the provided image geometry; let's use the provided diagram).
From the diagram,the angle between the vertical force $\overrightarrow{P}$ and rod $AC$ is $35^{\circ}$.
The component of force $\overrightarrow{P}$ along the direction of rod $AC$ is given by $P \cos(\theta)$,where $\theta$ is the angle between the force vector and the rod.
Thus,the magnitude of the component along $AC$ is $x = 100 \cos(35^{\circ}) \; N$.
Given $\cos(35^{\circ}) = 0.819$.
$x = 100 \times 0.819 = 81.9 \; N$.
Rounding to the nearest integer,we get $x = 82$.

(A) Let the magnitude of each vector be $\lambda$.
$\overrightarrow{ OA } = \lambda (\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}) = \lambda (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j})$
$\overrightarrow{ OB } = \lambda (\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j}) = \lambda (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})$
$\overrightarrow{ OC } = \lambda (\cos 45^{\circ} (-\hat{i}) + \sin 45^{\circ} \hat{j}) = \lambda (-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$
Now,calculate $\overrightarrow{ R } = \overrightarrow{ OA } + \overrightarrow{ OB } - \overrightarrow{ OC }$.
$\overrightarrow{ R } = \lambda [(\frac{\sqrt{3}}{2} + \frac{1}{2} - (-\frac{1}{\sqrt{2}})) \hat{i} + (\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}) \hat{j}]$
$\overrightarrow{ R } = \lambda [(\frac{\sqrt{3}+1+\sqrt{2}}{2}) \hat{i} + (\frac{1-\sqrt{3}-\sqrt{2}}{2}) \hat{j}]$
The angle $\theta$ with the $x$-axis is given by $\tan \theta = \frac{R_y}{R_x}$.
$\tan \theta = \frac{\frac{1-\sqrt{3}-\sqrt{2}}{2}}{\frac{\sqrt{3}+1+\sqrt{2}}{2}} = \frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}$
Therefore,$\theta = \tan ^{-1} \frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}$.

(A) To find the resultant force,we resolve each force into its $x$ and $y$ components:
$F_x = 10 \cos 30^\circ + 20 \sin 30^\circ + 20 \cos 45^\circ - 15 \cos 45^\circ - 15 \sin 60^\circ$
$F_x = 10(0.85) + 20(0.5) + 20(0.7) - 15(0.7) - 15(0.85) = 8.5 + 10 + 14 - 10.5 - 12.75 = 9.25 \text{ N}$
$F_y = 10 \sin 30^\circ + 20 \cos 30^\circ + 15 \cos 60^\circ - 20 \sin 45^\circ - 15 \sin 45^\circ$
$F_y = 10(0.5) + 20(0.85) + 15(0.5) - 20(0.7) - 15(0.7) = 5 + 17 + 7.5 - 14 - 10.5 = 5 \text{ N}$
Thus,the resultant force is $9.25 \hat{i} + 5 \hat{j} \text{ N}$.

(D) The person walks a displacement vector of magnitude $r = 3.18 \,km$ at an angle $\theta = 25.0^{\circ}$ north of east.
To find the equivalent path by walking due north and then due east,we resolve the displacement vector into its rectangular components.
The component along the east direction (x-axis) is given by $x = r \cos \theta = 3.18 \times \cos 25.0^{\circ} \approx 3.18 \times 0.9063 = 2.88 \,km$.
The component along the north direction (y-axis) is given by $y = r \sin \theta = 3.18 \times \sin 25.0^{\circ} \approx 3.18 \times 0.4226 = 1.34 \,km$.
Therefore,the person must walk $1.34 \,km$ due north and $2.88 \,km$ due east to reach the same location.
Thus,the correct option is $(d)$.

(A) Given,magnitude of displacement vector $|\vec{d}| = 4$ and angle $\theta = 30^{\circ}$ with the $x$-axis.
The rectangular components of a vector $\vec{d}$ are given by:
$d_x = |\vec{d}| \cos \theta$
$d_y = |\vec{d}| \sin \theta$
Substituting the values:
$d_x = 4 \cos 30^{\circ} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$
$d_y = 4 \sin 30^{\circ} = 4 \times \frac{1}{2} = 2$
Thus,the rectangular components are $2\sqrt{3}$ and $2$.

(D) For a vector of magnitude $F = 10 \, N$ to have rectangular components $F_x$ and $F_y$,it must satisfy the condition $F = \sqrt{F_x^2 + F_y^2}$,which implies $F^2 = F_x^2 + F_y^2 = 10^2 = 100$.
Checking each option:
$A$: $6^2 + 8^2 = 36 + 64 = 100$. (Possible)
$B$: $7^2 + (\sqrt{51})^2 = 49 + 51 = 100$. (Possible)
$C$: $(6\sqrt{2})^2 + (2\sqrt{7})^2 = (36 \times 2) + (4 \times 7) = 72 + 28 = 100$. (Possible)
$D$: $9^2 + 1^2 = 81 + 1 = 82 \neq 100$. (Not possible)
Therefore,the pair $9 \, N$ and $1 \, N$ cannot be the rectangular components.

(D) Let the vector be $\vec{A}$. The $y$-component is given by $A_y = A \cos 30^{\circ}$.
Given $A_y = 2 \sqrt{3}$,we have:
$A \cos 30^{\circ} = 2 \sqrt{3}$
$A \left( \frac{\sqrt{3}}{2} \right) = 2 \sqrt{3}$
$A = 4$
Now,the $x$-component is given by $A_x = A \sin 30^{\circ}$.
$A_x = 4 \times \sin 30^{\circ} = 4 \times \frac{1}{2} = 2$.
Thus,the magnitude of the $x$-component is $2$.

(D) Let the vector $\overrightarrow{OP}$ be along the $x$-axis,$\overrightarrow{OP} = A \hat{i}$.
Vector $\overrightarrow{OQ}$ is at $90^\circ$ to $\overrightarrow{OP}$,so $\overrightarrow{OQ} = A \hat{j}$.
Vector $\overrightarrow{OR}$ is at $45^\circ$ below the $x$-axis,so $\overrightarrow{OR} = A \cos(45^\circ) \hat{i} - A \sin(45^\circ) \hat{j} = \frac{A}{\sqrt{2}} \hat{i} - \frac{A}{\sqrt{2}} \hat{j}$.
The resultant vector $\vec{R} = \overrightarrow{OP} + \overrightarrow{OQ} + \overrightarrow{OR} = (A + \frac{A}{\sqrt{2}}) \hat{i} + (A - \frac{A}{\sqrt{2}}) \hat{j}$.
The magnitude of the resultant is $|\vec{R}| = \sqrt{(A + \frac{A}{\sqrt{2}})^2 + (A - \frac{A}{\sqrt{2}})^2}$.
$|\vec{R}| = \sqrt{A^2(1 + \frac{1}{\sqrt{2}})^2 + A^2(1 - \frac{1}{\sqrt{2}})^2} = A \sqrt{(1 + \frac{1}{2} + \sqrt{2}) + (1 + \frac{1}{2} - \sqrt{2})} = A \sqrt{1 + 0.5 + 1 + 0.5} = A \sqrt{3}$.
Comparing $A \sqrt{3}$ with $A \sqrt{x}$,we get $x = 3$.

(B) The vector is given by $\vec{A} = \sqrt{3} \hat{i} - \hat{j}$.
Comparing this with $\vec{A} = A_x \hat{i} + A_y \hat{j}$,we get $A_x = \sqrt{3}$ and $A_y = -1$.
The angle $\theta$ that the vector makes with the positive $x$-axis is given by $\tan \theta = \frac{A_y}{A_x}$.
Substituting the values,we have $\tan \theta = \frac{-1}{\sqrt{3}}$.
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,$\tan(-30^{\circ}) = -\frac{1}{\sqrt{3}}$.
Therefore,$\theta = -30^{\circ}$.

(D) The components of a force $F$ acting at an angle $\theta$ with the $x$-axis are given by $F_{x} = F \cos \theta$ and $F_{y} = F \sin \theta$.
Given $\theta = 30^{\circ}$,we have:
$F_{x} = F \cos 30^{\circ} = F \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} F$.
$F_{y} = F \sin 30^{\circ} = F \times \frac{1}{2} = \frac{1}{2} F$.
Therefore,the $X$ and $Y$ components are $\frac{\sqrt{3}}{2} F$ and $\frac{1}{2} F$ respectively.

(A) The component of a vector $\vec{r}$ along the $x$-axis is given by $r_x = r \cos \theta$,where $\theta$ is the angle between the vector $\vec{r}$ and the $x$-axis.
For the component $r_x$ to have a maximum value,the term $\cos \theta$ must be maximum.
The maximum value of $\cos \theta$ is $1$,which occurs when $\theta = 0^{\circ}$.
Therefore,the component of the vector $\vec{r}$ along the $x$-axis is maximum when the vector $\vec{r}$ is directed along the positive $x$-axis.

(B) For a force $F$ with rectangular components $F_x$ and $F_y$,the relationship is given by $F^2 = F_x^2 + F_y^2$.
Given that the resultant force $F = 40 \ N$ and one component $F_x = 20 \sqrt{3} \ N$.
Substituting these values into the equation:
$(40)^2 = (20 \sqrt{3})^2 + F_y^2$
$1600 = (400 \times 3) + F_y^2$
$1600 = 1200 + F_y^2$
$F_y^2 = 1600 - 1200 = 400$
$F_y = \sqrt{400} = 20 \ N$.
Thus,the other rectangular component is $20 \ N$.

(B) The initial vector is $\vec{A} = 4\hat{i} + 10\hat{j}$.
The magnitude of the vector is $|\vec{A}| = \sqrt{(4)^2 + (10)^2} = \sqrt{16 + 100} = \sqrt{116} \,m$.
When the vector is rotated, its magnitude remains constant.
Let the new vector be $\vec{A}' = 8\hat{i} + n\hat{j}$, where the $x$-component is doubled $(4 \times 2 = 8)$.
Since the magnitude is conserved, $|\vec{A}'| = |\vec{A}|$.
$\sqrt{8^2 + n^2} = \sqrt{116}$.
Squaring both sides, we get $64 + n^2 = 116$.
$n^2 = 116 - 64 = 52$.
$n = \sqrt{52} \approx 7.21 \,m$.
Thus, the new $y$-component is approximately $7.2 \,m$.