Addition and Subtraction of Vectors Questions in English

(B) Let the required vector be $\vec{R}$.
Given vectors are $\vec{A} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{B} = 3\hat{i} + 6\hat{j} - 7\hat{k}$.
The resultant vector is a unit vector along the $y$-axis,which is $\hat{j}$.
According to the problem,$\vec{A} + \vec{B} + \vec{R} = \hat{j}$.
First,calculate the sum of $\vec{A}$ and $\vec{B}$:
$\vec{A} + \vec{B} = (1+3)\hat{i} + (-3+6)\hat{j} + (2-7)\hat{k} = 4\hat{i} + 3\hat{j} - 5\hat{k}$.
Now,substitute this into the equation:
$4\hat{i} + 3\hat{j} - 5\hat{k} + \vec{R} = \hat{j}$.
$\vec{R} = \hat{j} - (4\hat{i} + 3\hat{j} - 5\hat{k})$.
$\vec{R} = -4\hat{i} + (1-3)\hat{j} + 5\hat{k}$.
$\vec{R} = -4\hat{i} - 2\hat{j} + 5\hat{k}$.

(B) For the resultant of a set of vectors to be zero,the vectors must form a closed polygon when placed head-to-tail.
If we have two vectors with different magnitudes,their resultant can never be zero because they cannot form a closed shape (they would form a line segment if they are collinear or a triangle with an open side if they are not).
With three coplanar vectors of different magnitudes,we can form a triangle such that the sum of the vectors is zero (e.g.,$\vec{A} + \vec{B} + \vec{C} = 0$).
Therefore,the minimum number of coplanar vectors with different magnitudes required to obtain a zero resultant is $3$.

(A) First,find the resultant vector $\vec R$ by adding vectors $\vec A$ and $\vec B$:
$\vec R = \vec A + \vec B = (4\hat i + 3\hat j + 6\hat k) + (- \hat i + 3\hat j - 8\hat k)$
$\vec R = (4 - 1)\hat i + (3 + 3)\hat j + (6 - 8)\hat k = 3\hat i + 6\hat j - 2\hat k$
Next,calculate the magnitude of the resultant vector $|\vec R|$:
$|\vec R| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$
Finally,the unit vector $\hat R$ is given by $\hat R = \frac{\vec R}{|\vec R|}$:
$\hat R = \frac{3\hat i + 6\hat j - 2\hat k}{7} = \frac{1}{7}(3\hat i + 6\hat j - 2\hat k)$

(A) Given vectors are $\vec a = 4\hat i - \hat j$,$\vec b = -3\hat i + 2\hat j$,and $\vec c = -\hat k$.
Let the sum of these vectors be $\vec r = \vec a + \vec b + \vec c$.
$\vec r = (4\hat i - \hat j) + (-3\hat i + 2\hat j) + (-\hat k)$.
Grouping the components,we get $\vec r = (4 - 3)\hat i + (-1 + 2)\hat j - \hat k = \hat i + \hat j - \hat k$.
The magnitude of vector $\vec r$ is $|\vec r| = \sqrt{(1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.
The unit vector $\hat r$ in the direction of $\vec r$ is given by $\hat r = \frac{\vec r}{|\vec r|}$.
Therefore,$\hat r = \frac{\hat i + \hat j - \hat k}{\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat i + \hat j - \hat k)$.

(B) The resultant vector $\vec{R}$ is given by the sum of vectors $\vec{A}$ and $\vec{B}$:
$\vec{R} = \vec{A} + \vec{B} = (4\hat{i} - 3\hat{j}) + (8\hat{i} + 8\hat{j})$
$\vec{R} = (4 + 8)\hat{i} + (-3 + 8)\hat{j} = 12\hat{i} + 5\hat{j}$
The magnitude of the resultant vector is:
$|\vec{R}| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$
The unit vector $\hat{R}$ parallel to the resultant is given by:
$\hat{R} = \frac{\vec{R}}{|\vec{R}|} = \frac{12\hat{i} + 5\hat{j}}{13}$

(A) The resultant $R$ of two vectors $A$ and $B$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
$1$. For $R = 17\, N$: Since $17 = 5 + 12$,the vectors must be in the same direction. Thus,$\theta = 0^o$.
$2$. For $R = 7\, N$: Since $7 = 12 - 5$,the vectors must be in opposite directions. Thus,$\theta = 180^o$.
$3$. For $R = 13\, N$: Using the Pythagorean theorem,$\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\, N$. This occurs when the vectors are perpendicular,so $\theta = 90^o$.

(B) Given vectors are $\overrightarrow A = 4\hat i - 3\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$.
First,find the resultant vector $\overrightarrow R = \overrightarrow A + \overrightarrow B$:
$\overrightarrow R = (4\hat i - 3\hat j) + (6\hat i + 8\hat j) = (4+6)\hat i + (-3+8)\hat j = 10\hat i + 5\hat j$.
The magnitude of the resultant vector is given by:
$|\overrightarrow R| = \sqrt{(10)^2 + (5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}$.
The direction $\theta$ with respect to the $x$-axis is given by:
$\tan \theta = \frac{R_y}{R_x} = \frac{5}{10} = \frac{1}{2}$.
Therefore,$\theta = \tan^{-1}(1/2)$.

(D) Let the initial velocity be $\vec{v}_1 = 20\hat{j} \, m/s$ (towards North).
After turning west,the final velocity is $\vec{v}_2 = -20\hat{i} \, m/s$ (towards West).
The change in velocity is given by $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$.
Substituting the values,$\Delta\vec{v} = -20\hat{i} - 20\hat{j} = -20(\hat{i} + \hat{j}) \, m/s$.
The magnitude of the change in velocity is $|\Delta\vec{v}| = \sqrt{(-20)^2 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, m/s$.
The direction is given by $\tan\theta = \frac{|\Delta v_y|}{|\Delta v_x|} = \frac{20}{20} = 1$,so $\theta = 45^\circ$.
Since both components are negative,the direction is South-West $(S-W)$.
Thus,the correct option is $(d)$.

(B) Let $\hat{n}_1$ and $\hat{n}_2$ be two unit vectors. The magnitude of their sum is given by $|\hat{n}_1 + \hat{n}_2|^2 = |\hat{n}_1|^2 + |\hat{n}_2|^2 + 2|\hat{n}_1||\hat{n}_2|\cos\theta$.
Since the sum is a unit vector,$|\hat{n}_1 + \hat{n}_2| = 1$. Substituting the magnitudes $|\hat{n}_1| = 1$ and $|\hat{n}_2| = 1$,we get $1^2 = 1^2 + 1^2 + 2(1)(1)\cos\theta$.
$1 = 2 + 2\cos\theta \implies 2\cos\theta = -1 \implies \cos\theta = -1/2$.
This implies $\theta = 120^\circ$.
The magnitude of the difference vector is given by $|\hat{n}_1 - \hat{n}_2| = \sqrt{|\hat{n}_1|^2 + |\hat{n}_2|^2 - 2|\hat{n}_1||\hat{n}_2|\cos\theta}$.
Substituting the values,$|\hat{n}_1 - \hat{n}_2| = \sqrt{1^2 + 1^2 - 2(1)(1)(-1/2)} = \sqrt{1 + 1 + 1} = \sqrt{3}$.

(B) Given vectors are $\overrightarrow A = 2\hat i + \hat j$,$\overrightarrow B = 3\hat j - \hat k$,and $\overrightarrow C = 6\hat i - 2\hat k$.
We need to calculate $\overrightarrow A - 2\overrightarrow B + 3\overrightarrow C$.
Substituting the given vectors:
$\overrightarrow A - 2\overrightarrow B + 3\overrightarrow C = (2\hat i + \hat j) - 2(3\hat j - \hat k) + 3(6\hat i - 2\hat k)$
$= 2\hat i + \hat j - 6\hat j + 2\hat k + 18\hat i - 6\hat k$
Grouping the components of $\hat i$,$\hat j$,and $\hat k$:
$= (2 + 18)\hat i + (1 - 6)\hat j + (2 - 6)\hat k$
$= 20\hat i - 5\hat j - 4\hat k$
Thus,the correct option is $B$.

(B) The resultant $R$ of two vectors $A$ and $B$ with an angle $\theta$ between them is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given that the magnitudes of the two forces are $A = F$ and $B = F$,and their resultant magnitude is $R = F$.
Substituting these values into the formula: $F = \sqrt{F^2 + F^2 + 2(F)(F) \cos \theta}$.
Squaring both sides: $F^2 = F^2 + F^2 + 2F^2 \cos \theta$.
$F^2 = 2F^2 + 2F^2 \cos \theta$.
Dividing by $F^2$: $1 = 2 + 2 \cos \theta$.
$-1 = 2 \cos \theta$.
$\cos \theta = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = 120^\circ$.

(D) When two forces $\vec{F}_1$ and $\vec{F}_2$ act on a particle,the magnitude of the resultant force $R$ depends on the angle $\theta$ between them.
The maximum resultant force occurs when the forces are in the same direction $(\theta = 0^\circ)$:
$R_{\text{max}} = F_1 + F_2 = 4\, N + 3\, N = 7\, N$.
The minimum resultant force occurs when the forces are in opposite directions $(\theta = 180^\circ)$:
$R_{\text{min}} = |F_1 - F_2| = |4\, N - 3\, N| = 1\, N$.
Therefore,the net force on the particle can take any value in the range between $1\, N$ and $7\, N$ inclusive,depending on the angle between the forces.

(D) Let the two forces be $\vec{F_1}$ and $\vec{F_2}$ with magnitudes $F_1$ and $F_2$,where $F_1 > F_2$.
The magnitude of the resultant force $R$ is given by $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$,where $\theta$ is the angle between the forces.
We are given that $R < F_1$.
Squaring both sides,we get $F_1^2 + F_2^2 + 2F_1F_2 \cos \theta < F_1^2$.
This simplifies to $F_2^2 + 2F_1F_2 \cos \theta < 0$.
Dividing by $F_2$ (since $F_2 > 0$),we get $F_2 + 2F_1 \cos \theta < 0$,which implies $\cos \theta < -\frac{F_2}{2F_1}$.
Since $F_1 > F_2$,the value of $\frac{F_2}{2F_1}$ is between $0$ and $0.5$.
Thus,$\cos \theta$ must be negative,meaning the angle $\theta$ is obtuse $(90^\circ < \theta \le 180^\circ)$.
Among the given options,the condition that the forces point in opposite directions (or generally have an obtuse angle between them) is the only one that satisfies the requirement for the resultant to be smaller than the larger force.

(C) When two vectors act at an angle $\theta$ to each other,their resultant magnitude is given by the formula $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
Since the forces are mutually perpendicular,the angle between them is $\theta = 90^\circ$.
Substituting $\cos 90^\circ = 0$ into the formula:
$R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2(0)}$
$R = \sqrt{F_1^2 + F_2^2}$.
Therefore,the correct option is $C$.

(C) Given that the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\theta = 120^{\circ}$.
The magnitude of the resultant $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$ is given by:
$|\overrightarrow{C}| = \sqrt{A^{2} + B^{2} + 2AB \cos 120^{\circ}} = \sqrt{A^{2} + B^{2} - AB}$.
The magnitude of the difference vector $|\overrightarrow{A} - \overrightarrow{B}|$ is given by:
$|\overrightarrow{A} - \overrightarrow{B}| = \sqrt{A^{2} + B^{2} - 2AB \cos 120^{\circ}} = \sqrt{A^{2} + B^{2} - 2AB (-1/2)} = \sqrt{A^{2} + B^{2} + AB}$.
Comparing the two expressions,since $A$ and $B$ are nonzero magnitudes,$\sqrt{A^{2} + B^{2} - AB} < \sqrt{A^{2} + B^{2} + AB}$.
Therefore,$|\overrightarrow{C}| < |\overrightarrow{A} - \overrightarrow{B}|$.

(C) Given that the magnitudes are $|\overrightarrow{A}| = 12$,$|\overrightarrow{B}| = 5$,and $|\overrightarrow{C}| = 13$.
Since $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$,we can check if the magnitudes satisfy the Pythagorean theorem: $|\overrightarrow{A}|^2 + |\overrightarrow{B}|^2 = 12^2 + 5^2 = 144 + 25 = 169 = 13^2 = |\overrightarrow{C}|^2$.
Since $|\overrightarrow{A}|^2 + |\overrightarrow{B}|^2 = |\overrightarrow{C}|^2$,the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ must be perpendicular to each other.
Therefore,the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\pi / 2$ radians (or $90^{\circ}$).

(C) Let the two vectors be $\vec{A} = 6\hat{i} + 7\hat{j}$ and $\vec{B} = 3\hat{i} + 4\hat{j}$.
The resultant vector $\vec{R}$ is the sum of $\vec{A}$ and $\vec{B}$:
$\vec{R} = \vec{A} + \vec{B} = (6\hat{i} + 7\hat{j}) + (3\hat{i} + 4\hat{j})$
$\vec{R} = (6 + 3)\hat{i} + (7 + 4)\hat{j} = 9\hat{i} + 11\hat{j}$
The magnitude of the resultant vector $\vec{R}$ is given by:
$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$
$|\vec{R}| = \sqrt{9^2 + 11^2}$
$|\vec{R}| = \sqrt{81 + 121}$
$|\vec{R}| = \sqrt{202}$

(C) First,check if the vectors form a triangle by checking if their sum is zero. $\vec A + \vec B + \vec C = (3+1+2)\hat i + (-2-3+1)\hat j + (1+5-4)\hat k = 6\hat i - 4\hat j + 2\hat k \neq 0$. Since the sum is not zero,these vectors do not form a closed triangle in the standard sense of vector addition. However,if we treat the magnitudes of these vectors as the sides of a triangle,we calculate the magnitudes:
$|\vec A| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$
$|\vec B| = \sqrt{1^2 + (-3)^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$
$|\vec C| = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$
Observe that $|\vec A|^2 + |\vec C|^2 = 14 + 21 = 35 = |\vec B|^2$. Since the square of one side equals the sum of the squares of the other two sides,these lengths satisfy the Pythagorean theorem,forming a right-angled triangle.

(C) According to the triangle law of vector addition,if two vectors are represented by two sides of a triangle in sequence,their sum is represented by the third side taken in the opposite order.
In the given figure,vectors $\overrightarrow C$ and $\overrightarrow A$ are arranged in sequence (head-to-tail).
The resultant vector is $\overrightarrow B$,which connects the tail of $\overrightarrow C$ to the head of $\overrightarrow A$.
Therefore,by the triangle law of vector addition,we have: $\overrightarrow C + \overrightarrow A = \overrightarrow B$.

(B) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by $|\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
The range of the magnitude of the resultant vector is $|A - B| \le |\vec{C}| \le A + B$.
If the angle $\theta$ between $\vec{A}$ and $\vec{B}$ is obtuse (i.e.,$90^\circ < \theta \le 180^\circ$),the resultant vector $\vec{C}$ can have a magnitude smaller than either $|\vec{A}|$ or $|\vec{B}|$. For example,if $\vec{A}$ and $\vec{B}$ are vectors of equal magnitude $A=B$ and the angle between them is $120^\circ$,then $|\vec{C}| = \sqrt{A^2 + A^2 + 2A^2 \cos(120^\circ)} = \sqrt{2A^2 - A^2} = A$. If the angle is greater than $120^\circ$,$|\vec{C}|$ will be less than $A$ and $B$. Thus,option $(b)$ is correct.

(A) The resultant vector $\vec{R}$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the vector addition rule: $\vec{R} = \vec{A} + \vec{B}$.
The magnitude of the resultant vector is calculated using the parallelogram law of vector addition.
The formula for the magnitude $R$ is given by:
$R = |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,where $A$ and $B$ are the magnitudes of vectors $\vec{A}$ and $\vec{B}$ respectively,and $\theta$ is the angle between them.

(A) Let the two forces be $A$ and $B$,where $A < B$. Given $A + B = 16$ (equation $i$).
Since the resultant $R = 8\, N$ is perpendicular to the smaller force $A$,the angle $\alpha$ between $R$ and $A$ is $90^\circ$.
The formula for the direction of the resultant is $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} = \tan 90^\circ$.
This implies $A + B \cos \theta = 0$,so $\cos \theta = -A/B$ (equation $ii$).
The magnitude of the resultant is $R^2 = A^2 + B^2 + 2AB \cos \theta$.
Substituting $\cos \theta = -A/B$ into the resultant formula: $8^2 = A^2 + B^2 + 2AB(-A/B) = A^2 + B^2 - 2A^2 = B^2 - A^2$.
Thus,$B^2 - A^2 = 64$,which factors to $(B - A)(B + A) = 64$.
Since $B + A = 16$,we have $(B - A)(16) = 64$,so $B - A = 4$.
Solving $A + B = 16$ and $B - A = 4$ gives $2B = 20$,so $B = 10\, N$ and $A = 6\, N$.

(C) Given magnitudes are $|\overrightarrow P| = 5$,$|\overrightarrow Q| = 12$,and $|\overrightarrow R| = 13$.
Since $\overrightarrow P + \overrightarrow Q = \overrightarrow R$,these vectors form a right-angled triangle because $5^2 + 12^2 = 25 + 144 = 169 = 13^2$.
In the triangle formed by these vectors,$\overrightarrow Q$ and $\overrightarrow R$ meet at an angle $\theta$.
Using trigonometry in the right-angled triangle,the cosine of the angle $\theta$ between $\overrightarrow Q$ and $\overrightarrow R$ is given by the ratio of the adjacent side to the hypotenuse.
$\cos \theta = \frac{|\overrightarrow Q|}{|\overrightarrow R|} = \frac{12}{13}$.
Therefore,$\theta = \cos^{-1}(\frac{12}{13})$.

(B) Let the required vector be $\vec{R}$.
According to the problem,the sum of the given vectors and $\vec{R}$ is equal to the unit vector along the $X$-axis,which is $\hat{i}$.
$(\hat{i} - 2\hat{j} + 2\hat{k}) + (2\hat{i} + \hat{j} - \hat{k}) + \vec{R} = \hat{i}$
Combine the given vectors:
$(1+2)\hat{i} + (-2+1)\hat{j} + (2-1)\hat{k} + \vec{R} = \hat{i}$
$3\hat{i} - \hat{j} + \hat{k} + \vec{R} = \hat{i}$
Now,solve for $\vec{R}$:
$\vec{R} = \hat{i} - (3\hat{i} - \hat{j} + \hat{k})$
$\vec{R} = \hat{i} - 3\hat{i} + \hat{j} - \hat{k}$
$\vec{R} = -2\hat{i} + \hat{j} - \hat{k}$

(A) Let the resultant vector be $\vec R$.
Given vectors are $\vec A = \vec P + \vec Q$ and $\vec B = \vec P - \vec Q$.
The resultant $\vec R$ is the sum of these two vectors:
$\vec R = \vec A + \vec B = (\vec P + \vec Q) + (\vec P - \vec Q) = 2\vec P$.
Since $\vec R = 2\vec P$,the resultant vector is in the same direction as $\vec P$.
Therefore,the angle between $\vec P$ and the resultant $\vec R$ is $0$ degrees.

(B) Let $\theta$ be the angle between vectors $\overrightarrow P$ and $\overrightarrow Q$.
The resultant vector $\overrightarrow R = \overrightarrow P + \overrightarrow Q$ makes an angle $\alpha$ with $\overrightarrow P$,given by $\tan \alpha = \frac{Q \sin \theta}{P + Q \cos \theta}$.
Since the resultant is perpendicular to $\overrightarrow P$,the angle $\alpha = 90^\circ$.
Therefore,$\tan 90^\circ = \frac{Q \sin \theta}{P + Q \cos \theta}$.
Since $\tan 90^\circ$ is undefined,the denominator must be zero:
$P + Q \cos \theta = 0$
$Q \cos \theta = -P$
$\cos \theta = -\frac{P}{Q}$
$\theta = \cos^{-1}\left(-\frac{P}{Q}\right)$.

(A) The maximum magnitude of the resultant of two vectors $P$ and $Q$ is given by $R_{max} = P + Q$.
The minimum magnitude of the resultant of two vectors $P$ and $Q$ is given by $R_{min} = |P - Q|$.
According to the problem,the ratio of the maximum to the minimum magnitude is $3:1$,so $\frac{P + Q}{P - Q} = \frac{3}{1}$.
By cross-multiplying,we get $P + Q = 3(P - Q)$.
Expanding the equation,we have $P + Q = 3P - 3Q$.
Rearranging the terms,we get $4Q = 2P$,which simplifies to $P = 2Q$.

(C) Let the smaller force be $F_1 = F$ and the greater force be $F_2 = 2F$.
Given that the resultant force $R$ is equal to the greater force,so $R = 2F$.
The formula for the resultant of two vectors is $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
Substituting the given values: $2F = \sqrt{F^2 + (2F)^2 + 2(F)(2F) \cos \theta}$.
Squaring both sides: $(2F)^2 = F^2 + 4F^2 + 4F^2 \cos \theta$.
$4F^2 = 5F^2 + 4F^2 \cos \theta$.
Subtracting $5F^2$ from both sides: $-F^2 = 4F^2 \cos \theta$.
Dividing by $4F^2$: $\cos \theta = -1/4$.
Therefore,the angle $\theta = \cos^{-1}(-1/4)$.

(C) Let the initial displacement be $\vec{d_1} = 25\hat{i} - 6\hat{j} \text{ m}$.
Let the required displacement to be added be $\vec{d_x}$.
The resultant displacement is given as $\vec{d_R} = 7.0\hat{i} \text{ m}$.
According to the vector addition rule: $\vec{d_1} + \vec{d_x} = \vec{d_R}$.
Therefore,$\vec{d_x} = \vec{d_R} - \vec{d_1}$.
Substituting the values: $\vec{d_x} = (7.0\hat{i}) - (25\hat{i} - 6\hat{j})$.
$\vec{d_x} = 7.0\hat{i} - 25\hat{i} + 6\hat{j}$.
$\vec{d_x} = -18\hat{i} + 6\hat{j} \text{ m}$.

(D) The body moves in two perpendicular directions: East and North.
Let the velocity towards the East be $\vec{v}_1 = 20 \, km/h$ (along the $x$-axis) and the velocity towards the North be $\vec{v}_2 = 15 \, km/h$ (along the $y$-axis).
The resultant velocity $\vec{v}_R$ is given by the vector sum: $\vec{v}_R = \vec{v}_1 + \vec{v}_2$.
Since the vectors are perpendicular,the magnitude of the resultant velocity is calculated using the Pythagorean theorem:
$v_R = \sqrt{v_1^2 + v_2^2}$
$v_R = \sqrt{20^2 + 15^2}$
$v_R = \sqrt{400 + 225}$
$v_R = \sqrt{625}$
$v_R = 25 \, km/h$.
Therefore,the correct option is $D$.

(A) Given that the magnitudes are $|vec A| = 3$,$|vec B| = 4$,and $|vec C| = 5$.
Since $\vec A + \vec B = \vec C$,we can check if the magnitudes satisfy the Pythagorean theorem: $|vec A|^2 + |\vec B|^2 = 3^2 + 4^2 = 9 + 16 = 25 = 5^2 = |\vec C|^2$.
Since $|vec A|^2 + |\vec B|^2 = |\vec C|^2$,the vectors $\vec A$ and $\vec B$ must be perpendicular to each other.
Therefore,the angle between $\vec A$ and $\vec B$ is $\frac{\pi}{2}$.

(C) Let the starting point be the origin $(0, 0)$.
First displacement: $75 \, km$ North $\rightarrow \vec{d_1} = 75 \hat{j}$.
Second displacement: $60 \, km$ North-east $\rightarrow \vec{d_2} = 60 \cos 45^{\circ} \hat{i} + 60 \sin 45^{\circ} \hat{j} = 30\sqrt{2} \hat{i} + 30\sqrt{2} \hat{j}$.
Third displacement: $20 \, km$ East $\rightarrow \vec{d_3} = 20 \hat{i}$.
The total displacement vector $\vec{R} = \vec{d_1} + \vec{d_2} + \vec{d_3} = (30\sqrt{2} + 20) \hat{i} + (75 + 30\sqrt{2}) \hat{j}$.
Using $\sqrt{2} \approx 1.414$,$x = 30(1.414) + 20 = 42.42 + 20 = 62.42 \, km$.
$y = 75 + 30(1.414) = 75 + 42.42 = 117.42 \, km$.
The minimum distance (magnitude of displacement) is $S = \sqrt{x^2 + y^2} = \sqrt{(62.42)^2 + (117.42)^2} \approx \sqrt{3896 + 13787} = \sqrt{17683} \approx 132.97 \, km$.
Rounding to the nearest integer,the distance is $132 \, km$.

(D) The resultant $R$ of two forces $A$ and $B$ lies in the range $(A - B) \le R \le (A + B)$.
For option $A$: $2 \, N$ and $2 \, N$,the range is $(2 - 2) \le R \le (2 + 2)$,i.e.,$0 \le R \le 4 \, N$. Since $2 \, N$ is in this range,it is possible.
For option $B$: $1 \, N$ and $1 \, N$,the range is $(1 - 1) \le R \le (1 + 1)$,i.e.,$0 \le R \le 2 \, N$. Since $2 \, N$ is in this range,it is possible.
For option $C$: $1 \, N$ and $3 \, N$,the range is $(3 - 1) \le R \le (3 + 1)$,i.e.,$2 \le R \le 4 \, N$. Since $2 \, N$ is in this range,it is possible.
For option $D$: $1 \, N$ and $4 \, N$,the range is $(4 - 1) \le R \le (4 + 1)$,i.e.,$3 \le R \le 5 \, N$. Since $2 \, N$ is not in this range,it is impossible to obtain a resultant of $2 \, N$.

(D) Let the two forces be $F_1 = 3\,N$ and $F_2 = 2\,N$. The resultant $R$ is given by the formula $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
Substituting the values,we get $R = \sqrt{3^2 + 2^2 + 2(3)(2) \cos \theta} = \sqrt{9 + 4 + 12 \cos \theta} = \sqrt{13 + 12 \cos \theta}$ ... $(i)$.
When the first force is increased to $F_1' = 6\,N$,the new resultant is $2R$. Thus,$(2R)^2 = (F_1')^2 + F_2^2 + 2F_1'F_2 \cos \theta$.
$4R^2 = 6^2 + 2^2 + 2(6)(2) \cos \theta = 36 + 4 + 24 \cos \theta = 40 + 24 \cos \theta$.
From $(i)$,$R^2 = 13 + 12 \cos \theta$. Substituting this into the equation for $4R^2$:
$4(13 + 12 \cos \theta) = 40 + 24 \cos \theta$.
$52 + 48 \cos \theta = 40 + 24 \cos \theta$.
$24 \cos \theta = -12$.
$\cos \theta = -\frac{12}{24} = -\frac{1}{2}$.
Therefore,$\theta = 120^\circ$.

(C) The resultant force $R$ of two vectors $A$ and $B$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
For the resultant force to be maximum,the angle $\theta$ between the two forces must be $0^\circ$.
Thus,$R_{\max} = A + B$.
Given $A = 12 \, N$ and $B = 8 \, N$.
Therefore,$R_{\max} = 12 + 8 = 20 \, N$.

(C) The magnitude of the resultant $R$ of two vectors $A$ and $B$ inclined at an angle $\theta$ is given by the formula:
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Given that the two forces are equal,$A = B = P$,and the angle between them is $\theta = 120^\circ$.
Substituting these values into the formula:
$R = \sqrt{P^2 + P^2 + 2(P)(P) \cos(120^\circ)}$
Since $\cos(120^\circ) = -1/2$:
$R = \sqrt{2P^2 + 2P^2(-1/2)}$
$R = \sqrt{2P^2 - P^2}$
$R = \sqrt{P^2} = P$
Therefore,the magnitude of the resultant force is $P$.

(A) Let the two vectors be $\vec{A} = 5\hat{i} + 8\hat{j}$ and $\vec{B} = 2\hat{i} + 7\hat{j}$.
Adding these vectors,we get the resultant vector $\vec{R} = \vec{A} + \vec{B} = (5+2)\hat{i} + (8+7)\hat{j} = 7\hat{i} + 15\hat{j}$.
The magnitude of the resultant vector $\vec{R}$ is given by $|\vec{R}| = \sqrt{R_x^2 + R_y^2}$.
Substituting the values,$|\vec{R}| = \sqrt{7^2 + 15^2} = \sqrt{49 + 225} = \sqrt{274}$.

(D) Given the equation: $\vec{A} + \vec{B} = \vec{A} - \vec{B}$.
Subtract $\vec{A}$ from both sides of the equation:
$(\vec{A} - \vec{A}) + \vec{B} = -\vec{B}$.
This simplifies to: $0 + \vec{B} = -\vec{B}$.
Add $\vec{B}$ to both sides:
$\vec{B} + \vec{B} = 0$.
$2\vec{B} = 0$.
Therefore,$\vec{B} = 0$.

(A) Given that $(\overrightarrow{A} + \overrightarrow{B})$ is perpendicular to $(\overrightarrow{A} - \overrightarrow{B})$.
Since the dot product of two perpendicular vectors is zero,we have:
$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} - \overrightarrow{B}) = 0$
Expanding the dot product:
$\overrightarrow{A} \cdot \overrightarrow{A} - \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} - \overrightarrow{B} \cdot \overrightarrow{B} = 0$
Using the commutative property of the dot product,$\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A}$,the middle terms cancel out:
$|\overrightarrow{A}|^2 - |\overrightarrow{B}|^2 = 0$
$|\overrightarrow{A}|^2 = |\overrightarrow{B}|^2$
$|\overrightarrow{A}| = |\overrightarrow{B}|$
Therefore,the ratio of their magnitudes is $\frac{|\overrightarrow{A}|}{|\overrightarrow{B}|} = 1$.

(C) Given: $|{\vec V_1} + {\vec V_2}| = |{\vec V_1} - {\vec V_2}|$
Squaring both sides:
$|{\vec V_1} + {\vec V_2}|^2 = |{\vec V_1} - {\vec V_2}|^2$
Using the property $|\vec A \pm \vec B|^2 = |\vec A|^2 + |\vec B|^2 \pm 2(\vec A \cdot \vec B)$:
$|{\vec V_1}|^2 + |{\vec V_2}|^2 + 2(\vec V_1 \cdot \vec V_2) = |{\vec V_1}|^2 + |{\vec V_2}|^2 - 2(\vec V_1 \cdot \vec V_2)$
$2(\vec V_1 \cdot \vec V_2) = -2(\vec V_1 \cdot \vec V_2)$
$4(\vec V_1 \cdot \vec V_2) = 0$
$\vec V_1 \cdot \vec V_2 = 0$
This implies that the vectors ${\vec V_1}$ and ${\vec V_2}$ are mutually perpendicular.

(A) Given that $|\vec{A}| = |\vec{B}|$.
Let $\vec{R} = \vec{A} + \vec{B}$ and $\vec{D} = \vec{A} - \vec{B}$.
Consider the dot product $\vec{R} \cdot \vec{D} = (\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B})$.
$= \vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{A} - \vec{B} \cdot \vec{B}$.
$= |\vec{A}|^2 - |\vec{B}|^2$.
Since $|\vec{A}| = |\vec{B}|$,the dot product is $0$.
Therefore,$\vec{A} + \vec{B}$ is perpendicular to $\vec{A} - \vec{B}$.

(C) The resultant of $2$ vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{R} = \vec{A} + \vec{B}$.
For the resultant to be zero,$\vec{A} + \vec{B} = 0$,which implies $\vec{A} = -\vec{B}$.
This means the two vectors must have the same magnitude $(|A| = |B|)$ but must be directed in opposite directions (an angle of $180^{\circ}$ or $\pi$ radians between them).
Therefore,the correct condition is that the $2$ vectors are same in magnitude but opposite in sense.

(C) Let $P$ be the smaller force and $Q$ be the greater force. According to the problem:
$P + Q = 18$......$(i)$
Given that the resultant $R = 12 \; N$ is perpendicular to the smaller force $P$,the angle between $R$ and $P$ is $90^{\circ}$.
Using the formula for the direction of the resultant: $\tan \alpha = \frac{Q \sin \theta}{P + Q \cos \theta}$.
Since $\alpha = 90^{\circ}$,$\tan 90^{\circ} = \infty$,which implies $P + Q \cos \theta = 0$,so $Q \cos \theta = -P$......$(ii)$
The magnitude of the resultant is given by $R^2 = P^2 + Q^2 + 2PQ \cos \theta$.
Substituting $R = 12$ and $Q \cos \theta = -P$ into the equation:
$12^2 = P^2 + Q^2 + 2P(-P)$
$144 = P^2 + Q^2 - 2P^2$
$144 = Q^2 - P^2$
$144 = (Q - P)(Q + P)$
Since $Q + P = 18$,we have $144 = (Q - P)(18)$,which gives $Q - P = 8$......$(iii)$
Adding $(i)$ and $(iii)$: $2Q = 26 \implies Q = 13 \; N$.
Subtracting $(iii)$ from $(i)$: $2P = 10 \implies P = 5 \; N$.
Thus,the magnitudes of the forces are $5 \; N$ and $13 \; N$.

(B) Let the vectors be represented in a Cartesian coordinate system with $O$ at the origin $(0,0)$.
Vector $\overrightarrow{OA}$ is along the positive $x$-axis: $\overrightarrow{OA} = R\hat{i}$.
Vector $\overrightarrow{OC}$ is along the positive $y$-axis: $\overrightarrow{OC} = R\hat{j}$.
Vector $\overrightarrow{OB}$ makes an angle of $45^{\circ}$ with the $x$-axis: $\overrightarrow{OB} = R\cos(45^{\circ})\hat{i} + R\sin(45^{\circ})\hat{j} = \frac{R}{\sqrt{2}}\hat{i} + \frac{R}{\sqrt{2}}\hat{j}$.
The resultant vector $\vec{R}_{net} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = (R + \frac{R}{\sqrt{2}})\hat{i} + (R + \frac{R}{\sqrt{2}})\hat{j}$.
The magnitude is $|\vec{R}_{net}| = \sqrt{(R + \frac{R}{\sqrt{2}})^2 + (R + \frac{R}{\sqrt{2}})^2} = \sqrt{2(R + \frac{R}{\sqrt{2}})^2} = \sqrt{2}(R + \frac{R}{\sqrt{2}}) = R\sqrt{2} + R = R(\sqrt{2} + 1)$.

(B) The resultant $R$ of two forces $F_1$ and $F_2$ with an angle $\theta$ between them is given by the formula: $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
Given $F_1 = F$,$F_2 = F$,and the resultant $R = F$.
Substituting these values into the formula: $F = \sqrt{F^2 + F^2 + 2F^2 \cos \theta}$.
Squaring both sides: $F^2 = 2F^2 + 2F^2 \cos \theta$.
Dividing by $F^2$: $1 = 2 + 2 \cos \theta$.
Rearranging the terms: $2 \cos \theta = 1 - 2 = -1$.
Thus,$\cos \theta = -1/2$.
Since $\cos(120^\circ) = -1/2$,the angle between the two forces is $\theta = 120^\circ$.

(A) The magnitude of the resultant force $R$ of two vectors $F_1$ and $F_2$ with an angle $\theta$ between them is given by the formula: $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$.
Given that $F_1 = F_2 = F$ and $R = F/3$,we substitute these values into the equation:
$(F/3)^2 = F^2 + F^2 + 2(F)(F)\cos\theta$.
$F^2/9 = 2F^2 + 2F^2\cos\theta$.
Dividing both sides by $F^2$:
$1/9 = 2 + 2\cos\theta$.
$1/9 - 2 = 2\cos\theta$.
$-17/9 = 2\cos\theta$.
$\cos\theta = -17/18$.
Therefore,$\theta = \cos^{-1}(-17/18)$.

(C) The resultant force $R$ of two vectors $A$ and $B$ lies in the range $|A - B| \le R \le |A + B|$.
Given $A = 5 \, N$ and $B = 10 \, N$.
The maximum resultant force is $F_{\max} = 10 + 5 = 15 \, N$.
The minimum resultant force is $F_{\min} = 10 - 5 = 5 \, N$.
Therefore,the resultant force must be in the range $5 \, N \le R \le 15 \, N$.
Since $4 \, N$ is less than the minimum possible resultant force of $5 \, N$,the resultant cannot be $4 \, N$.

(B) Let the angle between the two forces be $\theta$.
Using the law of parallelogram of vectors,the resultant $R$ is given by:
$R^2 = (3P)^2 + (2P)^2 + 2(3P)(2P) \cos \theta$
$R^2 = 9P^2 + 4P^2 + 12P^2 \cos \theta = 13P^2 + 12P^2 \cos \theta$ ... $(i)$
When the first force is doubled,it becomes $6P$. The new resultant is $2R$:
$(2R)^2 = (6P)^2 + (2P)^2 + 2(6P)(2P) \cos \theta$
$4R^2 = 36P^2 + 4P^2 + 24P^2 \cos \theta = 40P^2 + 24P^2 \cos \theta$ ... (ii)
Divide equation (ii) by $4$:
$R^2 = 10P^2 + 6P^2 \cos \theta$ ... (iii)
Equating $(i)$ and (iii):
$13P^2 + 12P^2 \cos \theta = 10P^2 + 6P^2 \cos \theta$
$3P^2 = -6P^2 \cos \theta$
$\cos \theta = -3/6 = -1/2$
Therefore,$\theta = 120^\circ$.

(B) Let the two forces be $F_1$ and $F_2$,where $F_1$ is the larger force and $F_2$ is the smaller force.
Given: $F_1 + F_2 = 18 \,N$ $(i)$
Let the resultant $R = 12 \,N$ be perpendicular to the smaller force $F_2$.
The angle $\alpha$ between the resultant $R$ and the force $F_2$ is $90^\circ$.
The formula for the direction of the resultant is $\tan \alpha = \frac{F_1 \sin \theta}{F_2 + F_1 \cos \theta} = \tan 90^\circ$.
This implies the denominator $F_2 + F_1 \cos \theta = 0$,so $\cos \theta = -\frac{F_2}{F_1}$.
The magnitude of the resultant is $R^2 = F_1^2 + F_2^2 + 2F_1 F_2 \cos \theta$.
Substituting $\cos \theta = -\frac{F_2}{F_1}$,we get $R^2 = F_1^2 + F_2^2 + 2F_1 F_2 (-F_2/F_1) = F_1^2 - F_2^2$.
Given $R = 12$,so $144 = F_1^2 - F_2^2 = (F_1 + F_2)(F_1 - F_2)$.
Using $(i)$,$144 = 18(F_1 - F_2)$,which gives $F_1 - F_2 = 8 \,N$ (ii).
Adding $(i)$ and (ii): $2F_1 = 26 \,N \implies F_1 = 13 \,N$.
Subtracting (ii) from $(i)$: $2F_2 = 10 \,N \implies F_2 = 5 \,N$.
Thus,the magnitudes are $13 \,N$ and $5 \,N$.

(D) According to the triangle law of vector addition,if two vectors are represented by two sides of a triangle in sequence (head-to-tail),then their sum (resultant) is represented by the third side of the triangle taken in the opposite order (from the tail of the first to the head of the second).
In option $D$,the vector $\overrightarrow {{F_2}} $ is placed such that its tail is at the head of $\overrightarrow {{F_1}} $. The resultant vector $\overrightarrow {{F_3}} $ starts from the tail of $\overrightarrow {{F_1}} $ and ends at the head of $\overrightarrow {{F_2}} $. This correctly represents $\overrightarrow {{F_3}} = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} $.
Therefore,the correct arrangement is $D$.