Addition and Subtraction of Vectors Questions in English

(C) Angular velocity is a vector quantity. Let the angular velocity of the first disc be $\vec{\omega}_1 = 3 \hat{i} \, rad/s$ and the angular velocity of the second disc be $\vec{\omega}_2 = 4 \hat{j} \, rad/s$.
Since the planes of the discs are perpendicular,their axes of rotation are also perpendicular.
The resultant angular velocity $\vec{\omega}_{res}$ is given by the vector sum of the two individual angular velocities:
$\vec{\omega}_{res} = \vec{\omega}_1 + \vec{\omega}_2$.
The magnitude of the resultant angular velocity is given by:
$|\vec{\omega}_{res}| = \sqrt{\omega_1^2 + \omega_2^2}$.
Substituting the given values:
$|\vec{\omega}_{res}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, rad/s$.

(C) Let the initial velocity be $\vec{v}_1 = 50 \; \text{km/h}$ due north.
After turning left by $90^{\circ}$,the final velocity is $\vec{v}_2 = 50 \; \text{km/h}$ due west.
The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = \vec{v}_2 + (-\vec{v}_1)$.
Here,$-\vec{v}_1 = 50 \; \text{km/h}$ due south.
The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v_2^2 + v_1^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} \approx 70.7 \; \text{km/h}$.
The direction of $\Delta \vec{v}$ is the resultant of a vector pointing west and a vector pointing south,which is in the south-west direction.

(A) The resultant vector $\vec{R}$ is given by $\vec{R} = \vec{A} + \vec{B}$.
Substituting the given vectors: $\vec{R} = (4\hat{i} + 3\hat{j} + 6\hat{k}) + (-\hat{i} + 3\hat{j} - 8\hat{k})$.
Combining the components: $\vec{R} = (4-1)\hat{i} + (3+3)\hat{j} + (6-8)\hat{k} = 3\hat{i} + 6\hat{j} - 2\hat{k}$.
The unit vector parallel to $\vec{R}$ is $\hat{R} = \frac{\vec{R}}{|\vec{R}|}$.
First,calculate the magnitude $|\vec{R}| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
Therefore,the unit vector is $\hat{R} = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{7} = \frac{1}{7}(3\hat{i} + 6\hat{j} - 2\hat{k})$.

(C) Given the condition $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.
Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.
Using the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$,we expand the expression:
$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$.
$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Canceling $|\vec{a}|^2$ and $|\vec{b}|^2$ from both sides,we get $2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})$.
This implies $4(\vec{a} \cdot \vec{b}) = 0$,so $\vec{a} \cdot \vec{b} = 0$.
Since the dot product of two non-zero vectors is zero,the vectors must be perpendicular to each other.
Therefore,the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.

(B) Given,two forces $F_1 = 3 \ N$ and $F_2 = 4 \ N$.
The angle between them is $\theta = 180^\circ$.
Since the angle is $180^\circ$,the forces are anti-parallel (acting in opposite directions).
The resultant force $R$ is given by the difference of the magnitudes of the two forces:
$R = |F_2 - F_1| = |4 \ N - 3 \ N| = 1 \ N$.
The direction of the resultant force is in the direction of the larger force (i.e.,the $4 \ N$ force).

(D) When the angle between two vectors is $\theta = 0^{\circ}$,the vectors are parallel and act in the same direction.
The resultant force $R$ is given by the sum of the magnitudes of the two forces:
$R = F_1 + F_2$
Given $F_1 = 3\;N$ and $F_2 = 4\;N$,
$R = 3\;N + 4\;N = 7\;N$.
The direction of the resultant force is the same as the direction of the individual forces.

(C) Let the two forces be $F_1 = F$ and $F_2 = F$.
The magnitude of the resultant force $R$ is given by $R^2 = F_1^2 + F_2^2 + 2F_1F_2 \cos \theta$,where $\theta$ is the angle between the forces.
Given that $R^2 = 3(F_1 \times F_2) = 3F^2$.
Substituting the values into the formula:
$F^2 + F^2 + 2F^2 \cos \theta = 3F^2$
$2F^2 + 2F^2 \cos \theta = 3F^2$
$2F^2 \cos \theta = F^2$
$\cos \theta = \frac{1}{2}$
Therefore,$\theta = 60^o$.

(D) The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ of two vectors with magnitudes $A$ and $B$ must lie in the range $|A - B| \leq R \leq A + B$.
Given $A = 10 \ N$ and $B = 6 \ N$,the range is $|10 - 6| \leq R \leq 10 + 6$,which simplifies to $4 \leq R \leq 16$.
This means the resultant magnitude $R$ must be between $4 \ N$ and $16 \ N$ (inclusive).
Among the given options,$2 \ N$ is less than $4 \ N$,so it cannot be the magnitude of the resultant vector.

(A) The resultant vector $\vec{R} = \vec{P} + \vec{Q} = (2\hat{i} + 7\hat{j} - 10\hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k})$.
$\vec{R} = (2+1)\hat{i} + (7+2)\hat{j} + (-10+3)\hat{k} = 3\hat{i} + 9\hat{j} - 7\hat{k}$.
Let the required vector be $\vec{S}$. We are given that $\vec{R} + \vec{S} = \hat{i}$ (the unit vector along the $X$-axis).
Therefore,$\vec{S} = \hat{i} - \vec{R} = \hat{i} - (3\hat{i} + 9\hat{j} - 7\hat{k})$.
$\vec{S} = (1-3)\hat{i} - 9\hat{j} + 7\hat{k} = -2\hat{i} - 9\hat{j} + 7\hat{k}$.

(B) The angle between vectors $\vec{A}$ and $\vec{B}$ is $\theta = 110^\circ - 20^\circ = 90^\circ$.
The magnitude of the resultant vector $\vec{R}$ is given by:
$R = \sqrt{A^2 + B^2 + 2AB \cos(90^\circ)} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ m$.
Let $\alpha$ be the angle that the resultant vector $\vec{R}$ makes with vector $\vec{A}$.
Using the formula for the direction of the resultant:
$\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} = \frac{12 \sin 90^\circ}{5 + 12 \cos 90^\circ} = \frac{12 \times 1}{5 + 12 \times 0} = \frac{12}{5}$.
Therefore,$\alpha = \tan^{-1}(12/5)$.
Since vector $\vec{A}$ itself makes an angle of $20^\circ$ with the $x$-axis,the angle of the resultant vector $\vec{R}$ with the $x$-axis is $\alpha + 20^\circ = \tan^{-1}(12/5) + 20^\circ$.

(A) Given: Magnitude of vectors $A = B = 5$ units,angle $\theta = 60^\circ$.
$(a)$ Magnitude of the sum of vectors: $R_s = \sqrt{A^2 + B^2 + 2AB \cos \theta} = \sqrt{5^2 + 5^2 + 2(5)(5) \cos 60^\circ} = \sqrt{25 + 25 + 50(0.5)} = \sqrt{75} = 5\sqrt{3}$ units.
$(b)$ Magnitude of the difference of vectors: $R_d = \sqrt{A^2 + B^2 - 2AB \cos \theta} = \sqrt{5^2 + 5^2 - 2(5)(5) \cos 60^\circ} = \sqrt{25 + 25 - 50(0.5)} = \sqrt{25} = 5$ units.

(C) Given that $\vec{R_1} = \vec{A} + \vec{B}$.
Taking the magnitude squared: $R_1^2 = |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB \cos \theta$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
When vector $\vec{B}$ is reversed,the new resultant is $\vec{R_2} = \vec{A} - \vec{B}$.
Taking the magnitude squared: $R_2^2 = |\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB \cos \theta$.
Adding the two equations:
$R_1^2 + R_2^2 = (A^2 + B^2 + 2AB \cos \theta) + (A^2 + B^2 - 2AB \cos \theta)$.
$R_1^2 + R_2^2 = 2A^2 + 2B^2 = 2(A^2 + B^2)$.

(B) Let the two forces be $\vec{F}_1$ and $\vec{F}_2$.
The vector sum is $(\vec{F}_1 + \vec{F}_2)$ and the vector difference is $(\vec{F}_1 - \vec{F}_2)$.
Given that the sum is perpendicular to the difference,their dot product must be zero:
$(\vec{F}_1 + \vec{F}_2) \cdot (\vec{F}_1 - \vec{F}_2) = 0$
Expanding the dot product:
$\vec{F}_1 \cdot \vec{F}_1 - \vec{F}_1 \cdot \vec{F}_2 + \vec{F}_2 \cdot \vec{F}_1 - \vec{F}_2 \cdot \vec{F}_2 = 0$
Since the dot product is commutative $(\vec{F}_1 \cdot \vec{F}_2 = \vec{F}_2 \cdot \vec{F}_1)$,the middle terms cancel out:
$|\vec{F}_1|^2 - |\vec{F}_2|^2 = 0$
Therefore,$|\vec{F}_1|^2 = |\vec{F}_2|^2$,which implies $|\vec{F}_1| = |\vec{F}_2|$.
Thus,the forces have equal magnitudes.

(A) Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$.
Given that $\vec{R} \perp \vec{A}$,the dot product $\vec{R} \cdot \vec{A} = 0$.
Substituting $\vec{R} = \vec{A} + \vec{B}$,we get $(\vec{A} + \vec{B}) \cdot \vec{A} = 0$.
Expanding the dot product: $\vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = 0$.
This simplifies to $A^2 + AB \cos \theta = 0$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
Solving for $\cos \theta$: $\cos \theta = -\frac{A^2}{AB} = -\frac{A}{B}$.
Therefore,$\theta = \cos^{-1}\left(-\frac{A}{B}\right)$.

(A) To obtain a zero resultant vector,the vectors must form a closed polygon when placed head-to-tail.
For vectors of equal magnitude,the minimum number of vectors required to form a closed polygon is $2$ (if they are in opposite directions,i.e.,$\vec{A} + (-\vec{A}) = 0$).
Therefore,the minimum number of vectors of equal magnitude required to produce a zero resultant is $2$.

(B) Let $\theta$ be the angle between $\vec{P}$ and $\vec{Q}$. The resultant $\vec{R} = \vec{P} + \vec{Q}$.
When $\vec{Q}$ is doubled,the new resultant is $\vec{R}' = \vec{P} + 2\vec{Q}$.
It is given that $\vec{R}'$ is perpendicular to $\vec{P}$,so their dot product is zero:
$(\vec{P} + 2\vec{Q}) \cdot \vec{P} = 0$
$\vec{P} \cdot \vec{P} + 2(\vec{Q} \cdot \vec{P}) = 0$
$P^2 + 2PQ \cos \theta = 0$
$2PQ \cos \theta = -P^2$
Now,the magnitude of the original resultant $\vec{R}$ is:
$R = |\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$
Substituting $2PQ \cos \theta = -P^2$ into the equation:
$R = \sqrt{P^2 + Q^2 - P^2}$
$R = \sqrt{Q^2} = Q$

(B) Let there be $n$ vectors $\vec{A}_1, \vec{A}_2, ..., \vec{A}_n$ with $x$-components $A_{1x}, A_{2x}, ..., A_{nx}$ and magnitudes $|\vec{A}_1|, |\vec{A}_2|, ..., |\vec{A}_n|$.
The resultant vector is $\vec{R} = \sum_{i=1}^{n} \vec{A}_i$.
The $x$-component of the resultant is $R_x = \sum_{i=1}^{n} A_{ix}$. Thus,statement $(a)$ is correct.
Since $|A_{ix}| \le |\vec{A}_i|$,the sum of $x$-components $\sum A_{ix} \le \sum |\vec{A}_i|$. Therefore,the $x$-component of the resultant is generally smaller than or equal to the sum of the magnitudes of the vectors. Thus,statement $(b)$ is correct.
Statement $(c)$ is incorrect because the $x$-component of a vector cannot exceed its own magnitude,and consequently,the sum of $x$-components cannot exceed the sum of magnitudes.
Statement $(d)$ is only true if all vectors are directed along the positive $x$-axis. In general,it is not true. Thus,statement $(d)$ is incorrect.
Therefore,the correct statements are $(a)$ and $(b)$.

(C) Given that the magnitudes are $P = 5$,$Q = 12$,and $R = 13$.
Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the vectors form a right-angled triangle where $\vec R$ is the hypotenuse.
In the triangle formed by vectors $\vec P$,$\vec Q$,and $\vec R$,the angle $\theta$ is between $\vec Q$ and $\vec R$.
Using trigonometry in the right-angled triangle,$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{Q}{R}$.
Substituting the values,$\cos \theta = \frac{12}{13}$.
Therefore,$\theta = \cos^{-1}(\frac{12}{13})$.

(B) The angle between vector $\vec{A}$ and vector $\vec{B}$ is $\theta = 110^\circ - 20^\circ = 90^\circ$.
The magnitude of the resultant vector $R$ is given by the formula:
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Substituting the given values $A = 5 \, m$,$B = 12 \, m$,and $\theta = 90^\circ$:
$R = \sqrt{5^2 + 12^2 + 2(5)(12) \cos 90^\circ}$
Since $\cos 90^\circ = 0$:
$R = \sqrt{25 + 144 + 0} = \sqrt{169} = 13 \, m$.
Thus,the magnitude of the resultant vector is $13 \, m$.

(B) Let the initial vector be $\vec{\ell}_1$ and the final vector be $\vec{\ell}_2$. Both have magnitude $\ell$,so $|\vec{\ell}_1| = |\vec{\ell}_2| = \ell$.
The change in the position vector of the tip is given by $\Delta \vec{\ell} = \vec{\ell}_2 - \vec{\ell}_1$.
The magnitude of the change is $|\Delta \vec{\ell}| = \sqrt{|\vec{\ell}_2|^2 + |\vec{\ell}_1|^2 - 2|\vec{\ell}_2||\vec{\ell}_1| \cos \theta}$.
Substituting the magnitudes,we get $|\Delta \vec{\ell}| = \sqrt{\ell^2 + \ell^2 - 2\ell^2 \cos \theta}$.
$|\Delta \vec{\ell}| = \sqrt{2\ell^2(1 - \cos \theta)}$.
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get:
$|\Delta \vec{\ell}| = \sqrt{2\ell^2 \cdot 2 \sin^2 \frac{\theta}{2}} = \sqrt{4\ell^2 \sin^2 \frac{\theta}{2}}$.
$|\Delta \vec{\ell}| = 2\ell \sin \frac{\theta}{2}$.

(C) According to the triangle law of vector addition,$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.
Substituting this into the given expression: $|\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{AC}| = |\overrightarrow{AC} + \overrightarrow{AC}|$.
This simplifies to $|2\overrightarrow{AC}| = 2|\overrightarrow{AC}|$.
Since the length of each side of the equilateral triangle is $a$,$|\overrightarrow{AC}| = a$.
Therefore,$|2\overrightarrow{AC}| = 2a$.
Comparing this with the given expression $na$,we get $na = 2a$,which implies $n = 2$.

(B) Given,the two forces are $A = 3 \, N$ and $B = 4 \, N$ with an angle $\theta = 90^\circ$ between them.
Since the forces are perpendicular,the resultant force $R$ is given by:
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
$R = \sqrt{3^2 + 4^2 + 2(3)(4) \cos 90^\circ}$
Since $\cos 90^\circ = 0$,we have:
$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, N$
The direction of the resultant force with respect to the $4 \, N$ force is given by:
$\tan \alpha = \frac{3}{4} \implies \alpha = \tan^{-1}(0.75) \approx 37^\circ$.

(A) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by the formula $C = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
Given that $C = A + B$,we substitute this into the equation:
$A + B = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Squaring both sides:
$(A + B)^2 = A^2 + B^2 + 2AB \cos \theta$
$A^2 + B^2 + 2AB = A^2 + B^2 + 2AB \cos \theta$
Subtracting $A^2 + B^2$ from both sides:
$2AB = 2AB \cos \theta$
Assuming $A, B \neq 0$,we divide by $2AB$:
$1 = \cos \theta$
Therefore,$\theta = 0$.

(A) Step $1$: Applying the triangle law of vector addition.
In $\triangle OPR$,by the triangle law of vector addition:
$\vec{a} = \vec{c} + \overrightarrow{RP} \quad (1)$
In $\triangle ORQ$,by the triangle law of vector addition:
$\vec{b} = \overrightarrow{RQ} - \vec{c} \implies \vec{c} + \vec{b} = \overrightarrow{RQ} \implies \vec{b} = \overrightarrow{RQ} - \vec{c}$ (Wait,let's re-evaluate the geometry).
Correct approach: In $\triangle OPR$,$\vec{a} = \vec{c} + \overrightarrow{RP}$.
In $\triangle ORQ$,$\vec{c} = \vec{b} + \overrightarrow{RQ} \implies \overrightarrow{RQ} = \vec{c} - \vec{b}$.
Since $R$ is the midpoint of $PQ$,$\overrightarrow{PR} + \overrightarrow{RQ} = 0$,so $\overrightarrow{RP} = -\overrightarrow{RQ}$.
From $\triangle OPR$,$\overrightarrow{RP} = \vec{a} - \vec{c}$.
From $\triangle ORQ$,$\overrightarrow{RQ} = \vec{c} - \vec{b}$.
Since $\overrightarrow{RP} + \overrightarrow{RQ} = 0$,we have $(\vec{a} - \vec{c}) + (\vec{c} - \vec{b}) = 0$ is incorrect. Let's use position vectors.
Let origin be $O$. Then $\vec{a} = \vec{P}$,$\vec{b} = \vec{Q}$,$\vec{c} = \vec{R}$.
Since $R$ is the midpoint of $PQ$,$\vec{R} = \frac{\vec{P} + \vec{Q}}{2}$.
Therefore,$\vec{c} = \frac{\vec{a} + \vec{b}}{2}$,which implies $\vec{a} + \vec{b} = 2\vec{c}$.

(B) Let $|\vec{A}| = |\vec{B}| = A$.
Given that $|\vec{A} + \vec{B}| = A$.
The magnitude of the resultant vector is given by the formula: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Substituting the given values: $A = \sqrt{A^2 + A^2 + 2A^2 \cos \theta}$.
Squaring both sides: $A^2 = 2A^2 + 2A^2 \cos \theta$.
Dividing by $A^2$: $1 = 2 + 2 \cos \theta$.
$-1 = 2 \cos \theta$.
$\cos \theta = -1/2$.
Therefore,$\theta = 120^o$.

(B) The angle between vectors $\vec{A}$ and $\vec{B}$ is $\theta = 110^\circ - 20^\circ = 90^\circ$.
The magnitude of the resultant vector $\vec{R}$ is given by:
$R = \sqrt{A^2 + B^2 + 2AB \cos 90^\circ} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, m$.
Let $\alpha$ be the angle that the resultant vector $\vec{R}$ makes with vector $\vec{A}$.
Using the formula $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$:
$\tan \alpha = \frac{12 \sin 90^\circ}{5 + 12 \cos 90^\circ} = \frac{12 \times 1}{5 + 12 \times 0} = \frac{12}{5}$.
Thus,$\alpha = \tan^{-1}(12/5)$.
Since $\vec{A}$ makes an angle of $20^\circ$ with the $X$-axis,the angle of the resultant vector $\vec{R}$ with the $X$-axis is $\alpha + 20^\circ = \tan^{-1}(12/5) + 20^\circ$.

(A) The magnitude of the resultant $R$ of two vectors $A$ and $B$ acting at an angle $\theta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given $A = (x + y)$,$B = (x - y)$,and $R = \sqrt{x^2 + y^2}$.
Squaring both sides: $R^2 = A^2 + B^2 + 2AB \cos \theta$.
Substituting the values: $x^2 + y^2 = (x + y)^2 + (x - y)^2 + 2(x + y)(x - y) \cos \theta$.
Expanding the terms: $x^2 + y^2 = (x^2 + y^2 + 2xy) + (x^2 + y^2 - 2xy) + 2(x^2 - y^2) \cos \theta$.
Simplifying: $x^2 + y^2 = 2x^2 + 2y^2 + 2(x^2 - y^2) \cos \theta$.
Rearranging: $x^2 + y^2 - 2x^2 - 2y^2 = 2(x^2 - y^2) \cos \theta$.
$-(x^2 + y^2) = 2(x^2 - y^2) \cos \theta$.
Therefore,$\cos \theta = \frac{-(x^2 + y^2)}{2(x^2 - y^2)}$.
Thus,$\theta = \cos^{-1} \left( \frac{-(x^2 + y^2)}{2(x^2 - y^2)} \right)$.

(C) Let the origin be $(0, 0)$.
The first displacement vector $\vec{d_1}$ is $6 \, km$ at $45^\circ$ with the East (x-axis).
$\vec{d_1} = 6 \cos 45^\circ \hat{i} + 6 \sin 45^\circ \hat{j} = 6 \left( \frac{1}{\sqrt{2}} \right) \hat{i} + 6 \left( \frac{1}{\sqrt{2}} \right) \hat{j} = 3\sqrt{2} \hat{i} + 3\sqrt{2} \hat{j}$.
The second displacement vector $\vec{d_2}$ is $4 \, km$ at $135^\circ$ with the East.
$\vec{d_2} = 4 \cos 135^\circ \hat{i} + 4 \sin 135^\circ \hat{j} = 4 \left( -\frac{1}{\sqrt{2}} \right) \hat{i} + 4 \left( \frac{1}{\sqrt{2}} \right) \hat{j} = -2\sqrt{2} \hat{i} + 2\sqrt{2} \hat{j}$.
The resultant displacement vector $\vec{s} = \vec{d_1} + \vec{d_2} = (3\sqrt{2} - 2\sqrt{2}) \hat{i} + (3\sqrt{2} + 2\sqrt{2}) \hat{j} = \sqrt{2} \hat{i} + 5\sqrt{2} \hat{j}$.
The magnitude of the displacement is $|\vec{s}| = \sqrt{(\sqrt{2})^2 + (5\sqrt{2})^2} = \sqrt{2 + 50} = \sqrt{52} \, km$.

(D) Let the magnitudes of the vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = x$ and $|\overrightarrow{C}| = x\sqrt{2}$.
Given $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,which implies $\overrightarrow{A} + \overrightarrow{B} = -\overrightarrow{C}$.
Squaring both sides: $|\overrightarrow{A}|^2 + |\overrightarrow{B}|^2 + 2|\overrightarrow{A}||\overrightarrow{B}|\cos(\theta_{AB}) = |\overrightarrow{C}|^2$.
Substituting the values: $x^2 + x^2 + 2x^2\cos(\theta_{AB}) = (x\sqrt{2})^2 = 2x^2$.
$2x^2 + 2x^2\cos(\theta_{AB}) = 2x^2$,which gives $\cos(\theta_{AB}) = 0$,so $\theta_{AB} = 90^\circ$.
Since the vectors form a closed triangle,the exterior angles between the vectors are determined by the geometry of the triangle. The angles between the vectors are $90^\circ, 135^\circ, 135^\circ$.

(B) Given vectors are $\overrightarrow A = 4\hat i - 3\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$.
Sum of vectors: $\overrightarrow A + \overrightarrow B = (4+6)\hat i + (-3+8)\hat j = 10\hat i + 5\hat j$.
Magnitude: $|\overrightarrow A + \overrightarrow B| = \sqrt{(10)^2 + (5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}$.
Direction: $\tan \theta = \frac{y}{x} = \frac{5}{10} = \frac{1}{2}$.
Therefore,$\theta = \tan^{-1}(1/2)$.

(D) Let the initial velocity be $\vec{v}_1 = 20\hat{j} \, m/s$ (towards North).
After turning,the final velocity is $\vec{v}_2 = -20\hat{i} \, m/s$ (towards West).
The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.
$\Delta \vec{v} = -20\hat{i} - 20\hat{j} = -20(\hat{i} + \hat{j}) \, m/s$.
The magnitude of 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 $(S-W)$.

(B) Let the two unit vectors be $\vec{A}$ and $\vec{B}$,where $|\vec{A}| = 1$ and $|\vec{B}| = 1$.
Given that their sum is also a unit vector: $|\vec{A} + \vec{B}| = 1$.
Squaring both sides: $|\vec{A} + \vec{B}|^2 = 1^2$.
$|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta = 1$,where $\theta$ is the angle between them.
$1 + 1 + 2(1)(1) \cos \theta = 1 \implies 2 + 2 \cos \theta = 1 \implies 2 \cos \theta = -1 \implies \cos \theta = -1/2$.
Now,we need to find the magnitude of their difference: $|\vec{A} - \vec{B}|$.
$|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos \theta$.
Substituting the values: $|\vec{A} - \vec{B}|^2 = 1 + 1 - 2(1)(1)(-1/2)$.
$|\vec{A} - \vec{B}|^2 = 2 + 1 = 3$.
Therefore,$|\vec{A} - \vec{B}| = \sqrt{3}$.

(D) The line $x = 0$ corresponds to the $y$-axis. Therefore,the force ${F_1} = 1\,N$ acting along the $y$-axis is represented as ${\overrightarrow F _1} = 1\hat j$.
The line $y = 0$ corresponds to the $x$-axis. Therefore,the force ${F_2} = 2\,N$ acting along the $x$-axis is represented as ${\overrightarrow F _2} = 2\hat i$.
The resultant force $\overrightarrow F$ is the vector sum of the two forces:
$\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} = 2\hat i + 1\hat j$.

(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 the expression $\overrightarrow A - 2\overrightarrow B + 3\overrightarrow C$.
Substituting the given vectors into the expression:
$= (2\hat i + \hat j) - 2(3\hat j - \hat k) + 3(6\hat i - 2\hat k)$
$= 2\hat i + \hat j - 6\hat j + 2\hat k + 18\hat i - 6\hat k$
Grouping the components of $\hat i$,$\hat j$,and $\hat k$:
$= (2 + 18)\hat i + (1 - 6)\hat j + (2 - 6)\hat k$
$= 20\hat i - 5\hat j - 4\hat k$.

(D) The magnitude of the difference of two vectors $A$ and $B$ is given by the formula:
$|A-B| = \sqrt{|A|^2 + |B|^2 - 2|A||B|\cos\theta}$
Given $|A| = 2$,$|B| = 4$,and $\theta = 60^{\circ}$.
Substituting the values into the formula:
$|A-B| = \sqrt{2^2 + 4^2 - 2(2)(4)\cos(60^{\circ})}$
$|A-B| = \sqrt{4 + 16 - 16 \times \frac{1}{2}}$
$|A-B| = \sqrt{20 - 8}$
$|A-B| = \sqrt{12}$
$|A-B| = 2\sqrt{3}$

(D) Let the magnitude of vectors $A$ and $B$ be $|A| = |B| = a$. Let the angle between $A$ and $B$ be $\theta$.
Consider the vector $R = A - B$. This can be written as $R = A + (-B)$.
The magnitude of vector $-B$ is equal to the magnitude of vector $B$,which is $a$.
The angle between $A$ and $-B$ is $(180^{\circ} - \theta)$.
Using the law of vector addition,the resultant vector $R$ forms a triangle with $A$ and $-B$.
Since $|A| = |-B| = a$,the triangle formed by $A$,$-B$,and $R$ is an isosceles triangle.
The angle $\alpha$ between $A$ and $R$ (which is $A - B$) is given by the base angle of this isosceles triangle.
The sum of angles in a triangle is $180^{\circ}$. The angle between $A$ and $-B$ is $(180^{\circ} - \theta)$.
Thus,$2\alpha + (180^{\circ} - \theta) = 180^{\circ}$.
Solving for $\alpha$,we get $2\alpha = \theta$,or $\alpha = \frac{\theta}{2}$.
Since the angle $\alpha$ depends on $\theta$ (the angle between $A$ and $B$),the correct option is $(d)$.

(B) The resultant force $R$ of two forces $F_1$ and $F_2$ acting at an angle $\theta$ is given by $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$.
For two forces of $6\,N$ and $8\,N$,the maximum resultant is $F_1 + F_2 = 6 + 8 = 14\,N$ (when $\theta = 0^\circ$).
The minimum resultant is $|F_1 - F_2| = |6 - 8| = 2\,N$ (when $\theta = 180^\circ$).
Therefore,the resultant force must lie in the range $[2\,N, 14\,N]$.
Among the given options,only $11\,N$ lies within this range.
Thus,the correct option is $B$.

(A) Let the two vectors be $\vec{P}$ and $\vec{Q}$ such that $|\vec{P}| = |\vec{Q}| = A$.
The magnitude of the resultant vector $\vec{R} = \vec{P} + \vec{Q}$ is given by $R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$.
Substituting the values,we get $R = \sqrt{A^2 + A^2 + 2A^2 \cos \theta} = \sqrt{2A^2(1 + \cos \theta)}$.
Using the trigonometric identity $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$,we get $R = \sqrt{2A^2 \cdot 2 \cos^2 \frac{\theta}{2}} = \sqrt{4A^2 \cos^2 \frac{\theta}{2}} = 2A \cos \frac{\theta}{2}$.
Since the magnitudes of the two vectors are equal,the resultant vector bisects the angle between them. Thus,the direction is along the angle bisector.

(B) Given the vector equation $P + Q + R = 0$.
We can rewrite this as $P + Q = -R$.
Taking the magnitude on both sides,we get $|P + Q| = |-R|$.
Since the magnitude of a vector is always non-negative,$|-R| = |R|$.
Therefore,$|P + Q| = |R|$.
Thus,the correct statement is $|P + Q| = |R|$.

(C) The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by the formula:
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Given: $A = 3$,$B = 4$,and $\theta = 60^{\circ}$.
Substituting the values:
$R = \sqrt{3^2 + 4^2 + 2(3)(4) \cos 60^{\circ}}$
Since $\cos 60^{\circ} = 0.5$:
$R = \sqrt{9 + 16 + 24(0.5)}$
$R = \sqrt{25 + 12}$
$R = \sqrt{37} \text{ units}$.

(C) According to the triangle law of vector addition,when 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.
From the provided figure,if we place the tail of vector $\overrightarrow e$ at the head of vector $\overrightarrow d$,the resultant vector $\overrightarrow f$ connects the tail of $\overrightarrow d$ to the head of $\overrightarrow e$.
Therefore,$\overrightarrow d + \overrightarrow e = \overrightarrow f$.

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$.
The magnitude of the sum of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
The magnitude of the difference of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta}$.
Given that $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$,we square both sides:
$A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta$.
Subtracting $A^2 + B^2$ from both sides,we get:
$2AB \cos \theta = -2AB \cos \theta$.
$4AB \cos \theta = 0$.
Since $A$ and $B$ are non-zero vectors,$4AB \neq 0$,therefore $\cos \theta = 0$.
This implies $\theta = 90^o$.

(C) Let the starting point be the origin $(0,0)$.
First displacement vector $\vec{d_1} = 6 \cos(45^o) \hat{i} + 6 \sin(45^o) \hat{j} = 6(\frac{1}{\sqrt{2}}) \hat{i} + 6(\frac{1}{\sqrt{2}}) \hat{j} = 3\sqrt{2} \hat{i} + 3\sqrt{2} \hat{j}$.
Second displacement vector $\vec{d_2} = 4 \cos(135^o) \hat{i} + 4 \sin(135^o) \hat{j} = 4(-\frac{1}{\sqrt{2}}) \hat{i} + 4(\frac{1}{\sqrt{2}}) \hat{j} = -2\sqrt{2} \hat{i} + 2\sqrt{2} \hat{j}$.
Resultant displacement $\vec{R} = \vec{d_1} + \vec{d_2} = (3\sqrt{2} - 2\sqrt{2}) \hat{i} + (3\sqrt{2} + 2\sqrt{2}) \hat{j} = \sqrt{2} \hat{i} + 5\sqrt{2} \hat{j}$.
Magnitude $|\vec{R}| = \sqrt{(\sqrt{2})^2 + (5\sqrt{2})^2} = \sqrt{2 + 50} = \sqrt{52} \, km$.
Angle with east direction $\theta = \tan^{-1}(\frac{R_y}{R_x}) = \tan^{-1}(\frac{5\sqrt{2}}{\sqrt{2}}) = \tan^{-1}(5)$.

(A) The magnitude of the resultant $R$ of two vectors $A$ and $B$ acting at an angle $\theta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given $A = (x + y)$,$B = (x - y)$,and $R = \sqrt{x^2 + y^2}$.
Squaring both sides,we get $R^2 = A^2 + B^2 + 2AB \cos \theta$.
Substituting the values: $x^2 + y^2 = (x + y)^2 + (x - y)^2 + 2(x + y)(x - y) \cos \theta$.
Expanding the terms: $x^2 + y^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) + 2(x^2 - y^2) \cos \theta$.
Simplifying: $x^2 + y^2 = 2x^2 + 2y^2 + 2(x^2 - y^2) \cos \theta$.
Rearranging: $x^2 + y^2 - 2x^2 - 2y^2 = 2(x^2 - y^2) \cos \theta$.
$- (x^2 + y^2) = 2(x^2 - y^2) \cos \theta$.
Therefore,$\cos \theta = -\frac{x^2 + y^2}{2(x^2 - y^2)}$.
Thus,$\theta = \cos^{-1}\left(-\frac{x^2 + y^2}{2(x^2 - y^2)}\right)$.

(C) Given,$\vec{R}_1 = \vec{A} + \vec{B}$ and $\vec{R}_2 = \vec{A} - \vec{B}$.
Using the parallelogram law of vector addition,the magnitude of the resultant is given by $R^2 = A^2 + B^2 + 2AB \cos \theta$.
For $\vec{R}_1$,the magnitude squared is $R_1^2 = A^2 + B^2 + 2AB \cos \theta$.
For $\vec{R}_2$,the angle between $\vec{A}$ and $-\vec{B}$ is $(180^\circ - \theta)$,so $R_2^2 = A^2 + B^2 + 2AB \cos(180^\circ - \theta) = A^2 + B^2 - 2AB \cos \theta$.
Adding these two equations:
$R_1^2 + R_2^2 = (A^2 + B^2 + 2AB \cos \theta) + (A^2 + B^2 - 2AB \cos \theta)$
$R_1^2 + R_2^2 = 2(A^2 + B^2)$.

(B) Given vectors are $A = 2 \hat{i} + \hat{j}$,$B = 3 \hat{j} - \hat{k}$,and $C = 6 \hat{i} - 2 \hat{k}$.
We need to calculate $A - 2B + 3C$.
Substitute the given vectors into the expression:
$A - 2B + 3C = (2 \hat{i} + \hat{j}) - 2(3 \hat{j} - \hat{k}) + 3(6 \hat{i} - 2 \hat{k})$
Expand the terms:
$= 2 \hat{i} + \hat{j} - 6 \hat{j} + 2 \hat{k} + 18 \hat{i} - 6 \hat{k}$
Group the components of $\hat{i}$,$\hat{j}$,and $\hat{k}$:
$= (2 + 18) \hat{i} + (1 - 6) \hat{j} + (2 - 6) \hat{k}$
$= 20 \hat{i} - 5 \hat{j} - 4 \hat{k}$

(C) The resultant force $R$ of two vectors $F_1$ 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 $F_1 = 3 \ N$,$F_2 = 5 \ N$,and $\theta = 60^o$.
Substituting the values: $R = \sqrt{3^2 + 5^2 + 2(3)(5) \cos 60^o}$.
Since $\cos 60^o = 0.5$,we have $R = \sqrt{9 + 25 + 30(0.5)} = \sqrt{34 + 15} = \sqrt{49} = 7 \ N$.
$A$ force that balances these two forces must be equal in magnitude and opposite in direction to the resultant force.
Therefore,the magnitude of the balancing force is $7 \ N$.

(A) Let the two forces be $F_1 = P + Q$ and $F_2 = P - Q$. The angle between them is $2 \alpha$. The bisector divides this into two angles of $\alpha$ each.
If the resultant makes an angle $\theta$ with the bisector,then the angle between the resultant and $F_1$ is $(\alpha - \theta)$ and the angle between the resultant and $F_2$ is $(\alpha + \theta)$.
Resolving the forces perpendicular to the resultant,the components must cancel out:
$(P + Q) \sin (\alpha - \theta) = (P - Q) \sin (\alpha + \theta)$
Expanding the terms:
$P \sin (\alpha - \theta) + Q \sin (\alpha - \theta) = P \sin (\alpha + \theta) - Q \sin (\alpha + \theta)$
Rearranging terms to group $P$ and $Q$:
$Q [\sin (\alpha + \theta) + \sin (\alpha - \theta)] = P [\sin (\alpha + \theta) - \sin (\alpha - \theta)]$
Using trigonometric identities $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$ and $\sin(A+B) - \sin(A-B) = 2 \cos A \sin B$:
$Q [2 \sin \alpha \cos \theta] = P [2 \cos \alpha \sin \theta]$
Dividing both sides by $2 \cos \alpha \cos \theta$:
$Q \tan \alpha = P \tan \theta$
Thus,$P \tan \theta = Q \tan \alpha$.

(C) Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A = 3$ and $B = 4$. Let $\vec{R}$ be their resultant.
If we resolve both vectors along the direction perpendicular to the resultant $\vec{R}$,the sum of the components must be zero because the resultant has no component perpendicular to itself.
Thus,the component of vector $3$ perpendicular to $\vec{R}$ must be equal in magnitude to the component of vector $4$ perpendicular to $\vec{R}$ in the opposite direction.
Therefore,$3 \sin \alpha = 4 \sin \beta$.
Rearranging the terms,we get $\frac{\sin \alpha}{\sin \beta} = \frac{4}{3}$.

(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
The angle between them is $\theta = 60^{\circ}$.
For the sum of vectors,the magnitude is given by $|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}| |\vec{b}| \cos \theta}$.
Substituting the values: $|\vec{a} + \vec{b}| = \sqrt{1^2 + 1^2 + 2(1)(1) \cos 60^{\circ}} = \sqrt{1 + 1 + 2(0.5)} = \sqrt{3} \approx 1.732$.
Since $1.732 > 1$,it follows that $|\vec{a} + \vec{b}| > 1$.
For the difference of vectors,the magnitude is given by $|\vec{a} - \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}| |\vec{b}| \cos \theta}$.
Substituting the values: $|\vec{a} - \vec{b}| = \sqrt{1^2 + 1^2 - 2(1)(1) \cos 60^{\circ}} = \sqrt{1 + 1 - 2(0.5)} = \sqrt{1} = 1$.
Thus,the correct statement is $|\vec{a} + \vec{b}| > 1$.