Fundamentals of Vectors Questions in English

(B) The position vector $\vec{r}$ of a point $P(x, y, z)$ in a Cartesian coordinate system is given by the formula $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
Given the coordinates of the particle are $(3, 2, 5)$,where $x = 3$,$y = 2$,and $z = 5$.
Substituting these values into the formula,we get $\vec{r} = 3\hat{i} + 2\hat{j} + 5\hat{k}$.

(C) The displacement vector $\vec{d}$ is given by the difference between the position vector of the final point $Q$ and the initial point $P$.
Position vector of $P$ is $\vec{r}_P = 2\hat{i} + 3\hat{j} + 5\hat{k}$.
Position vector of $Q$ is $\vec{r}_Q = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
Displacement vector $\vec{d} = \vec{r}_Q - \vec{r}_P$.
$\vec{d} = (3 - 2)\hat{i} + (4 - 3)\hat{j} + (5 - 5)\hat{k}$.
$\vec{d} = 1\hat{i} + 1\hat{j} + 0\hat{k} = \hat{i} + \hat{j}$.

(D) First,calculate the magnitude of vector $B$:
$|B| = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.
Next,find the unit vector of $A$ to determine its direction:
$|A| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
$\hat{A} = \frac{A}{|A|} = \frac{3\hat{i} + 4\hat{j}}{5}$.
The required vector $V$ has magnitude $|B|$ and direction $\hat{A}$:
$V = |B| \cdot \hat{A} = 25 \cdot \left( \frac{3\hat{i} + 4\hat{j}}{5} \right)$.
$V = 5 \cdot (3\hat{i} + 4\hat{j}) = 15\hat{i} + 20\hat{j}$.

(A) The magnitude of the vector $\overrightarrow{A} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ is given by:
$|\overrightarrow{A}| = \sqrt{(2)^2 + (4)^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45}$.
The direction cosines $(l, m, n)$ are defined as the ratios of the components of the vector to its magnitude:
$l = \cos \alpha = \frac{A_x}{|\overrightarrow{A}|} = \frac{2}{\sqrt{45}}$
$m = \cos \beta = \frac{A_y}{|\overrightarrow{A}|} = \frac{4}{\sqrt{45}}$
$n = \cos \gamma = \frac{A_z}{|\overrightarrow{A}|} = \frac{-5}{\sqrt{45}}$
Thus,the direction cosines are $\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}, \text{and } \frac{-5}{\sqrt{45}}$.

(B) The resultant vector $R = A + B$ is always closer to the vector with the larger magnitude.
If $A > B$,the resultant is closer to $A$,meaning the angle $\alpha$ with $A$ is smaller than the angle $\beta$ with $B$ (i.e.,$\alpha < \beta$).
If $A < B$,the resultant is closer to $B$,meaning the angle $\beta$ with $B$ is smaller than the angle $\alpha$ with $A$ (i.e.,$\beta < \alpha$,or $\alpha > \beta$).
If $A = B$,the resultant bisects the angle between them,so $\alpha = \beta$.
Therefore,the statement $\alpha > \beta$ if $A < B$ is correct.

(A) The resultant force $R$ of two vectors $A$ and $B$ acting at an angle $\theta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Given $A = 2P$,$B = \sqrt{2}P$,and $R = P\sqrt{10}$.
Substituting these values into the formula:
$P\sqrt{10} = \sqrt{(2P)^2 + (\sqrt{2}P)^2 + 2(2P)(\sqrt{2}P) \cos \theta}$.
Squaring both sides:
$10P^2 = 4P^2 + 2P^2 + 4\sqrt{2}P^2 \cos \theta$.
$10P^2 = 6P^2 + 4\sqrt{2}P^2 \cos \theta$.
$4P^2 = 4\sqrt{2}P^2 \cos \theta$.
$1 = \sqrt{2} \cos \theta$.
$\cos \theta = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = 45^{\circ}$.

(B) vector $\vec{A} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ is a unit vector if its magnitude is $1$,i.e.,$|\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = 1$.
Given vector is $\vec{A} = 0.4\hat{i} + 0.8\hat{j} + c\hat{k}$.
Calculating the magnitude: $\sqrt{(0.4)^2 + (0.8)^2 + c^2} = 1$.
Squaring both sides: $(0.4)^2 + (0.8)^2 + c^2 = 1^2$.
$0.16 + 0.64 + c^2 = 1$.
$0.80 + c^2 = 1$.
$c^2 = 1 - 0.80 = 0.20$.
Therefore,$c = \sqrt{0.2}$.

(C) Let the vector be $\vec{R} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.
Comparing this with $\vec{R} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k}$,we get $R_x = 1, R_y = 1, R_z = \sqrt{2}$.
The magnitude of the vector is $|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2} = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.
The direction cosines are given by $\cos \alpha = \frac{R_x}{|\vec{R}|}, \cos \beta = \frac{R_y}{|\vec{R}|}, \cos \gamma = \frac{R_z}{|\vec{R}|}$.
For the $X$-axis: $\cos \alpha = \frac{1}{2} \Rightarrow \alpha = 60^\circ$.
For the $Y$-axis: $\cos \beta = \frac{1}{2} \Rightarrow \beta = 60^\circ$.
For the $Z$-axis: $\cos \gamma = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow \gamma = 45^\circ$.
Thus,the angles are $60^\circ, 60^\circ, 45^\circ$.

(D) The given vector is $\vec A = 5\hat i - 12\hat j$.
The magnitude of the vector $\vec A$ is calculated as $|\vec A| = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
$A$ unit vector $\hat A$ in the direction of $\vec A$ is defined as $\hat A = \frac{\vec A}{|\vec A|}$.
Substituting the values,we get $\hat A = \frac{5\hat i - 12\hat j}{13}$.

(A) vector is a physical quantity that has both magnitude and direction.
When we represent a vector as $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$,the components $(A_x, A_y, A_z)$ depend on the orientation of the coordinate axes.
However,the vector sum $\vec{P} + \vec{Q} + \vec{R}$ represents a resultant vector,which is a physical entity independent of the coordinate system used to describe it.
Options $B$ and $C$ represent components or partial sums that change if the axes are rotated.
Therefore,the vector sum $\vec{P} + \vec{Q} + \vec{R}$ is invariant under coordinate transformation.

(D) Let the magnitudes of the vectors be $|A| = x$,$|B| = x$,and $|C| = \sqrt{2}x$.
Since $\overrightarrow A + \overrightarrow B + \overrightarrow C = 0$,these three vectors form a closed triangle.
According to the law of cosines for a triangle with sides $x, x, \sqrt{2}x$,we observe that $x^2 + x^2 = (\sqrt{2}x)^2$,which satisfies the Pythagorean theorem.
Thus,the vectors form a right-angled isosceles triangle.
To find the angles between the vectors,we place them head-to-tail.
The angle between $\overrightarrow A$ and $\overrightarrow B$ is $90^{\circ}$ (as they are the legs of the right triangle).
The angle between $\overrightarrow B$ and $\overrightarrow C$ is $180^{\circ} - 45^{\circ} = 135^{\circ}$.
The angle between $\overrightarrow A$ and $\overrightarrow C$ is $180^{\circ} - 45^{\circ} = 135^{\circ}$.
Therefore,the angles are $90^{\circ}, 135^{\circ}, 135^{\circ}$.

(D) The line $x = 0$ represents the $y$-axis. Since the force $F_1 = 1\,N$ acts along this line,it can be represented as $\vec{F}_1 = 1\hat{j} = \hat{j}$.
The line $y = 0$ represents the $x$-axis. Since the force $F_2 = 2\,N$ acts along this line,it can be represented as $\vec{F}_2 = 2\hat{i}$.
The resultant force $\vec{F}$ is the vector sum of $\vec{F}_1$ and $\vec{F}_2$.
Therefore,$\vec{F} = \vec{F}_1 + \vec{F}_2 = 2\hat{i} + \hat{j}$.

(C) Two vectors $P = A_1\hat{i} + B_1\hat{j} + C_1\hat{k}$ and $Q = A_2\hat{i} + B_2\hat{j} + C_2\hat{k}$ are parallel if their components are proportional,i.e.,$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.
Given $P = 2\hat{i} + b\hat{j} + 2\hat{k}$ and $Q = 1\hat{i} + 1\hat{j} + 1\hat{k}$.
Comparing the components,we get $\frac{2}{1} = \frac{b}{1} = \frac{2}{1}$.
From the middle term,we find $b = 2$.

(A) The magnitude of the resultant $R$ of two vectors $A$ and $B$ inclined at an angle $\theta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
Since the magnitudes of $A$ and $B$ and the angle $\theta$ between them remain unchanged when their directions are interchanged,the magnitude of the resultant $R$ remains the same.
However,the direction of the resultant vector is given by $\tan \phi = \frac{B \sin \theta}{A + B \cos \theta}$.
When $A$ and $B$ are interchanged,the new angle $\phi'$ becomes $\tan \phi' = \frac{A \sin \theta}{B + A \cos \theta}$.
Since $\phi \neq \phi'$,the direction of the resultant changes.
Therefore,only the magnitude remains the same.

(C) The correct option is $C$.
According to the triangle inequality for vectors,for any two vectors $a$ and $b$,the magnitude of their difference satisfies the inequality $| a - b | \geq | | a | - | b | |$.
Since $| | a | - | b | | \geq | a | - | b |$ is always true for any real numbers $| a |$ and $| b |$,it follows that $| a - b | \geq | a | - | b |$.
This represents the lower bound of the magnitude of the resultant vector when subtracting two vectors.

(B) According to the triangle inequality for vectors,the magnitude of the sum of two vectors is always less than or equal to the sum of their individual magnitudes.
Mathematically,this is expressed as $| \vec{a} + \vec{b} | \leq | \vec{a} | + | \vec{b} |$.
This equality holds when the two vectors are in the same direction (parallel).
Therefore,option $B$ is the correct statement.

(C) Let the resultant force be $\vec{R} = \overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF}$.
In a regular hexagon $ABCDEF$,the center $O$ is the midpoint of the diagonals $AD$,$BE$,and $CF$. Thus,$\overrightarrow{AB} + \overrightarrow{AE} = 2\overrightarrow{AO} = \overrightarrow{AD}$ (since $\overrightarrow{AD} = 2\overrightarrow{AO}$).
Similarly,$\overrightarrow{AC} + \overrightarrow{AF} = 2\overrightarrow{AO} = \overrightarrow{AD}$.
Substituting these into the expression for $\vec{R}$:
$\vec{R} = (\overrightarrow{AB} + \overrightarrow{AE}) + (\overrightarrow{AC} + \overrightarrow{AF}) + \overrightarrow{AD}$
$\vec{R} = \overrightarrow{AD} + \overrightarrow{AD} + \overrightarrow{AD} = 3\overrightarrow{AD}$.

(D) Given the condition: $|A + B| = 2|A - B|$.
Squaring both sides,we get: $|A + B|^2 = 4|A - B|^2$.
Using the vector identity $|u \pm v|^2 = |u|^2 + |v|^2 \pm 2|u||v| \cos \theta$,we have:
$A^2 + B^2 + 2AB \cos \theta = 4(A^2 + B^2 - 2AB \cos \theta)$.
$A^2 + B^2 + 2AB \cos \theta = 4A^2 + 4B^2 - 8AB \cos \theta$.
$10AB \cos \theta = 3A^2 + 3B^2$.
$\cos \theta = \frac{3(A^2 + B^2)}{10AB}$.
Since the value of $\cos \theta$ depends on the magnitudes of $A$ and $B$,and they are not provided,the angle $\theta$ cannot be uniquely determined. Thus,the data is insufficient.

(B) Let the plane containing $\overrightarrow{A}$ and $\overrightarrow{B}$ be the $xy$-plane. Thus,$\overrightarrow{A} = A_x \hat{i} + A_y \hat{j}$ and $\overrightarrow{B} = B_x \hat{i} + B_y \hat{j}$.
Since $\overrightarrow{C}$ lies outside this plane,it must have a non-zero component in the $z$-direction: $\overrightarrow{C} = C_x \hat{i} + C_y \hat{j} + C_z \hat{k}$,where $C_z \neq 0$.
The resultant vector is $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = (A_x + B_x + C_x) \hat{i} + (A_y + B_y + C_y) \hat{j} + C_z \hat{k}$.
For the resultant to be zero,all components must be zero. However,the $z$-component is $C_z$,which is given as non-zero.
Therefore,the resultant $\overrightarrow{R}$ cannot be zero.

(C) Given that $\hat{a}$ and $\hat{b}$ are unit vectors,so $|\hat{a}| = 1$ and $|\hat{b}| = 1$.
For option $(A)$: $|\frac{\hat{a} + \hat{b}}{\sqrt{3}}| = \frac{1}{\sqrt{3}} \sqrt{|\hat{a}|^2 + |\hat{b}|^2 + 2|\hat{a}||\hat{b}| \cos 60^{\circ}} = \frac{1}{\sqrt{3}} \sqrt{1 + 1 + 2(1)(1)(0.5)} = \frac{1}{\sqrt{3}} \sqrt{3} = 1$.
For option $(B)$: $|\hat{a} + \hat{b}| = \sqrt{1 + 1 + 2(1)(1)(0.5)} = \sqrt{3} \neq 1$.
For option $(C)$: $|\hat{a}| = 1$.
For option $(D)$: $|\hat{b}| = 1$.
Thus,vectors in $(A), (C),$ and $(D)$ have magnitude one.

(A) The dot product of two vectors $a$ and $b$ is defined as $a \cdot b = |a| |b| \cos \theta$,where $\theta$ is the angle between the vectors.
Squaring both sides,we get $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 \theta$.
Given the condition $(a \cdot b)^2 = a^2 b^2$,we substitute the expression:
$|a|^2 |b|^2 \cos^2 \theta = |a|^2 |b|^2$.
Assuming $a$ and $b$ are non-zero vectors,we divide by $|a|^2 |b|^2$ to get $\cos^2 \theta = 1$.
This implies $\cos \theta = \pm 1$,which means $\theta = 0^{\circ}$ or $\theta = 180^{\circ}$.
In both cases,the vectors $a$ and $b$ are parallel (or anti-parallel). Thus,the condition is satisfied when $a$ is parallel to $b$.

(A) When a vector $\vec{A}$ is rotated by a small angle $\Delta \theta$,the magnitude of the vector remains unchanged,so $|\vec{A}| = |\vec{B}| = A$.
The difference vector $\vec{B} - \vec{A}$ represents the chord length between the tips of the two vectors.
For a very small angle $\Delta \theta$,the chord length is approximately equal to the arc length of the circle traced by the tip of the vector.
Using the relation $\text{Arc length} = \text{radius} \times \text{angle}$,we have:
$|\vec{B} - \vec{A}| \approx |\vec{A}| \Delta \theta$.

(B) unit vector is defined as a vector whose magnitude is equal to $1$.
Let us check the magnitude of the given options:
For option $A$: $|\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \neq 1$.
For option $B$: $|\cos \theta \hat{i} - \sin \theta \hat{j}| = \sqrt{\cos^2 \theta + (-\sin \theta)^2} = \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt{1} = 1$.
For option $C$: $|\sin \theta \hat{i} + 2 \cos \theta \hat{j}| = \sqrt{\sin^2 \theta + 4 \cos^2 \theta} \neq 1$.
For option $D$: $|\frac{1}{\sqrt{3}}(\hat{i} + \hat{j})| = \frac{1}{\sqrt{3}} \sqrt{1^2 + 1^2} = \frac{\sqrt{2}}{\sqrt{3}} \neq 1$.
Therefore,the vector in option $B$ is a unit vector.

(A) The component of a vector $\vec{A}$ along a direction making an angle $\theta$ with it is $A_\theta = A \cos \theta$.
Since the range of the cosine function is $-1 \le \cos \theta \le 1$,the magnitude of the component is $|A \cos \theta| = |A| |\cos \theta|$.
Because $|\cos \theta| \le 1$,it follows that $|A \cos \theta| \le |A|$.
Therefore,the component of a vector along any direction is always less than or equal to its magnitude.

(D) The three vectors along the coordinate axes representing the adjacent sides of a cube of length $b$ are $b\hat{i}$,$b\hat{j}$,and $b\hat{k}$.
The diagonal vector $\vec{A}$ passing through the origin is the resultant of these three vectors: $\vec{A} = b\hat{i} + b\hat{j} + b\hat{k}$.
The magnitude of this diagonal vector is $|\vec{A}| = \sqrt{b^2 + b^2 + b^2} = \sqrt{3b^2} = b\sqrt{3}$.
The unit vector $\hat{A}$ along the diagonal is given by $\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{b\hat{i} + b\hat{j} + b\hat{k}}{b\sqrt{3}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

(B) Two vectors $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$ are parallel if their components are proportional,i.e.,$\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z}$.
Given $\vec{A} = 6\hat{i} + x\hat{j} - 2\hat{k}$ and $\vec{B} = 5\hat{i} - 6\hat{j} - y\hat{k}$.
Equating the ratios: $\frac{6}{5} = \frac{x}{-6} = \frac{-2}{-y}$.
First,solve for $x$: $\frac{6}{5} = \frac{x}{-6} \implies x = \frac{6 \times (-6)}{5} = -\frac{36}{5}$.
Next,solve for $y$: $\frac{6}{5} = \frac{-2}{-y} \implies \frac{6}{5} = \frac{2}{y} \implies 6y = 10 \implies y = \frac{10}{6} = \frac{5}{3}$.
Thus,$x = -\frac{36}{5}$ and $y = \frac{5}{3}$.

(A) Two vectors $\vec{A}$ and $\vec{B}$ are parallel if their components are proportional,meaning $\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z} = k$.
For option $A$: $\frac{3}{1} = 3$,$\frac{6}{2} = 3$,$\frac{9}{3} = 3$. Since all ratios are equal to $3$,$\vec{A} = 3\vec{B}$,so they are parallel.
For option $B$: $\frac{3}{1} = 3$,$\frac{-6}{2} = -3$. Ratios are not equal.
For option $C$: $\frac{2}{1} = 2$,$\frac{6}{2} = 3$. Ratios are not equal.
For option $D$: $\frac{2}{1} = 2$,$\frac{3}{-2} = -1.5$. Ratios are not equal.
Thus,the correct option is $A$.

(D) Given the equation $|\vec{v}_1 + \vec{v}_2| = |\vec{v}_1 - \vec{v}_2|$.
Squaring both sides,we get $|\vec{v}_1 + \vec{v}_2|^2 = |\vec{v}_1 - \vec{v}_2|^2$.
Using the property $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$,we expand the terms:
$(\vec{v}_1 + \vec{v}_2) \cdot (\vec{v}_1 + \vec{v}_2) = (\vec{v}_1 - \vec{v}_2) \cdot (\vec{v}_1 - \vec{v}_2)$.
$|\vec{v}_1|^2 + |\vec{v}_2|^2 + 2(\vec{v}_1 \cdot \vec{v}_2) = |\vec{v}_1|^2 + |\vec{v}_2|^2 - 2(\vec{v}_1 \cdot \vec{v}_2)$.
Subtracting $|\vec{v}_1|^2 + |\vec{v}_2|^2$ from both sides,we get $2(\vec{v}_1 \cdot \vec{v}_2) = -2(\vec{v}_1 \cdot \vec{v}_2)$.
This implies $4(\vec{v}_1 \cdot \vec{v}_2) = 0$,which means $\vec{v}_1 \cdot \vec{v}_2 = 0$.
Since the dot product of two non-zero vectors is zero,they must be mutually perpendicular.

(B) physical quantity is defined as a vector only if it possesses both magnitude and direction $AND$ follows the laws of vector addition (e.g.,the parallelogram law).
Physical quantities like time,pressure,surface tension,and electric current possess a specific direction but do not follow the laws of vector addition.
Therefore,a physical quantity having a direction does not necessarily have to be a vector; it may or may not be a vector.

(A) Two vectors $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$ are parallel if their components are proportional,i.e.,$\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z}$.
Given $\vec{A} = 2\hat{i} + p\hat{j} + q\hat{k}$ and $\vec{B} = 5\hat{i} + 7\hat{j} + 3\hat{k}$.
Equating the ratios of the components: $\frac{2}{5} = \frac{p}{7} = \frac{q}{3}$.
For $p$: $\frac{2}{5} = \frac{p}{7} \Rightarrow p = \frac{2 \times 7}{5} = \frac{14}{5}$.
For $q$: $\frac{2}{5} = \frac{q}{3} \Rightarrow q = \frac{2 \times 3}{5} = \frac{6}{5}$.
Thus,the values are $p = \frac{14}{5}$ and $q = \frac{6}{5}$.

(B) Two vectors $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$ are parallel if their components are proportional,i.e.,$\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z}$.
Given $\vec{A} = 6 \hat{i} + x \hat{j} - 2 \hat{k}$ and $\vec{B} = 5 \hat{i} + 6 \hat{j} - y \hat{k}$.
Equating the ratios: $\frac{6}{5} = \frac{x}{6} = \frac{-2}{-y}$.
First,solve for $x$: $\frac{6}{5} = \frac{x}{6} \implies x = \frac{6 \times 6}{5} = \frac{36}{5}$.
Next,solve for $y$: $\frac{6}{5} = \frac{-2}{-y} \implies \frac{6}{5} = \frac{2}{y} \implies 6y = 10 \implies y = \frac{10}{6} = \frac{5}{3}$.
Thus,the values are $x = \frac{36}{5}$ and $y = \frac{5}{3}$.

(D) The angle between two vectors $\vec A$ and $\vec B$ is defined as the angle between their directions.
Multiplying a vector by a positive scalar does not change its direction.
Multiplying a vector by a negative scalar reverses its direction (i.e.,rotates it by $\pi$ radians or $180^\circ$).
Given the angle between $\vec a$ and $\vec b$ is $\theta = \frac{\pi }{6}$.
The vector $-3\vec a$ is in the opposite direction to $\vec a$.
The vector $2\vec b$ is in the same direction as $\vec b$.
Therefore,the angle between $-3\vec a$ and $2\vec b$ is $\pi - \theta = \pi - \frac{\pi }{6} = \frac{5\pi }{6}$.

(C) For three vectors to have a resultant of zero,they must form a closed triangle. $A$ triangle is a planar figure,so the vectors must be coplanar. This makes statement $(C)$ correct.
Statement $(A)$ is incorrect because three vectors of different magnitudes can also sum to zero (e.g.,sides of a $3-4-5$ triangle).
Statement $(B)$ is incorrect because four non-zero forces can sum to zero in three dimensions (e.g.,the four vectors pointing from the center of a regular tetrahedron to its vertices).
Therefore,only statement $(C)$ is correct.

(B) The unit vector along the $Z$-axis is $\hat{k} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.
Let $\theta$ be the angle between $\vec{A}$ and the $Z$-axis.
The dot product formula is $\vec{A} \cdot \hat{k} = |\vec{A}| |\hat{k}| \cos \theta$.
Calculating the dot product: $\vec{A} \cdot \hat{k} = (3)(0) + (-1)(0) + (4)(1) = 4$.
The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{3^2 + (-1)^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$.
The magnitude of $\hat{k}$ is $1$.
Thus,$\cos \theta = \frac{4}{\sqrt{26} \times 1} = \frac{4}{\sqrt{26}}$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we find $\sin \theta = \sqrt{1 - \left(\frac{4}{\sqrt{26}}\right)^2} = \sqrt{1 - \frac{16}{26}} = \sqrt{\frac{10}{26}} = \frac{\sqrt{10}}{\sqrt{26}}$.
Therefore,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{10}/\sqrt{26}}{4/\sqrt{26}} = \frac{\sqrt{10}}{4}$.
Hence,$\theta = \tan^{-1}\left(\frac{\sqrt{10}}{4}\right)$.

(B) The velocity vector $\vec{v}$ is given by the product of its speed and the unit vector in the direction of motion.
First,find the unit vector $\hat{A}$ in the direction of $\vec{A} = 2\hat{i} + 2\hat{j} - \hat{k}$:
$|\vec{A}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,$\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$.
The velocity vector is $\vec{v} = |\vec{v}| \hat{A} = 6 \left( \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3} \right)$.
Simplifying the expression:
$\vec{v} = 2(2\hat{i} + 2\hat{j} - \hat{k}) = 4\hat{i} + 4\hat{j} - 2\hat{k} \, m/s$.

(C) Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$.
From the triangle law of vector addition,the angle $\alpha$ between $\vec{R}$ and $\vec{A}$ is given by $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
Similarly,the angle $\beta$ between $\vec{R}$ and $\vec{B}$ is given by $\tan \beta = \frac{A \sin \theta}{B + A \cos \theta}$.
Geometrically,the resultant vector $\vec{R}$ is always closer to the vector with the larger magnitude.
If $A > B$,then the resultant $\vec{R}$ will be closer to $\vec{A}$,which means the angle $\alpha$ will be smaller than the angle $\beta$ (i.e.,$\alpha < \beta$).
Conversely,if $A < B$,then $\alpha > \beta$.
Therefore,$\alpha < \beta$ if $A > B$.

(A) The component of a vector $\vec{A}$ along a direction defined by unit vector $\hat{n}$ is given by $\vec{A} \cdot \hat{n}$.
$(A)$ Component along $x$-axis: $\vec{A} \cdot \hat{i} = (3 \hat{i} + 4 \hat{j} - 5 \hat{k}) \cdot \hat{i} = 3$. This is not in the options,so $(A \rightarrow s)$.
$(B)$ Component along $\vec{B} = (2 \hat{i} + \hat{j} + 2 \hat{k})$: $\vec{A} \cdot \frac{\vec{B}}{|B|} = \frac{(3)(2) + (4)(1) + (-5)(2)}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{6 + 4 - 10}{3} = 0$. Thus,$(B \rightarrow r)$.
$(C)$ Component along $\vec{C} = (6 \hat{i} + 8 \hat{j} - 10 \hat{k}) = 2\vec{A}$. Since it is parallel,the component is the magnitude $|\vec{A}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}$. This is not in the options,so $(C \rightarrow s)$.
$(D)$ Component along $\vec{D} = (-3 \hat{i} - 4 \hat{j} + 5 \hat{k}) = -\vec{A}$. Since it is anti-parallel,the component is $-|\vec{A}| = -5\sqrt{2}$. This is not in the options,so $(D \rightarrow s)$.
Therefore,the correct matching is $(A \rightarrow s, B \rightarrow r, C \rightarrow s, D \rightarrow s)$.

(A) The component of a vector $\vec{A}$ along the direction of a vector $\vec{B}$ is given by the scalar projection multiplied by the unit vector of $\vec{B}$,or simply the scalar projection if only the magnitude is asked. The question asks for the component (magnitude) of $\vec{A}$ along $\vec{B}$.
Given $\vec{A} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = \hat{i} + \hat{j}$.
The scalar projection of $\vec{A}$ along $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
Calculate the dot product: $\vec{A} \cdot \vec{B} = (2)(1) + (3)(1) = 2 + 3 = 5$.
Calculate the magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Therefore,the component of $\vec{A}$ along $\vec{B}$ is $\frac{5}{\sqrt{2}}$.

(C) Given: $|\vec{A} + \vec{B}| = \frac{1}{2} |\vec{A} - \vec{B}|$ and $|\vec{A}| = 2|\vec{B}|$.
Squaring both sides of the first equation:
$|\vec{A} + \vec{B}|^2 = \frac{1}{4} |\vec{A} - \vec{B}|^2$
$A^2 + B^2 + 2AB \cos \theta = \frac{1}{4} (A^2 + B^2 - 2AB \cos \theta)$
$4(A^2 + B^2 + 2AB \cos \theta) = A^2 + B^2 - 2AB \cos \theta$
$4A^2 + 4B^2 + 8AB \cos \theta = A^2 + B^2 - 2AB \cos \theta$
$3A^2 + 3B^2 + 10AB \cos \theta = 0$
Substitute $A = 2B$ into the equation:
$3(2B)^2 + 3B^2 + 10(2B)(B) \cos \theta = 0$
$3(4B^2) + 3B^2 + 20B^2 \cos \theta = 0$
$12B^2 + 3B^2 + 20B^2 \cos \theta = 0$
$15B^2 + 20B^2 \cos \theta = 0$
$20B^2 \cos \theta = -15B^2$
$\cos \theta = -\frac{15}{20} = -\frac{3}{4}$
$\theta = \cos^{-1}\left(-\frac{3}{4}\right)$

(A) The magnitude of the change in a vector $\vec{A}$ of magnitude $x$ when rotated by an angle $\theta$ is given by $|\Delta \vec{A}| = |\vec{A}' - \vec{A}| = 2x \sin(\theta/2)$.
Given $|\Delta \vec{A}| = nx$, we have $nx = 2x \sin(\theta/2)$, which implies $n = 2 \sin(\theta/2)$.
For $(A)$ $\theta=60^{\circ}$: $n = 2 \sin(30^{\circ}) = 2(1/2) = 1$. Thus, $(A \rightarrow q)$.
For $(B)$ $\theta=90^{\circ}$: $n = 2 \sin(45^{\circ}) = 2(1/\sqrt{2}) = \sqrt{2}$. Thus, $(B \rightarrow r)$.
For $(C)$ $\theta=120^{\circ}$: $n = 2 \sin(60^{\circ}) = 2(\sqrt{3}/2) = \sqrt{3}$. Thus, $(C \rightarrow p)$.
For $(D)$ $\theta=180^{\circ}$: $n = 2 \sin(90^{\circ}) = 2(1) = 2$. Thus, $(D \rightarrow s)$.
Therefore, the correct matching is $(A \rightarrow q, B \rightarrow r, C \rightarrow p, D \rightarrow s)$.

(N/A) Meaningful: The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.
$(b)$ Not Meaningful: The addition of a vector quantity with a scalar quantity is not meaningful because they represent different types of physical entities.
$(c)$ Meaningful: $A$ scalar can be multiplied with a vector. For example,force is multiplied with time to give impulse $(I = F \cdot \Delta t)$.
$(d)$ Meaningful: $A$ scalar,irrespective of the physical quantity it represents,can be multiplied with another scalar having the same or different dimensions.
$(e)$ Meaningful: The addition of two vector quantities is meaningful only if they both represent the same physical quantity.
$(f)$ Meaningful: $A$ component of a vector can be added to the same vector as they both have the same dimensions and represent the same physical quantity.

(A) True: The magnitude of a vector is a real number representing its length. Hence,it is a scalar.
$(b)$ False: Each component of a vector is a vector quantity directed along the coordinate axes.
$(c)$ False: Total path length is the actual distance traveled,which is a scalar,while the magnitude of displacement is the shortest distance between the initial and final positions. Total path length is greater than or equal to the magnitude of displacement.
$(d)$ True: Since total path length is always greater than or equal to the magnitude of displacement,dividing both by the same time interval leads to the average speed being greater than or equal to the magnitude of average velocity.
$(e)$ True: Three vectors not lying in a plane cannot form a closed triangle,which is a necessary condition for their sum to be a null vector.

(N/A) Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram $OMNP$. In $\Delta OMN$,the side $ON$ represents $\vec{a} + \vec{b}$. By the triangle inequality,the length of any side is less than or equal to the sum of the lengths of the other two sides: $|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$. Equality holds when $\vec{a}$ and $\vec{b}$ are in the same direction.
$(b)$ In $\Delta OMN$,the difference of two sides is less than or equal to the third side: $|\vec{a} + \vec{b}| \geq ||\vec{a}| - |\vec{b}||$. Equality holds when $\vec{a}$ and $\vec{b}$ are in the same direction.
$(c)$ Similarly,for the vector difference $\vec{a} - \vec{b}$,using the triangle formed by $\vec{a}$ and $-\vec{b}$,we get $|\vec{a} - \vec{b}| \leq |\vec{a}| + |-\vec{b}| = |\vec{a}| + |\vec{b}|$. Equality holds when $\vec{a}$ and $\vec{b}$ are in opposite directions.
$(d)$ Using the triangle inequality for the difference,we have $|\vec{a} - \vec{b}| \geq ||\vec{a}| - |\vec{b}||$. Equality holds when $\vec{a}$ and $\vec{b}$ are in opposite directions.

(B, C, D) Incorrect: To satisfy $a+b+c+d=0,$ it is not necessary for all four vectors to be null vectors. Many other combinations can result in a zero sum.
$(b)$ Correct: Given $a+b+c+d=0,$ we can write $a+c = -(b+d).$ Taking the magnitude on both sides,$|a+c| = |-(b+d)| = |b+d|.$ Thus,the magnitude of $(a+c)$ equals the magnitude of $(b+d).$
$(c)$ Correct: From $a+b+c+d=0,$ we have $a = -(b+c+d).$ Taking the magnitude on both sides,$|a| = |-(b+c+d)| = |b+c+d|.$ By the triangle inequality,$|b+c+d| \leq |b| + |c| + |d|.$ Therefore,$|a| \leq |b| + |c| + |d|,$ meaning the magnitude of $a$ can never be greater than the sum of the magnitudes of $b, c,$ and $d.$
$(d)$ Correct: For $a+b+c+d=0,$ we can write $(b+c) = -(a+d).$ This implies that the vector $(b+c)$ is equal and opposite to the resultant of $a$ and $d.$ If $a$ and $d$ are not collinear,they define a plane,and $(b+c)$ must lie in that plane. If $a$ and $d$ are collinear,$(b+c)$ must lie along the same line to satisfy the equilibrium condition.

(N/A) No; Yes; No.
Generally speaking,a vector does not have a definite location in space. This is because a vector remains invariant when displaced in such a way that its magnitude and direction remain the same. However,a position vector has a definite location in space.
$A$ vector can vary with time. For example,the displacement vector of a particle moving with a certain velocity varies with time.
Two equal vectors located at different locations in space need not produce the same physical effect. For example,two equal forces acting on an object at different points can cause the body to rotate,whereas their combination might produce a different turning effect (torque) depending on the point of application.

(N/A) No; No.
$A$ physical quantity having both magnitude and direction is not necessarily a vector. For example,electric current has both magnitude and direction but is a scalar quantity because it does not follow the laws of vector addition.
Similarly,the rotation of a body about an axis is specified by the direction of the axis and the angle of rotation. However,finite rotations do not follow the commutative law of vector addition (i.e.,$\vec{A} + \vec{B} \neq \vec{B} + \vec{A}$ for finite rotations). Therefore,finite rotations are not vectors. Only infinitesimal rotations follow the laws of vector addition and are considered vectors.

(A) No; Yes; No.
$(a)$ The length of a wire bent into a loop is a scalar quantity. It has magnitude but no specific direction associated with it,so it cannot be represented as a vector.
$(b)$ $A$ plane area can be associated with a vector. The magnitude of the area vector is equal to the area of the plane,and its direction is defined as normal to the surface (either inward or outward).
$(c)$ $A$ sphere is a three-dimensional object defined by its volume,which is a scalar. Therefore,one cannot associate a vector with the volume of a sphere. However,an area vector can be associated with the surface area of a sphere at any specific point,directed radially outward.

(N/A) Scalar physical quantity: $A$ physical quantity that requires only magnitude to be fully described is called a scalar physical quantity.
Examples: $Mass$, $distance$, $time$, $density$, $temperature$, $power$, etc.
Vector physical quantity: $A$ physical quantity that requires both magnitude and direction to be fully described is called a vector physical quantity.
Examples: $Displacement$, $velocity$, $acceleration$, $force$, etc.

(A)

Vector Quantities	Scalar Quantities
$(1)$ Physical quantities that require both magnitude and direction for their complete specification are called vector quantities.	$(1)$ Physical quantities that can be described by their magnitude only are called scalar quantities.
$(2)$ Examples: velocity,acceleration,force,weight,displacement,momentum,etc.	$(2)$ Examples: speed,mass,volume,size,temperature,amount of substance,power,work done,pressure,time,etc.
$(3)$ While representing these quantities,both magnitude and direction must be stated.	$(3)$ While representing these quantities,only their magnitude is required,i.e.,the numerical value with units.
$(4)$ These quantities cannot be added algebraically; they follow vector addition laws.	$(4)$ These quantities are added or subtracted using simple algebraic rules.

(B) vector quantity is represented by a symbol in boldface (e.g.,$v$ for velocity).
Practically,to represent a vector quantity,an arrow sign is placed above the symbol.
For example,$\vec{v}$ represents velocity,where the arrow indicates the direction of the vector.