Fundamentals of Vectors Questions in English

(D) The vector is given by $\vec{A} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
To find the projection of a vector $\vec{A}$ on a unit vector $\hat{n}$,we use the dot product $\vec{A} \cdot \hat{n}$.
The $Y$-axis is represented by the unit vector $\hat{j}$.
Therefore,the projection is $(3\hat{i} + 0\hat{j} + 4\hat{k}) \cdot \hat{j} = 3(\hat{i} \cdot \hat{j}) + 0(\hat{j} \cdot \hat{j}) + 4(\hat{k} \cdot \hat{j})$.
Since $\hat{i} \cdot \hat{j} = 0$,$\hat{j} \cdot \hat{j} = 1$,and $\hat{k} \cdot \hat{j} = 0$,the projection is $0 + 0 + 0 = 0$.

(B) The position vector $\vec{r}$ of a point $P(x, y, z)$ in a three-dimensional 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 the position vector: $\vec{r} = 3\hat{i} + 2\hat{j} + 5\hat{k}$.
Therefore,the correct option is $B$.

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

(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 in the direction of $A$:
$|A| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
$\hat A = \frac{A}{|A|} = \frac{3\hat i + 4\hat j}{5}$.
The required vector has the same magnitude as $B$ $(25)$ and is parallel to $A$ (in the direction of $\hat A$):
$\text{Required Vector} = |B| \cdot \hat A = 25 \cdot \left( \frac{3\hat i + 4\hat j}{5} \right) = 5(3\hat i + 4\hat j) = 15\hat i + 20\hat j$.

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

(A) Given vector $\overrightarrow A = 2\hat i + 4\hat j - 5\hat k$.
First,calculate the magnitude of the vector $\overrightarrow A$:
$|\overrightarrow A| = \sqrt{(2)^2 + (4)^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45}$.
The direction cosines $(l, m, n)$ are given by the ratios of the components to the 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 $\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}, \text{and } \frac{-5}{\sqrt{45}}$.

(D) Let the initial point be $P(4, -4, 0)$ and the terminal point be $Q(-2, -2, 0)$.
The vector $\vec{r}$ is given by $\vec{r} = \vec{Q} - \vec{P}$.
$\vec{r} = (-2 - 4)\hat{i} + (-2 - (-4))\hat{j} + (0 - 0)\hat{k}$.
$\vec{r} = -6\hat{i} + 2\hat{j} + 0\hat{k}$.
The magnitude of the vector is $|\vec{r}| = \sqrt{(-6)^2 + (2)^2 + (0)^2}$.
$|\vec{r}| = \sqrt{36 + 4} = \sqrt{40}$.
$|\vec{r}| = \sqrt{4 \times 10} = 2\sqrt{10}$.

(A) Let the given vector be $\vec{P} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.
The magnitude of a vector $\vec{A} = x\hat{i} + y\hat{j}$ is given by $|\vec{A}| = \sqrt{x^2 + y^2}$.
Substituting the values,we get $|\vec{P}| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2}$.
$|\vec{P}| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1$.
Since the magnitude of the vector is $1$,it is a unit vector.

(C) unit vector $\hat{u}$ in the direction of a vector $\vec{A}$ is given by the formula $\hat{u} = \frac{\vec{A}}{|\vec{A}|}$.
Given the vector $\vec{A} = \hat{i} + \hat{j}$.
The magnitude of the vector is $|\vec{A}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,the unit vector is $\hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.

(D) vector quantity is a physical quantity that has both magnitude and direction,and follows the laws of vector addition.
Pressure is defined as force per unit area. Although force is a vector,pressure is a scalar quantity because it acts equally in all directions at a point in a fluid.
Surface tension is defined as force per unit length,but it is a scalar quantity.
Moment of inertia is a tensor quantity (specifically a second-rank tensor) that describes the resistance of an object to rotational motion.
Since none of the given options are vector quantities,the correct answer is $D$.

(D) Given $\vec P = \vec Q$.
$A$. $\hat P = \hat Q$: Since the vectors are equal,they have the same direction,so their unit vectors are equal. This is correct.
$B$. $|\vec P| = |\vec Q|$: Equal vectors must have the same magnitude. This is correct.
$C$. $P\hat Q = Q\hat P$: Since $\vec P = \vec Q$,we have $P = Q$ and $\hat P = \hat Q$. Thus,$P\hat Q = Q\hat P$ is equivalent to $\vec P = \vec Q$. This is correct.
$D$. $\vec P + \vec Q = \hat P + \hat Q$: The left side is a vector of magnitude $2P$,while the right side is a vector of magnitude $2$ (since the sum of two unit vectors has a magnitude between $0$ and $2$). These are generally not equal unless $P = 1$. Therefore,this is $NOT$ correct.

(B) unit vector has a magnitude of $1$.
Given vector $\vec{A} = 0.5\hat i + 0.8\hat j + c\hat k$.
The magnitude is given by $|\vec{A}| = \sqrt{(0.5)^2 + (0.8)^2 + c^2} = 1$.
Squaring both sides,we get $(0.5)^2 + (0.8)^2 + c^2 = 1^2$.
$0.25 + 0.64 + c^2 = 1$.
$0.89 + c^2 = 1$.
$c^2 = 1 - 0.89 = 0.11$.
Therefore,$c = \sqrt{0.11}$.

(B) In physics,surface area is considered a vector quantity. The magnitude of the area vector is equal to the surface area,and its direction is defined as being normal (perpendicular) to the surface. Therefore,it is a vector.

(B) The angle $\theta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos \theta = \frac{\vec A \cdot \vec B}{|A||B|}$.
Since $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j + 5\hat k$,we see that $\vec A = \vec B$.
The dot product is $\vec A \cdot \vec B = (3)(3) + (4)(4) + (5)(5) = 9 + 16 + 25 = 50$.
The magnitudes are $|A| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ and $|B| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}$.
Substituting these values into the formula: $\cos \theta = \frac{50}{\sqrt{50} \cdot \sqrt{50}} = \frac{50}{50} = 1$.
Since $\cos \theta = 1$,the angle $\theta = \cos^{-1}(1) = 0^o$.

(A) The resultant $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,where $\theta$ is the angle between the vectors.
For the resultant $R$ to be maximum,the value of $\cos \theta$ must be maximum.
The maximum value of $\cos \theta$ is $1$,which occurs when $\theta = 0^o$.
Therefore,the angle between the two vectors must be $0^o$ for their resultant to be maximum.

(B) Let the plane containing $\overrightarrow{A}$ and $\overrightarrow{B}$ be the $xy$-plane. Thus,$\overrightarrow{A} + \overrightarrow{B}$ also lies in the $xy$-plane.
Since $\overrightarrow{C}$ lies outside this plane,it must have a non-zero component perpendicular to the $xy$-plane (i.e.,along the $z$-axis).
Let $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}$.
For the resultant $\overrightarrow{R}$ to be zero,the sum of the components along the $z$-axis must be zero.
Since $\overrightarrow{A}$ and $\overrightarrow{B}$ have zero components along the $z$-axis,the $z$-component of the resultant is equal to the $z$-component of $\overrightarrow{C}$.
Because $\overrightarrow{C}$ is outside the plane,its $z$-component is non-zero,so the resultant $\overrightarrow{R}$ cannot be zero.

(A) For two vectors $\vec{A}$ and $\vec{B}$,let the angle between them be $\theta$. The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by:
$|\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta}$
Given the condition $|\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}|$,we square both sides:
$|\vec{A} + \vec{B}|^2 = (|\vec{A}| + |\vec{B}|)^2$
$|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|$
Subtracting $|\vec{A}|^2 + |\vec{B}|^2$ from both sides:
$2|\vec{A}||\vec{B}| \cos \theta = 2|\vec{A}||\vec{B}|$
Assuming $\vec{A}$ and $\vec{B}$ are non-zero vectors,we divide by $2|\vec{A}||\vec{B}|$:
$\cos \theta = 1$
Since $\cos \theta = 1$,the angle $\theta = 0^o$.

(C) The displacement vector can be represented as $\vec{r} = 12\hat{i} + 5\hat{j} + 6\hat{k}$.
The magnitude of the resultant displacement $R$ is given by the formula $R = \sqrt{x^2 + y^2 + z^2}$.
Substituting the values: $R = \sqrt{12^2 + 5^2 + 6^2}$.
$R = \sqrt{144 + 25 + 36}$.
$R = \sqrt{205}$.
$R \approx 14.31 \, m$.

(C) The magnitude of the resultant of two vectors $\vec A$ and $\vec B$ is given by the formula:
$|\vec R| = |\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,where $\theta$ is the angle between the two vectors.
Given the condition $|\vec A + \vec B| = |\vec A| + |\vec B|$,we substitute the magnitude formula:
$\sqrt{A^2 + B^2 + 2AB \cos \theta} = A + B$
Squaring both sides:
$A^2 + B^2 + 2AB \cos \theta = (A + B)^2$
$A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 + 2AB$
$2AB \cos \theta = 2AB$
$\cos \theta = 1$
Therefore,$\theta = 0^o$.

(B) Let $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = -4\hat{i} - 6\hat{j} + \lambda\hat{k}$.
If two vectors $\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are parallel,then their components must be proportional:
$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.
Substituting the given values:
$\frac{2}{-4} = \frac{3}{-6} = \frac{-1}{\lambda}$.
Simplifying the ratios:
$-\frac{1}{2} = -\frac{1}{2} = -\frac{1}{\lambda}$.
Equating the last two parts:
$-\frac{1}{2} = -\frac{1}{\lambda} \implies \lambda = 2$.

(C) Two vectors are normal (perpendicular) to each other if their dot product is zero.
Let $\overrightarrow{B} = \hat{i}B_x + \hat{j}B_y$.
For $\overrightarrow{A} \cdot \overrightarrow{B} = 0$:
$(A\cos\theta)(B_x) + (A\sin\theta)(B_y) = 0$.
If we choose $\overrightarrow{B} = B(\hat{i}\sin\theta - \hat{j}\cos\theta)$,then:
$\overrightarrow{A} \cdot \overrightarrow{B} = (A\cos\theta)(B\sin\theta) + (A\sin\theta)(-B\cos\theta) = AB\sin\theta\cos\theta - AB\sin\theta\cos\theta = 0$.
Thus,the vector $\hat{i}B\sin\theta - \hat{j}B\cos\theta$ is normal to $\overrightarrow{A}$.

(B) The displacement vector $\overrightarrow{AB}$ is calculated as: $\overrightarrow{AB} = \vec{B} - \vec{A} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (3\hat{i} + 4\hat{j} + 5\hat{k}) = \hat{i} + \hat{j} + \hat{k}$.
The displacement vector $\overrightarrow{CD}$ is calculated as: $\overrightarrow{CD} = \vec{D} - \vec{C} = (4\hat{i} + 6\hat{j}) - (7\hat{i} + 9\hat{j} + 3\hat{k}) = -3\hat{i} - 3\hat{j} - 3\hat{k}$.
We can rewrite $\overrightarrow{CD}$ as: $\overrightarrow{CD} = -3(\hat{i} + \hat{j} + \hat{k}) = -3\overrightarrow{AB}$.
Since $\overrightarrow{CD} = k\overrightarrow{AB}$ where $k = -3$ (a negative scalar),the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are antiparallel.

(C) The dot product of two vectors is defined as $\vec A \cdot \vec B = |A||B| \cos \theta$,where $\theta$ is the angle between the vectors.
Given that $\vec A \cdot \vec B = - |A||B|$,we can equate the two expressions:
$|A||B| \cos \theta = - |A||B|$
Dividing both sides by $|A||B|$ (assuming non-zero vectors),we get:
$\cos \theta = - 1$
This implies that $\theta = 180^\circ$.
Therefore,the vectors $\vec A$ and $\vec B$ act in the opposite direction.

(C) To obtain a zero resultant,the vectors must form a closed polygon.
If vectors are in the same plane,the minimum number of vectors required is $3$ (forming a triangle).
However,the question specifies that the vectors are in different planes.
If we have $3$ vectors in different planes,their resultant cannot be zero because the resultant of any two vectors lies in the plane containing them,and the third vector (being in a different plane) cannot cancel this resultant.
Therefore,we need at least $4$ vectors. For example,consider a tetrahedron where the four vectors represent the four faces; their sum is zero. Thus,the minimum number of non-zero vectors in different planes required to give a zero resultant is $4$.

(C) For any vector,the direction cosines satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
We know the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$.
Substituting this into the expression:
$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = (1 - \cos^2 \alpha) + (1 - \cos^2 \beta) + (1 - \cos^2 \gamma)$
$= 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma)$
$= 3 - 1$
$= 2$

(B) From the figure,let the initial vector be $\overrightarrow{OA}$ and the final vector be $\overrightarrow{OB}$. Since the length does not change,$|\overrightarrow{OA}| = |\overrightarrow{OB}| = a$.
By the triangle law of vector addition,$\overrightarrow{OB} = \overrightarrow{OA} + \Delta \overrightarrow{a}$,which implies $\Delta \overrightarrow{a} = \overrightarrow{OB} - \overrightarrow{OA}$.
The magnitude $|\Delta \overrightarrow{a}|$ represents the length of the chord $AB$ in the circular arc of radius $a$ subtending an angle $d\theta$ at the center $O$. For a small angle $d\theta$,the chord length is approximately equal to the arc length.
Using the relation $\text{angle} = \frac{\text{arc}}{\text{radius}}$,we have $d\theta = \frac{AB}{a}$,which gives $AB = a\,d\theta$.
Therefore,$|\Delta \overrightarrow{a}| = a\,d\theta$.
$\Delta a$ represents the change in the magnitude of the vector,which is $\Delta a = |\overrightarrow{OB}| - |\overrightarrow{OA}| = a - a = 0$.
Thus,the values are $a\,d\theta$ and $0$ respectively.

(D) In a regular hexagon $ABCDEF$ with center $O$,we have the following properties:
$\overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO}$ (by triangle law in $\triangle AOF$ is not correct,rather consider the vectors from center $O$)
Alternatively,using the property of the center $O$:
$\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} = (\overrightarrow{AO} + \overrightarrow{OB}) + (\overrightarrow{AO} + \overrightarrow{OC}) + (\overrightarrow{AO} + \overrightarrow{OD}) + (\overrightarrow{AO} + \overrightarrow{OE}) + (\overrightarrow{AO} + \overrightarrow{OF})$
$= 5\overrightarrow{AO} + (\overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF})$
Since $O$ is the center of the regular hexagon,the sum of vectors from the center to the vertices is zero: $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} = 0$.
Therefore,$\overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} = -\overrightarrow{OA} = \overrightarrow{AO}$.
Substituting this back:
$= 5\overrightarrow{AO} + \overrightarrow{AO} = 6\overrightarrow{AO}$.

(A) The man moves $10\,m$ along the North direction (y-axis) and $20\,m$ along the East direction (x-axis).
Let the initial position be the origin $(0,0)$.
The final position vector is $\vec{r} = 20\hat{i} + 10\hat{j}$.
The magnitude of displacement is given by $r = |\vec{r}| = \sqrt{x^2 + y^2}$.
$r = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500}$.
$r = 10\sqrt{5} \approx 10 \times 2.236 = 22.36\,m$.

(C) For an object to be in equilibrium,the net force acting on it must be zero.
If a force $\vec{F}_1$ acts in the north-east direction,a second force $\vec{F}_2$ must be equal in magnitude and opposite in direction to balance it.
Mathematically,$\vec{F}_1 + \vec{F}_2 = 0$,which implies $\vec{F}_2 = -\vec{F}_1$.
The opposite direction of north-east is south-west.
Therefore,the second force should be applied in the south-west direction.

(C) For any vector,the direction cosines satisfy the relation: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$,we can rewrite the expression as:
$(1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1$.
Rearranging the terms:
$3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1$.
Therefore,$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2$.

(A) The vectors $\overrightarrow{AB}$,$\overrightarrow{BC}$,and $\overrightarrow{CA}$ represent the sides of a triangle taken in the same cyclic order.
According to the polygon law of vector addition,the sum of vectors forming a closed loop is zero.
Therefore,$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{0}$.

(C) Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$.
By the law of sines in the triangle formed by vectors $\vec{A}$,$\vec{B}$,and $\vec{R}$,we have $\frac{A}{\sin \beta} = \frac{B}{\sin \alpha}$.
This implies $\frac{\sin \alpha}{\sin \beta} = \frac{B}{A}$.
If $A > B$,then $\frac{B}{A} < 1$,which means $\sin \alpha < \sin \beta$.
Since $\alpha$ and $\beta$ are angles within a triangle,$\sin \alpha < \sin \beta$ implies $\alpha < \beta$.
Therefore,the resultant vector is closer to the vector with the larger magnitude. Thus,if $A > B$,then $\alpha < \beta$.

(C) The direction cosines of a vector $\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}$ are given by the formula:
$l = \frac{a}{|\vec{A}|}, m = \frac{b}{|\vec{A}|}, n = \frac{c}{|\vec{A}|}$
where $|\vec{A}| = \sqrt{a^2 + b^2 + c^2}$.
For the given vector $\vec{A} = 1\hat{i} + 1\hat{j} + \sqrt{2}\hat{k}$,the magnitude is:
$|\vec{A}| = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.
Now,calculating the direction cosines:
$l = \frac{1}{2}$
$m = \frac{1}{2}$
$n = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
Thus,the direction cosines are $\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}}$.

(B) Given the equation: $\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$.
Subtracting $\vec{P}$ from both sides,we get: $\vec{Q} = -\vec{Q}$.
Adding $\vec{Q}$ to both sides,we get: $2\vec{Q} = 0$.
This implies that $\vec{Q} = 0$,which means $\vec{Q}$ is a null vector.

(B) The projection of a vector $\vec{A}$ onto another vector $\vec{B}$ represents the scalar component of $\vec{A}$ along the direction of $\vec{B}$.
Mathematically,the projection of $\vec{A}$ on $\vec{B}$ is given by the dot product of $\vec{A}$ with the unit vector in the direction of $\vec{B}$.
Since the unit vector $\hat{B} = \frac{\vec{B}}{|\vec{B}|}$,the projection is $\vec{A} \cdot \hat{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.

(A) In vector algebra,for unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the $x, y, z$ axes respectively:
$1$. Cross product properties: $\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,and $\hat{k} \times \hat{i} = \hat{j}$. Also,$\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$.
$2$. Dot product properties: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$.
Comparing these with the options:
Option $A$: $\hat{j} \times \hat{k} = \hat{i}$ is correct.
Option $B$: $\hat{i} \cdot \hat{i} = 1$ (given $0$,so incorrect).
Option $C$: $\hat{j} \times \hat{j} = 0$ (given $1$,so incorrect).
Option $D$: $\hat{k} \cdot \hat{i} = 0$ (given $1$,so incorrect).
Therefore,the correct relation is $\hat{j} \times \hat{k} = \hat{i}$.

(C) Let the vector be $\vec{A} = \hat{i} + \hat{j}$.
Here,the components are $A_x = 1$ and $A_y = 1$.
The magnitude of the vector is $A = \sqrt{A_x^2 + A_y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The angle $\alpha$ with the $x$-axis is given by $\cos \alpha = \frac{A_x}{A} = \frac{1}{\sqrt{2}}$.
Therefore,$\alpha = 45^\circ$.
The angle $\beta$ with the $y$-axis is given by $\cos \beta = \frac{A_y}{A} = \frac{1}{\sqrt{2}}$.
Therefore,$\beta = 45^\circ$.

(D) Given that $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$.
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 have:
$(\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B})$
$A^2 + B^2 + 2\vec{A} \cdot \vec{B} = A^2 + B^2 - 2\vec{A} \cdot \vec{B}$
$4\vec{A} \cdot \vec{B} = 0$
Since $\vec{A} \cdot \vec{B} = AB \cos \theta$,we have:
$4AB \cos \theta = 0$
Assuming $A \neq 0$ and $B \neq 0$,we get $\cos \theta = 0$.
Therefore,$\theta = 90^\circ$.

(A) The magnitude of a vector $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ is given by $|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$.
Given vector is $\vec{A} = 2\hat{i} - 1\hat{j} - 5\hat{k}$.
Here,$A_x = 2$,$A_y = -1$,and $A_z = -5$.
Substituting these values into the formula:
$|\vec{A}| = \sqrt{(2)^2 + (-1)^2 + (-5)^2}$
$|\vec{A}| = \sqrt{4 + 1 + 25}$
$|\vec{A}| = \sqrt{30}$.

(D) We know that for any triangle $ABC$ with centroid $O$,the vector sum $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}$.
We can express the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ in terms of the position vectors relative to the centroid $O$:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \overrightarrow{AO} + \overrightarrow{OB}$
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \overrightarrow{AO} + \overrightarrow{OC}$
Adding these two equations:
$\overrightarrow{AB} + \overrightarrow{AC} = (\overrightarrow{AO} + \overrightarrow{OB}) + (\overrightarrow{AO} + \overrightarrow{OC})$
$\overrightarrow{AB} + \overrightarrow{AC} = 2\overrightarrow{AO} + (\overrightarrow{OB} + \overrightarrow{OC})$
From the centroid property $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}$,we have:
$\overrightarrow{OB} + \overrightarrow{OC} = -\overrightarrow{OA} = \overrightarrow{AO}$
Substituting this into the sum:
$\overrightarrow{AB} + \overrightarrow{AC} = 2\overrightarrow{AO} + \overrightarrow{AO} = 3\overrightarrow{AO}$
Comparing this with $\overrightarrow{AB} + \overrightarrow{AC} = n\overrightarrow{AO}$,we get $n = 3$.

(C) The magnitude of a vector $\vec{P} = P_x\hat{i} + P_y\hat{j} + P_z\hat{k}$ is given by the formula $|\vec{P}| = \sqrt{P_x^2 + P_y^2 + P_z^2}$.
Given $\vec{P} = 3\hat{i} + 4\hat{j} + 12\hat{k}$,we have $P_x = 3$,$P_y = 4$,and $P_z = 12$.
Substituting these values into the formula:
$|\vec{P}| = \sqrt{3^2 + 4^2 + 12^2}$
$|\vec{P}| = \sqrt{9 + 16 + 144}$
$|\vec{P}| = \sqrt{169}$
$|\vec{P}| = 13$.

(D) Given that $\vec{A} + \vec{B} = \vec{C}$ and $|\vec{A}| = |\vec{B}| = |\vec{C}| = A$ (let).
Squaring both sides of the vector equation $\vec{A} + \vec{B} = \vec{C}$,we get:
$|\vec{A} + \vec{B}|^2 = |\vec{C}|^2$
$|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta = |\vec{C}|^2$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
Substituting the magnitudes: $A^2 + A^2 + 2A^2 \cos \theta = A^2$.
$2A^2 + 2A^2 \cos \theta = A^2$.
$2A^2 \cos \theta = A^2 - 2A^2 = -A^2$.
$\cos \theta = -A^2 / 2A^2 = -1/2$.
Since $\cos \theta = -1/2$,the angle $\theta = 120^o$.

(A) In an equilateral triangle $ABC$,the centroid $O$ is the point where the medians intersect.
Since the triangle is equilateral,the vectors $\overrightarrow{OA}$,$\overrightarrow{OB}$,and $\overrightarrow{OC}$ have equal magnitudes and are directed from the centroid to the vertices.
The angles between these vectors are $120^{\circ}$ each.
According to the principle of vector addition for symmetric systems,the resultant of three vectors of equal magnitude separated by $120^{\circ}$ is zero.
Therefore,$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \vec{0}$.

(A) Given vector $\vec{A} = \hat{i} + \hat{j} + \hat{k}$.
Comparing with $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$,we get $A_x = 1, A_y = 1, A_z = 1$.
The magnitude of the vector is $|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
The direction cosines are given by $\cos \alpha = \frac{A_x}{|\vec{A}|} = \frac{1}{\sqrt{3}}$,$\cos \beta = \frac{A_y}{|\vec{A}|} = \frac{1}{\sqrt{3}}$,and $\cos \gamma = \frac{A_z}{|\vec{A}|} = \frac{1}{\sqrt{3}}$.
Thus,the cosine values are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

(B) For the sum of vectors to be zero,they must form a closed polygon when placed head-to-tail.
If there are only $2$ vectors of different magnitudes,say $\vec{A}$ and $\vec{B}$,their sum $\vec{A} + \vec{B} = 0$ implies $\vec{A} = -\vec{B}$. This means they must have the same magnitude and opposite directions,which contradicts the condition that they have different magnitudes.
If there are $3$ vectors of different magnitudes,they can form a triangle (a closed polygon) if the sum of the lengths of any two sides is greater than the third side. For example,vectors of magnitudes $3, 4,$ and $5$ can form a triangle and their sum can be zero.
Therefore,the minimum number of coplanar vectors of different magnitudes required to have a zero resultant is $3$.

(B) Since the forces are mutually perpendicular,we can represent them as vectors along the $x$,$y$,and $z$ axes:
$\vec{F}_1 = 3\hat{i} \ N$,$\vec{F}_2 = 4\hat{j} \ N$,and $\vec{F}_3 = 12\hat{k} \ N$.
The resultant force vector is given by the sum of these vectors:
$\vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 3\hat{i} + 4\hat{j} + 12\hat{k}$.
The magnitude of the resultant force is calculated using the formula $F = \sqrt{F_x^2 + F_y^2 + F_z^2}$:
$F = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \ N$.

(C) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by $C^2 = A^2 + B^2 + 2AB \cos \theta$,where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.
Given that $C^2 = A^2 + B^2$,we substitute this into the equation:
$A^2 + B^2 = A^2 + B^2 + 2AB \cos \theta$.
Subtracting $A^2 + B^2$ from both sides,we get:
$0 = 2AB \cos \theta$.
Since the magnitudes $A$ and $B$ are non-zero,we must have $\cos \theta = 0$.
This implies $\theta = 90^\circ$.
Therefore,$\vec{A}$ is perpendicular to $\vec{B}$.

(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).
For example,current has both magnitude and direction but is a scalar because it does not follow the laws of vector addition.
Therefore,having a direction is a necessary but not sufficient condition for a quantity to be a vector.
Thus,a physical quantity with a direction 'can be called' a vector,but it is not necessarily a vector.

(B) Since the three forces are mutually perpendicular,they can be represented along the $x$,$y$,and $z$ axes as vectors $\vec{F_1} = 3\hat{i}$,$\vec{F_2} = 4\hat{j}$,and $\vec{F_3} = 12\hat{k}$.
The resultant force vector is $\vec{F} = 3\hat{i} + 4\hat{j} + 12\hat{k}$.
The magnitude of the resultant force is given by $|\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2}$.
Substituting the values: $|\vec{F}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169}$.
Therefore,the magnitude of the resultant force is $13 \ N$.

(B) Let the initial vector be $\overrightarrow{OA}$ and the final vector after rotation be $\overrightarrow{OB}$.
Given that the magnitude of the vector remains constant,$|\overrightarrow{OA}| = |\overrightarrow{OB}| = a$.
The change in the vector is defined as $\Delta \overrightarrow{a} = \overrightarrow{OB} - \overrightarrow{OA}$.
From the geometry of the triangle $OAB$,the magnitude of the change in the vector is the length of the chord $AB$.
For a small angle $d\theta$,the arc length $AB$ is given by $AB = a \cdot d\theta$.
Therefore,$|\Delta \overrightarrow{a}| = a \cdot d\theta$.
Now,consider the change in the magnitude of the vector,$\Delta a = |\overrightarrow{OB}| - |\overrightarrow{OA}|$.
Since the magnitude is constant,$\Delta a = a - a = 0$.
Thus,$|\Delta \overrightarrow{a}| = a \cdot d\theta$ and $\Delta a = 0$.