Vardaan Learning Institute

Solution: (c) Angular Velocity. Electric current and electric potential are scalars (current does not follow vector addition laws). Pressure is a scalar/tensor. Angular velocity is an axial vector whose direction is given by the right-hand rule.

Solution: The magnitude of a vector $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ is given by $|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$.
$|\vec{A}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = \mathbf{13 \text{ units}}$.

Solution: First, find the resultant vector $\vec{R} = \vec{P} + \vec{Q}$.
$\vec{R} = (2+1)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k} = 3\hat{i} + 6\hat{j} - 2\hat{k}$.
Magnitude $|\vec{R}| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
Unit vector $\hat{R} = \frac{\vec{R}}{|\vec{R}|} = \mathbf{\frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{7}}$.

Solution: The magnitude of a unit vector is exactly 1.
$\sqrt{(0.4)^2 + (0.8)^2 + c^2} = 1 \implies 0.16 + 0.64 + c^2 = 1^2 \implies 0.80 + c^2 = 1$
$c^2 = 1 - 0.80 = 0.20 \implies c = \mathbf{\pm \sqrt{0.2}}$ (or $\approx \pm 0.447$).

Solution: For two vectors to be equal, their corresponding scalar components must be strictly equal.
Equating components: $\mathbf{x = 2}$, $\mathbf{y = -3}$, and $\mathbf{z = 1}$.

Solution: The position vector $\vec{r}$ of a point $(x,y,z)$ from origin $(0,0,0)$ is simply $x\hat{i} + y\hat{j} + z\hat{k}$.
Therefore, $\vec{r} = \mathbf{2\hat{i} - \hat{j} + 4\hat{k}}$.

Solution: Displacement vector $\vec{AB} = \vec{r}_B - \vec{r}_A$.
$\vec{AB} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = \mathbf{3\hat{i} + 3\hat{j} + 3\hat{k}}$.
Magnitude $|\vec{AB}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = \mathbf{3\sqrt{3} \text{ units}}$.

Solution: A null vector is a vector with zero magnitude and an arbitrary (undefined) direction.
Physical Situations:
1. The velocity vector of a stationary object.
2. The acceleration vector of an object moving with a uniform velocity.

Solution: Two vectors are collinear (parallel) if one is a scalar multiple of the other ($\vec{B} = \lambda\vec{A}$).
Here, $\vec{B} = 4\hat{i} + 6\hat{j} - 8\hat{k} = 2(2\hat{i} + 3\hat{j} - 4\hat{k}) = 2\vec{A}$.
Since $\vec{B} = 2\vec{A}$ (where $\lambda = 2$), they point in the same direction. Yes, they are collinear.

Solution:
Polar Vectors: Have a starting point or a point of application. They describe translation. Example: Displacement, Force.
Axial Vectors: Represent rotational effects and act along the axis of rotation in accordance with the right-hand rule. Example: Torque, Angular Velocity.

Solution: Maximum resultant $R_{max} = A + B = 17$ (when $\theta = 0^\circ$).
Minimum resultant $R_{min} = A - B = 7$ (when $\theta = 180^\circ$).
Adding the two equations: $2A = 24 \implies \mathbf{A = 12 \text{ units}}$.
Substituting $A$ in first equation: $12 + B = 17 \implies \mathbf{B = 5 \text{ units}}$.

Solution: Using the law of cosines: $R^2 = P^2 + Q^2 + 2PQ\cos\theta$.
Given $R=F$, $P=F$, $Q=F$.
$F^2 = F^2 + F^2 + 2F^2\cos\theta \implies F^2 = 2F^2(1 + \cos\theta) \implies \frac{1}{2} = 1 + \cos\theta$
$\cos\theta = -\frac{1}{2}$. Since $\cos(120^\circ) = -0.5$, the angle $\mathbf{\theta = 120^\circ}$.

Solution: Squaring both sides:
$|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2$
$A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta$
$4AB\cos\theta = 0 \implies \cos\theta = 0$. Therefore, $\mathbf{\theta = 90^\circ}$. The vectors are perpendicular.

Solution: Magnitude $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos(90^\circ)} = \sqrt{5^2 + 12^2 + 0} = \sqrt{25+144} = \sqrt{169} = \mathbf{13 \text{ N}}$.
Direction $\tan\alpha = \frac{F_2\sin\theta}{F_1+F_2\cos\theta} = \frac{12\sin 90^\circ}{5+12\cos 90^\circ} = \frac{12}{5}$.
$\alpha = \mathbf{\tan^{-1}(12/5)}$ (approx $67.38^\circ$) with respect to the $5 \text{ N}$ force.

Solution: Expand both terms using dot product properties:
$|\vec{A} + \vec{B}|^2 = (\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B}) = A^2 + B^2 + 2\vec{A}\cdot\vec{B}$
$|\vec{A} - \vec{B}|^2 = (\vec{A}-\vec{B})\cdot(\vec{A}-\vec{B}) = A^2 + B^2 - 2\vec{A}\cdot\vec{B}$
Adding them: $(A^2 + B^2 + 2\vec{A}\cdot\vec{B}) + (A^2 + B^2 - 2\vec{A}\cdot\vec{B}) = \mathbf{2(A^2 + B^2)}$. (Proved)

Solution: From relation, $\vec{A} + \vec{B} = -\vec{C}$.
Squaring both sides: $|\vec{A} + \vec{B}|^2 = |-\vec{C}|^2 \implies A^2 + B^2 + 2AB\cos\theta = C^2$
$(3)^2 + (4)^2 + 2(3)(4)\cos\theta = (5)^2 \implies 9 + 16 + 24\cos\theta = 25$
$25 + 24\cos\theta = 25 \implies 24\cos\theta = 0 \implies \cos\theta = 0$. So, $\mathbf{\theta = 90^\circ}$.

Solution: Let East be $+\hat{i}$ and North be $+\hat{j}$.
Displacement 1 ($\vec{S_1}$) = $10\hat{j}$.
Displacement 2 ($\vec{S_2}$): South-West means $225^\circ$ or $-45^\circ$ in 3rd quadrant.
$\vec{S_2} = -10\sqrt{2}\cos(45^\circ)\hat{i} - 10\sqrt{2}\sin(45^\circ)\hat{j} = -10\sqrt{2}(\frac{1}{\sqrt{2}})\hat{i} - 10\sqrt{2}(\frac{1}{\sqrt{2}})\hat{j} = -10\hat{i} - 10\hat{j}$.
Net Displacement $\vec{S} = \vec{S_1} + \vec{S_2} = 10\hat{j} + (-10\hat{i} - 10\hat{j}) = \mathbf{-10\hat{i}}$.
This is a magnitude of $\mathbf{10 \text{ km} \text{ due West}}$.

Solution: Let resultant be $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$.
Given $R = (A^2 + B^2 + AB)^{1/2}$. Squaring both equates to:
$A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 + AB \implies 2AB\cos\theta = AB \implies \cos\theta = \frac{1}{2}$.
Therefore, $\mathbf{\theta = 60^\circ}$. The condition is that the angle between them must be $60^\circ$.

Solution: Initial velocity $\vec{v_1} = 5\hat{i}$ m/s. Final velocity $\vec{v_2} = 5\hat{j}$ m/s. Time $t = 10$ s.
Change in velocity $\Delta\vec{v} = \vec{v_2} - \vec{v_1} = 5\hat{j} - 5\hat{i}$.
Avg Acceleration $\vec{a} = \frac{\Delta\vec{v}}{t} = \frac{5\hat{j} - 5\hat{i}}{10} = \mathbf{-0.5\hat{i} + 0.5\hat{j} \text{ m/s}^2}$.
Magnitude: $|\vec{a}| = \sqrt{(-0.5)^2 + (0.5)^2} = \sqrt{0.25+0.25} = \sqrt{0.5} = \mathbf{\frac{1}{\sqrt{2}} \text{ m/s}^2}$.
Direction: $(-\hat{i}, +\hat{j})$ is the 2nd quadrant, which corresponds to North-West.

Solution:
(a) False. 4 vectors summing to zero form a closed polygon in 3D space, they need not be coplanar. (Only 3 non-collinear vectors summing to 0 must be coplanar).
(b) True. Rearranging gives $\vec{a} = -(\vec{b} + \vec{c} + \vec{d})$. By the polygon inequality (extension of triangle inequality), the magnitude of one side of a polygon is always less than or equal to the sum of the magnitudes of the other sides: $|\vec{a}| \le |\vec{b}| + |\vec{c}| + |\vec{d}|$.

Solution: $R = \sqrt{P^2 + (2P)^2 + 2(P)(2P)\cos(120^\circ)}$. $\cos(120^\circ) = -1/2$.
$R = \sqrt{P^2 + 4P^2 + 4P^2(-0.5)} = \sqrt{5P^2 - 2P^2} = \sqrt{3P^2} = \mathbf{P\sqrt{3}}$.
Angle $\alpha$ with force $P$: $\tan\alpha = \frac{2P\sin 120^\circ}{P + 2P\cos 120^\circ} = \frac{2P(\sqrt{3}/2)}{P + 2P(-1/2)} = \frac{P\sqrt{3}}{0} = \infty$.
So $\alpha = 90^\circ$ relative to $P$. (Or $30^\circ$ relative to $2P$).

Solution: Consider the magnitude squared of their difference:
$|\hat{n}_1 - \hat{n}_2|^2 = |\hat{n}_1|^2 + |\hat{n}_2|^2 - 2|\hat{n}_1||\hat{n}_2|\cos\theta$
Since they are unit vectors, $|\hat{n}_1| = |\hat{n}_2| = 1$.
$|\hat{n}_1 - \hat{n}_2|^2 = 1 + 1 - 2(1)(1)\cos\theta = 2 - 2\cos\theta = 2(1 - \cos\theta)$.
Using half-angle identity: $1 - \cos\theta = 2\sin^2(\theta/2)$.
$|\hat{n}_1 - \hat{n}_2|^2 = 2(2\sin^2(\theta/2)) = 4\sin^2(\theta/2)$.
Taking the square root: $|\hat{n}_1 - \hat{n}_2| = 2\sin(\theta/2) \implies \mathbf{\sin(\theta/2) = \frac{1}{2}|\hat{n}_1 - \hat{n}_2|}$. (Proved)

Solution: X-component $F_x = F\cos\theta = 100\cos 60^\circ = 100(1/2) = \mathbf{50 \text{ N}}$.
Y-component $F_y = F\sin\theta = 100\sin 60^\circ = 100(\sqrt{3}/2) = \mathbf{50\sqrt{3} \text{ N}}$.

Solution: Let vector be $\vec{V} = V\cos\theta\hat{i} + V\sin\theta\hat{j}$.
Given $V = 10, \theta = 37^\circ$. (Note: $\sin 37^\circ \approx 3/5$, $\cos 37^\circ \approx 4/5$).
$\vec{V} = 10(4/5)\hat{i} + 10(3/5)\hat{j} = \mathbf{8\hat{i} + 6\hat{j}}$.

Solution: First, find the magnitude $|\vec{R}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Direction cosines are $l = \frac{R_x}{R}$, $m = \frac{R_y}{R}$, $n = \frac{R_z}{R}$.
$l = \mathbf{2/7}$, $m = \mathbf{-3/7}$, $n = \mathbf{6/7}$.

Solution: We know the property of direction cosines: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.
Substitute $\cos^2\theta = 1 - \sin^2\theta$ for each term:
$(1 - \sin^2\alpha) + (1 - \sin^2\beta) + (1 - \sin^2\gamma) = 1$
$3 - (\sin^2\alpha + \sin^2\beta + \sin^2\gamma) = 1$
Rearranging gives: $\mathbf{\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2}$. (Proved)

Solution: The true weight is $W = Mg$ acting vertically downward.
Geometry shows the angle between the weight vector and the normal (perpendicular) to the incline is also $\theta$.
Component perpendicular to incline = $\mathbf{Mg\cos\theta}$.
Component parallel (down) the incline = $\mathbf{Mg\sin\theta}$.

Solution: Magnitude $|\vec{A}| = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.
The direction cosine for the Z-axis is $\cos\gamma = \frac{A_z}{A} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Since $\cos\gamma = 1/\sqrt{2}$, the angle $\mathbf{\gamma = 45^\circ}$.

Solution: Vector projection of $\vec{A}$ on $\vec{B}$ is given by $\left(\frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\right)\vec{B}$.
$\vec{A}\cdot\vec{B} = (2)(1) + (3)(1) = 5$.
$|\vec{B}|^2 = (1)^2 + (1)^2 = 2$.
Projection = $\frac{5}{2}(\hat{i} + \hat{j}) = \mathbf{2.5\hat{i} + 2.5\hat{j}}$.

Solution: Let Force be $F$. Horizontal component $F_x = F\cos 60^\circ = 50$.
$F(1/2) = 50 \implies \mathbf{F = 100 \text{ N}}$.
Vertical component $F_y = F\sin 60^\circ = 100(\sqrt{3}/2) = \mathbf{50\sqrt{3} \text{ N}}$.

Solution: Required vector $\vec{V} = (\text{Magnitude of } \vec{B}) \times (\text{Unit vector of } \vec{A})$.
$|\vec{B}| = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.
$\hat{A} = \frac{3\hat{i} + 4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i} + 4\hat{j}}{5}$.
$\vec{V} = 25 \left( \frac{3\hat{i} + 4\hat{j}}{5} \right) = 5(3\hat{i} + 4\hat{j}) = \mathbf{15\hat{i} + 20\hat{j}}$.

Solution: For angles to represent direction cosines, they must satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.
Let's check: $\cos^2(45^\circ) + \cos^2(60^\circ) + \cos^2(120^\circ)$
$= (1/\sqrt{2})^2 + (1/2)^2 + (-1/2)^2 = 1/2 + 1/4 + 1/4 = 1$.
Since the sum equals 1, Yes, such a vector is physically possible.

Solution: $\vec{A}\cdot\vec{B} = (2)(1) + (-5)(2) + (3)(-1) = 2 - 10 - 3 = \mathbf{-11}$.

Solution: $\cos\theta = \frac{\vec{P}\cdot\vec{Q}}{|\vec{P}||\vec{Q}|}$.
$\vec{P}\cdot\vec{Q} = (1)(-1) + (1)(-1) + (1)(2) = -1 - 1 + 2 = 0$.
Since dot product is $0$, the vectors are orthogonal. $\mathbf{\theta = 90^\circ}$.

Solution: For perpendicular vectors, $\vec{A}\cdot\vec{B} = 0$.
$(3)(2) + (-2)(1) + (m)(4) = 0$
$6 - 2 + 4m = 0 \implies 4 + 4m = 0 \implies 4m = -4 \implies \mathbf{m = -1}$.

Solution: Displacement $\Delta\vec{r} = \vec{r}_f - \vec{r}_i = (2-1)\hat{i} + (3-1)\hat{j} + (5-1)\hat{k} = \hat{i} + 2\hat{j} + 4\hat{k}$.
Work $W = \vec{F}\cdot\Delta\vec{r} = (2)(1) + (3)(2) + (4)(4) = 2 + 6 + 16 = \mathbf{24 \text{ J}}$.

Solution: Expand the product: $\vec{A}\cdot\vec{A} - \vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{A} - \vec{B}\cdot\vec{B}$. Since dot product is commutative ($\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$), middle terms cancel.
$= |\vec{A}|^2 - |\vec{B}|^2 = \mathbf{A^2 - B^2}$.
If this equals zero, $A^2 - B^2 = 0 \implies A^2 = B^2 \implies \mathbf{|\vec{A}| = |\vec{B}|}$. (The vectors have equal magnitudes, meaning the sum and difference vectors form diagonals of a rhombus).

Solution: Given $|\vec{A}|=1, |\vec{B}|=1, |\vec{A}+\vec{B}|=1$.
$|\vec{A}+\vec{B}|^2 = 1 \implies A^2 + B^2 + 2\vec{A}\cdot\vec{B} = 1$
$1^2 + 1^2 + 2(1)(1)\cos\theta = 1 \implies 2 + 2\cos\theta = 1$
$2\cos\theta = -1 \implies \cos\theta = -1/2$. Therefore, $\mathbf{\theta = 120^\circ}$.

Solution: $P = (10)(5) + (10)(-3) + (20)(6) = 50 - 30 + 120 = \mathbf{140 \text{ W}}$.

Solution: Scalar projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$.
$\vec{a}\cdot\vec{b} = (3)(1) + (-1)(2) + (-2)(-3) = 3 - 2 + 6 = 7$.
$|\vec{b}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Projection = $\mathbf{\frac{7}{\sqrt{14}}}$ (or $\sqrt{3.5}$).

Solution: Unit Vectors Check:
$|\vec{a}| = \frac{1}{7}\sqrt{2^2+3^2+6^2} = \frac{1}{7}\sqrt{49} = 1$. Same for $|\vec{b}|$ and $|\vec{c}|$ ($\sqrt{3^2+6^2+2^2}=7$). Thus they are unit vectors.
Perpendicularity Check (Dot Products):
$\vec{a}\cdot\vec{b} = \frac{1}{49}[(2)(3) + (3)(-6) + (6)(2)] = \frac{1}{49}[6 - 18 + 12] = 0$.
$\vec{b}\cdot\vec{c} = \frac{1}{49}[(3)(6) + (-6)(2) + (2)(-3)] = \frac{1}{49}[18 - 12 - 6] = 0$.
$\vec{a}\cdot\vec{c} = \frac{1}{49}[(2)(6) + (3)(2) + (6)(-3)] = \frac{1}{49}[12 + 6 - 18] = 0$.
Since all dot products are 0 and magnitudes are 1, they are mutually perpendicular unit vectors. (Proved).

Solution: $AB\cos\theta = AB\sin\theta$.
Dividing by $AB\cos\theta$ gives: $\frac{\sin\theta}{\cos\theta} = 1 \implies \tan\theta = 1$.
Therefore, $\mathbf{\theta = 45^\circ}$.

Solution: First, find the vectors representing the sides of the triangle: $\vec{A} = \vec{r}_2 - \vec{r}_1$, $\vec{B} = \vec{r}_3 - \vec{r}_2$, $\vec{C} = \vec{r}_1 - \vec{r}_3$.
$\vec{A} = (2-1)\hat{i} + (3 - (-2))\hat{j} + (-4-3)\hat{k} = \hat{i} + 5\hat{j} - 7\hat{k}$.
$\vec{B} = (0-2)\hat{i} + (-7-3)\hat{j} + (10 - (-4))\hat{k} = -2\hat{i} - 10\hat{j} + 14\hat{k}$.
Notice that $\vec{B} = -2(\hat{i} + 5\hat{j} - 7\hat{k}) = -2\vec{A}$.
Since side $\vec{B}$ is a scalar multiple of side $\vec{A}$, the points are collinear (lie on a straight line). They do not form a triangle at all, let alone a right-angled one.

Solution: Area magnitude $A = 100 \text{ cm}^2 = 100 \times 10^{-4} \text{ m}^2 = 10^{-2} \text{ m}^2$.
Since the surface lies in the Y-Z plane, its normal vector (area vector direction) points along the X-axis ($\hat{i}$).
Area vector $\vec{A} = 10^{-2} \hat{i} \text{ m}^2$.
Flux $\Phi = \vec{E}\cdot\vec{A} = (2\times 10^3 \hat{i}) \cdot (10^{-2} \hat{i}) = (2\times 10^3) \times (10^{-2}) = \mathbf{20 \text{ N m}^2\text{/C}}$.

Solution: Using determinant method:
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & -1 & 1 \end{vmatrix}$
$= \hat{i}( (2)(1) - (-1)(-1) ) - \hat{j}( (3)(1) - (-1)(1) ) + \hat{k}( (3)(-1) - (2)(1) )$
$= \hat{i}(2 - 1) - \hat{j}(3 + 1) + \hat{k}(-3 - 2) = \mathbf{\hat{i} - 4\hat{j} - 5\hat{k}}$.

Solution: A vector perpendicular to both is their cross product $\vec{C} = \vec{P} \times \vec{Q}$.
$\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(6 - 2) - \hat{j}(-4 - 1) + \hat{k}(4 - (-3)) = 4\hat{i} + 5\hat{j} + 7\hat{k}$.
Magnitude $|\vec{C}| = \sqrt{4^2 + 5^2 + 7^2} = \sqrt{16 + 25 + 49} = \sqrt{90}$.
Unit vector $\hat{C} = \frac{\vec{C}}{|\vec{C}|} = \mathbf{\frac{4\hat{i} + 5\hat{j} + 7\hat{k}}{\sqrt{90}}}$.

Solution: Area = $|\vec{a} \times \vec{b}|$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = \hat{i}(-1 - (-21)) - \hat{j}(1 - 6) + \hat{k}(-7 - (-2))$
$= 20\hat{i} + 5\hat{j} - 5\hat{k}$.
Area = $\sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450} = \mathbf{15\sqrt{2} \text{ sq. units}}$.

Solution: $|\vec{A} \times \vec{B}| = AB\sin\theta = 0$.
Since $A \neq 0$ and $B \neq 0$, it must be that $\sin\theta = 0 \implies \theta = 0^\circ \text{ or } 180^\circ$. Hence they are parallel (or anti-parallel). (Proved).
The dot product will be $\vec{A} \cdot \vec{B} = AB\cos\theta = AB(\pm 1) = \mathbf{\pm AB}$.

Solution: $\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 3 & 1 & 5 \end{vmatrix}$
$= \hat{i}(15 - (-1)) - \hat{j}(10 - (-3)) + \hat{k}(2 - 9)$
$= \mathbf{16\hat{i} - 13\hat{j} - 7\hat{k} \text{ N m}}$.

Solution: Area of Triangle = $\frac{1}{2}|\vec{A} \times \vec{B}|$.
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ -3 & 7 & 0 \end{vmatrix} = \hat{k}((3)(7) - (-3)(4)) = \hat{k}(21 - (-12)) = 33\hat{k}$.
Magnitude = $33$. Area = $\frac{33}{2} = \mathbf{16.5 \text{ sq. units}}$.

Solution: Expand using distributive property:
$= \vec{A} \times \vec{A} + \vec{A} \times \vec{B} - \vec{B} \times \vec{A} - \vec{B} \times \vec{B}$.
We know $\vec{A} \times \vec{A} = 0$ and $\vec{B} \times \vec{B} = 0$. Also, $\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})$.
$= 0 + \vec{A} \times \vec{B} - (-\vec{A} \times \vec{B}) - 0 = \vec{A} \times \vec{B} + \vec{A} \times \vec{B} = \mathbf{2(\vec{A} \times \vec{B})}$.

Solution: For vectors to be parallel, their scalar components must be proportional.
$\frac{a_x}{b_x} = \frac{a_y}{b_y} = \frac{a_z}{b_z} \implies \frac{2}{-4} = \frac{3}{-6} = \frac{-1}{p}$.
$-\frac{1}{2} = -\frac{1}{2} = -\frac{1}{p} \implies \mathbf{p = 2}$.

Solution: Use cyclic properties of cross product: $\hat{j} \times \hat{k} = \hat{i}$, $\hat{i} \times \hat{k} = -\hat{j}$, $\hat{i} \times \hat{j} = \hat{k}$.
Substitute these back: $\hat{i}\cdot(\hat{i}) + \hat{j}\cdot(-\hat{j}) + \hat{k}\cdot(\hat{k})$
$= 1 + (-1) + 1 = \mathbf{1}$.

Solution: Let $\theta$ be the angle between $\vec{A}$ and $\vec{B}$.
$|\vec{A} \times \vec{B}|^2 = (AB\sin\theta)^2 = A^2 B^2 \sin^2\theta$.
$(\vec{A} \cdot \vec{B})^2 = (AB\cos\theta)^2 = A^2 B^2 \cos^2\theta$.
Adding them: $A^2 B^2 \sin^2\theta + A^2 B^2 \cos^2\theta = A^2 B^2 (\sin^2\theta + \cos^2\theta)$.
Since $\sin^2\theta + \cos^2\theta = 1$, it equals $A^2 B^2 = \mathbf{|\vec{A}|^2 |\vec{B}|^2}$. (Proved).

Solution: Linear velocity $\vec{v} = \vec{\omega} \times \vec{r}$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 0.4 & -0.3 \end{vmatrix}$
$= \hat{i}( (-2)(-0.3) - (2)(0.4) ) - \hat{j}( (1)(-0.3) - (2)(0) ) + \hat{k}( (1)(0.4) - (-2)(0) )$
$= \hat{i}(0.6 - 0.8) - \hat{j}(-0.3 - 0) + \hat{k}(0.4 - 0)$
$= \mathbf{-0.2\hat{i} + 0.3\hat{j} + 0.4\hat{k} \text{ units}}$.