Step 1: Definition of Scalar (Dot) Product

The scalar product (dot product) of two non-zero vectors $\vec{a}$ and $\vec{b}$ is defined by the geometric relation: $$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$$ where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$ respectively, and $\theta$ is the angle between them such that $0 \le \theta \le \pi$. [Per Definition of Scalar Product in Euclidean space].

Step 2: Analysis of Magnitudes

The magnitudes $|\vec{a}|$ and $|\vec{b}|$ represent the lengths of the vectors. By definition, the magnitude of any vector is a non-negative real number: $$|\vec{a}| \ge 0 \quad \text{and} \quad |\vec{b}| \ge 0$$ Given the condition $\vec{a} \cdot \vec{b} \ge 0$, we substitute the definition into the inequality:

Step 3: Determining the Sign of the Cosine Function

Since the product of magnitudes $|\vec{a}| |\vec{b}|$ is always non-negative, the sign of the dot product depends solely on the value of $\cos \theta$. To satisfy $\vec{a} \cdot \vec{b} \ge 0$, we must have: $$\cos \theta \ge 0$$ In the domain $0 \le \theta \le \pi$, the cosine function is non-negative within the interval $[0, \frac{\pi}{2}]$.

Step 4: Graphical Representation

Step 5: Logical Synthesis

• If $\theta = 0$, $\cos \theta = 1 > 0 \implies \vec{a} \cdot \vec{b} > 0$ (Vectors are parallel/codirectional).
• If $0 < \theta < \frac{\pi}{2}$, $\cos \theta > 0 \implies \vec{a} \cdot \vec{b} > 0$.
• If $\theta = \frac{\pi}{2}$, $\cos \theta = 0 \implies \vec{a} \cdot \vec{b} = 0$ (Vectors are orthogonal).

Final Solution: $0 \le \theta \le \frac{\pi}{2}$

Step 1: Determine the Components of the Vector $\vec{PQ}$

To find the vector $\vec{PQ}$ given the initial point $P(x_1, y_1, z_1) = (1, 2, 3)$ and terminal point $Q(x_2, y_2, z_2) = (4, 5, 6)$, we use the formula for the displacement vector [Per Vector Displacement Axiom]:

$\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

Substituting the given coordinates:

$\vec{PQ} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k}$

$\vec{PQ} = 3\hat{i} + 3\hat{j} + 3\hat{k}$

Step 2: Calculate the Magnitude of $\vec{PQ}$

The magnitude of a vector $\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}$ is defined as $|\vec{A}| = \sqrt{a^2 + b^2 + c^2}$ [By Euclidean Norm Definition].

$|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2}$

$|\vec{PQ}| = \sqrt{9 + 9 + 9} = \sqrt{27}$

$|\vec{PQ}| = 3\sqrt{3}$

Step 3: Derivation of the Unit Vector

The unit vector $\hat{u}$ in the direction of vector $\vec{A}$ is defined by the relation $\hat{u} = \frac{\vec{A}}{|\vec{A}|}$ [Per Unit Vector Normalization Theorem].

$\hat{PQ} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}}$

Distributing the scalar denominator across the components:

$\hat{PQ} = \frac{3}{3\sqrt{3}}\hat{i} + \frac{3}{3\sqrt{3}}\hat{j} + \frac{3}{3\sqrt{3}}\hat{k}$

$\hat{PQ} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$

Final Solution: The unit vector in the direction of $\vec{PQ}$ is $\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$

Analytical Definition of Collinear Vectors

Two vectors $\vec{a}$ and $\vec{b}$ are defined as collinear (or parallel) if there exists a non-zero scalar $\lambda$ such that:

$$\vec{b} = \lambda \vec{a}$$

[Per the definition of collinearity in Vector Algebra, vectors are collinear if they are parallel to the same line, irrespective of their magnitudes or directions.]

Step 1: Evaluating the Relationship Between Collinearity and Magnitude

Consider the magnitudes of the two vectors. If $\vec{b} = \lambda \vec{a}$, then by the property of vector magnitude:

$$|\vec{b}| = |\lambda \vec{a}| = |\lambda| |\vec{a}|$$

The condition for the two vectors to be equal in magnitude is $|\vec{b}| = |\vec{a}|$, which implies $|\lambda| = 1$. This only occurs if $\lambda = 1$ or $\lambda = -1$.

Step 2: Counter-Example Construction

To determine if the statement "Two collinear vectors are always equal in magnitude" is true, we test for a counter-example where $\lambda \neq \pm 1$.

• Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
• Let $\lambda = 2$.
• Define $\vec{b} = \lambda \vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.

Calculations:

Since $\sqrt{3} \neq 2\sqrt{3}$, the vectors are collinear but their magnitudes are not equal. [By the property of scalar multiplication of vectors].

Visual Representation of Collinear Vectors with Different Magnitudes

Conclusion

The statement is False because collinear vectors are defined by parallelism (the existence of a scalar multiple), not by identity of magnitude. Only if the scalar multiple $\lambda$ is restricted to $\pm 1$ would the magnitudes be equal.

Final Solution: False

Step 1: Identifying the Geometric Foundation

The area of a parallelogram defined by two adjacent vectors $\vec{a}$ and $\vec{b}$ is equal to the magnitude of their cross product. [By Geometric Interpretation of Vector Product].

Mathematically, the Area $A = |\vec{a} \times \vec{b}|$.

Step 2: Calculating the Cross Product ($\vec{a} \times \vec{b}$)

Given: $\vec{a} = 1\hat{i} - 1\hat{j} + 3\hat{k}$ $\vec{b} = 2\hat{i} - 7\hat{j} + 1\hat{k}$

The cross product is computed using the determinant of the matrix involving unit vectors $\hat{i}, \hat{j}, \hat{k}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$

Expanding along the first row: $= \hat{i} [(-1)(1) - (3)(-7)] - \hat{j} [(1)(1) - (3)(2)] + \hat{k} [(1)(-7) - (-1)(2)]$

$= \hat{i} [-1 + 21] - \hat{j} [1 - 6] + \hat{k} [-7 + 2]$

$= 20\hat{i} + 5\hat{j} - 5\hat{k}$

Step 3: Calculating the Magnitude of the Resultant Vector

The area is the magnitude $|\vec{a} \times \vec{b}| = \sqrt{x^2 + y^2 + z^2}$.

$|\vec{a} \times \vec{b}| = \sqrt{(20)^2 + (5)^2 + (-5)^2}$

$|\vec{a} \times \vec{b}| = \sqrt{400 + 25 + 25}$

$|\vec{a} \times \vec{b}| = \sqrt{450}$

Simplifying the radical: $\sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2}$

Visual Representation

Analytical Summary

Final Solution: The area of the parallelogram is $15\sqrt{2}$ square units.

1. Definitions & Conceptual Framework

To classify a physical quantity, we must distinguish between scalars and vectors based on their mathematical properties:

• Scalar: A physical quantity characterized solely by its magnitude. It requires a numerical value and a unit of measurement. It is independent of direction [Per Scalar Field Theory].
• Vector: A physical quantity characterized by both magnitude and a specific direction in space. It must satisfy the laws of vector addition [Per Euclidean Vector Space definitions].

2. Analysis of the Physical Quantity

The quantity provided is $40\text{ watt}$.

• Magnitude: $40$
• Unit: $\text{watt (W)}$, which is the SI unit of power.
• Dimensional Analysis: Power $P$ is defined as the rate of energy transfer or work done per unit time: $P = \frac{dW}{dt}$ [Per the Definition of Power in Classical Mechanics].

3. Logical Deduction

Power is calculated as the scalar product of force $\vec{F}$ and velocity $\vec{v}$ ($P = \vec{F} \cdot \vec{v}$). Although force and velocity are vectors, the result of a dot product is always a scalar quantity [By the Properties of the Scalar Product]. Since $40\text{ watt}$ describes the rate of energy conversion without reference to a directional component, it satisfies the criteria for a scalar quantity.

4. Visual Representation of Scalar Power

Final Solution: The measure 40 watt is a scalar quantity.

Step 1: Establishing the Position Vectors

To demonstrate the collinearity of points $A$, $B$, and $C$, we define their position vectors relative to the origin $O(0, 0, 0)$ as follows:

• $\vec{OA} = \mathbf{a} = 1\hat{i} + 2\hat{j} + 7\hat{k}$
• $\vec{OB} = \mathbf{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$
• $\vec{OC} = \mathbf{c} = 3\hat{i} + 10\hat{j} - 1\hat{k}$

Step 2: Calculating Directional Vectors

Collinearity implies that the vectors $\vec{AB}$ and $\vec{BC}$ are parallel (scalar multiples of each other). We calculate these vectors by subtracting the initial point from the terminal point:

$\vec{AB} = \vec{OB} - \vec{OA}$

$\vec{AB} = (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k} = 1\hat{i} + 4\hat{j} - 4\hat{k}$

$\vec{BC} = \vec{OC} - \vec{OB}$

$\vec{BC} = (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k} = 1\hat{i} + 4\hat{j} - 4\hat{k}$

Step 3: Determining Collinearity via Scalar Multiplicity

For points to be collinear, the vector connecting two segments must be a scalar multiple of the other. Here, we observe:

$\vec{AB} = \lambda \vec{BC}$

$1\hat{i} + 4\hat{j} - 4\hat{k} = \lambda(1\hat{i} + 4\hat{j} - 4\hat{k})$

Equating components:

• $1 = \lambda(1) \implies \lambda = 1$
• $4 = \lambda(4) \implies \lambda = 1$
• $-4 = \lambda(-4) \implies \lambda = 1$

Since $\lambda = 1$ satisfies all components, $\vec{AB} = \vec{BC}$.

Step 4: Geometrical Representation

Conclusion

Because $\vec{AB}$ and $\vec{BC}$ share a common point $B$ and possess the same direction vector (they are parallel), the points $A$, $B$, and $C$ necessarily lie on the same straight line.

Final Solution: Since $\vec{AB} = \vec{BC}$ and the segments share a common point $B$, the points $A$, $B$, and $C$ are collinear.

Step 1: Determining the Side Vectors

To determine if the points $A$, $B$, and $C$ form a right-angled triangle, we must calculate the vectors representing the sides of the triangle: $\vec{AB}$, $\vec{BC}$, and $\vec{CA}$.

Given position vectors:

• $\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}$
• $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}$

Calculating the displacement vectors [By Vector Subtraction Rule]:

• $\vec{AB} = \vec{b} - \vec{a} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$
• $\vec{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
• $\vec{CA} = \vec{a} - \vec{c} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$

Step 2: Calculating Magnitudes of the Sides

The squared magnitudes of the vectors represent the squares of the side lengths $|\vec{AB}|^2, |\vec{BC}|^2, |\vec{CA}|^2$ [Per the Definition of Magnitude of a Vector]:

Step 3: Verifying the Pythagorean Theorem

According to the Converse of the Pythagorean Theorem, if the sum of the squares of two sides equals the square of the third side, the triangle is right-angled.

Summing the two smaller magnitudes: $|\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41$

Comparing with the largest magnitude: $|\vec{BC}|^2 = 41$

Since $|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$, the triangle satisfies the condition for a right-angled triangle, specifically with the right angle at vertex $A$ (as $\vec{AB} \cdot \vec{CA} = -2 - 3 + 5 = 0$).

Final Solution: Since the magnitudes satisfy the relation $|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$, the points A, B, and C form a right-angled triangle with the right angle at vertex A.

1. Conceptual Definition of Velocity

In classical mechanics, velocity is defined as the rate of change of displacement with respect to time. Mathematically, it is expressed as the derivative of the position vector $\vec{r}$ with respect to time $t$:

$$\vec{v} = \frac{d\vec{r}}{dt}$$

[Per the definitions of Kinematics in Vector Algebra]

2. Analysis of Dimensional Requirements

To classify a physical quantity as either scalar or vector, we must evaluate its adherence to the following criteria:

• Scalar: A quantity defined solely by its magnitude (a real number) and unit. It is invariant under coordinate rotation.
• Vector: A quantity defined by both magnitude and a specific direction in space, satisfying the parallelogram law of vector addition.

3. Classification Logic

Velocity possesses two fundamental attributes:

1. Magnitude: The speed of the object, which is the scalar component $|\vec{v}|$.
2. Direction: The orientation of motion, denoted by the unit vector $\hat{v}$, such that $\vec{v} = |\vec{v}|\hat{v}$.

Since velocity explicitly requires direction to be fully defined—moving at $50 \, \text{km/h}$ is physically incomplete without specifying the cardinal direction (e.g., North) or vector orientation—it behaves as a directed line segment in $\mathbb{R}^3$ space.

4. Visual Representation of Velocity Vector

5. Comparison Table

Final Solution: Velocity is a vector quantity because it possesses both magnitude (speed) and a defined direction in space.

Step 1: Defining the Measurement

The given measure is $40^\circ$. This represents a magnitude of temperature or angular measurement depending on the physical context, but inherently, it signifies a scalar value.

Step 2: Theoretical Criteria for Classification

To distinguish between scalar and vector quantities, we apply the following definitions:

• Scalar Quantity: A physical quantity that is completely described by its magnitude (a numerical value along with units) and possesses no specific direction in space [Per the definition of Scalar fields in Euclidean space].
• Vector Quantity: A physical quantity that requires both magnitude and a defined direction to be completely specified. It must also satisfy the parallelogram law of vector addition [Per the axiomatic definition of Vectors in $\mathbb{R}^n$].

Step 3: Logical Analysis of the Measure

The measurement $40^\circ$ is a pure magnitude. Whether it represents an angle between two lines or a temperature on a scale (such as Celsius), it lacks an associated directional component (a vector direction such as $\hat{i}, \hat{j}, \hat{k}$ or an orientation in 3D space) required to qualify as a vector [By the principles of Scalar Algebra].

Visual Representation of a Scalar Scalar Representation

Step 4: Classification

Final Solution: The measure $40^\circ$ is classified as a Scalar quantity.

Step 1: Identification of Vector Components

Given the vector $\vec{c} = \frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{1}{3}\hat{k}$, we identify the vector in its Cartesian form $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, where:

• $x = \frac{1}{3}$
• $y = \frac{1}{3}$
• $z = -\frac{1}{3}$

Step 2: Definition of Magnitude

The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is defined by the Euclidean norm formula [Per the definition of the magnitude of a vector in 3D space]:

$|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$

Step 3: Substitution and Calculation

Substituting the identified components into the magnitude formula:

$|\vec{c}| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2}$

$|\vec{c}| = \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{9}}$

$|\vec{c}| = \sqrt{\frac{3}{9}}$

$|\vec{c}| = \sqrt{\frac{1}{3}}$

Step 4: Rationalization and Simplification

Simplifying the radical expression:

$|\vec{c}| = \frac{1}{\sqrt{3}}$

To rationalize the denominator [Standard mathematical practice for radical simplification]:

$|\vec{c}| = \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3}$

Visual Representation of Vector Components

Final Solution: The magnitude of the vector $\vec{c}$ is $\frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$ units.

Step 1: Analyzing the Dot Product Constraint

Given the scalar (dot) product: $\vec{a} \cdot \vec{b} = 0$ [Per the definition of the dot product: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ is the angle between the vectors].

This equation holds true if and only if:

• $|\vec{a}| = 0$ (i.e., $\vec{a} = \vec{0}$), or
• $|\vec{b}| = 0$ (i.e., $\vec{b} = \vec{0}$), or
• $\cos \theta = 0 \implies \theta = 90^\circ$ ($\vec{a} \perp \vec{b}$).

Step 2: Analyzing the Cross Product Constraint

Given the vector (cross) product: $\vec{a} \times \vec{b} = \vec{0}$ [Per the definition of the cross product: $\vec{a} \times \vec{b} = (|\vec{a}| |\vec{b}| \sin \theta) \hat{n}$].

This equation holds true if and only if:

• $|\vec{a}| = 0$ (i.e., $\vec{a} = \vec{0}$), or
• $|\vec{b}| = 0$ (i.e., $\vec{b} = \vec{0}$), or
• $\sin \theta = 0 \implies \theta = 0^\circ$ or $180^\circ$ ($\vec{a} \parallel \vec{b}$).

Step 3: Synthesis and Logical Conclusion

We are given that both conditions must be satisfied simultaneously:

If we assume neither $\vec{a}$ nor $\vec{b}$ is a zero vector, we face a contradiction: the vectors cannot be simultaneously perpendicular ($\theta = 90^\circ$) and parallel ($\theta = 0^\circ$ or $180^\circ$).

Therefore, the only mathematically consistent conclusion is that at least one of the vectors must be a zero vector. If $\vec{a} \neq \vec{0}$, then $|\vec{a}| \neq 0$; if $\vec{b} \neq \vec{0}$, then $|\vec{b}| \neq 0$. Since $\sin \theta$ and $\cos \theta$ cannot be zero at the same value of $\theta$, it forces the magnitude of at least one vector to be zero.

Visual Representation of Logical Contradiction

Final Solution: Either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ (or both).

Step 1: Establishing Given Vector Variables

Let the two adjacent sides of the parallelogram be represented by vectors $\vec{a}$ and $\vec{b}$:

• $\vec{a} = 2\hat{i} - 4\hat{j} + 5\hat{k}$
• $\vec{b} = \hat{i} - 2\hat{j} - 3\hat{k}$

Step 2: Calculating the Diagonals

Per the parallelogram law of vector addition, the diagonals $\vec{d_1}$ and $\vec{d_2}$ are given by the sum and difference of the adjacent sides [By Parallelogram Law of Vectors]:

$\vec{d_1} = \vec{a} + \vec{b} = (2+1)\hat{i} + (-4-2)\hat{j} + (5-3)\hat{k} = 3\hat{i} - 6\hat{j} + 2\hat{k}$

$\vec{d_2} = \vec{a} - \vec{b} = (2-1)\hat{i} + (-4-(-2))\hat{j} + (5-(-3))\hat{k} = \hat{i} - 2\hat{j} + 8\hat{k}$

Step 3: Determining Unit Vectors Parallel to Diagonals

The unit vector $\hat{d}$ is defined as $\frac{\vec{d}}{|\vec{d}|}$ [Per the definition of normalized vectors].

For $\vec{d_1}$: $|\vec{d_1}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.

Unit vector parallel to $\vec{d_1}$ is $\hat{n_1} = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}$.

For $\vec{d_2}$: $|\vec{d_2}| = \sqrt{1^2 + (-2)^2 + 8^2} = \sqrt{1 + 4 + 64} = \sqrt{69}$.

Unit vector parallel to $\vec{d_2}$ is $\hat{n_2} = \frac{1}{\sqrt{69}}\hat{i} - \frac{2}{\sqrt{69}}\hat{j} + \frac{8}{\sqrt{69}}\hat{k}$.

Step 4: Calculating the Area of the Parallelogram

The area $A$ of a parallelogram with sides $\vec{a}$ and $\vec{b}$ is $|\vec{a} \times \vec{b}|$ [Per the Geometric Interpretation of Cross Product].

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}$

$= \hat{i}((-4)(-3) - (5)(-2)) - \hat{j}((2)(-3) - (5)(1)) + \hat{k}((2)(-2) - (-4)(1))$

$= \hat{i}(12 + 10) - \hat{j}(-6 - 5) + \hat{k}(-4 + 4) = 22\hat{i} + 11\hat{j} + 0\hat{k}$

Area $A = \sqrt{22^2 + 11^2 + 0^2} = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5} \text{ square units}$.

Visual Representation

Final Solution: The unit vectors parallel to the diagonals are $\frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}$ and $\frac{1}{\sqrt{69}}\hat{i} - \frac{2}{\sqrt{69}}\hat{j} + \frac{8}{\sqrt{69}}\hat{k}$. The area is $11\sqrt{5}$ square units.

Step 1: Establishing Geometric Definitions

Given two vectors $\vec{a}$ and $\vec{b}$ with an angle $\theta$ between them, we utilize the fundamental definitions of the scalar (dot) product and the vector (cross) product.

• The dot product is defined as: $|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta|$ [Per the Definition of Scalar Product].
• The magnitude of the cross product is defined as: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$, where $\theta \in [0, \pi]$ [Per the Definition of Vector Product].

Step 2: Setting up the Equivalence Equation

The problem states the condition: $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$. Substituting the definitions from Step 1:

$|\vec{a}| |\vec{b}| |\cos \theta| = |\vec{a}| |\vec{b}| \sin \theta$

Step 3: Algebraic Simplification

Assuming $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$ (as the angle $\theta$ is undefined for zero vectors), we can divide both sides by the product $|\vec{a}| |\vec{b}|$:

$|\cos \theta| = \sin \theta$

Since $\theta$ is the angle between two vectors, by definition $\theta \in [0, \pi]$. In this interval, $\sin \theta$ is always non-negative ($\sin \theta \geq 0$). Thus:

$\cos \theta = \sin \theta$ or $-\cos \theta = \sin \theta$

Step 4: Solving for $\theta$

Case 1: $\cos \theta = \sin \theta$

Dividing by $\cos \theta$ (given $\cos \theta \neq 0$):

$\tan \theta = 1$

$\theta = \frac{\pi}{4}$ or $45^\circ$

Case 2: $-\cos \theta = \sin \theta$

$\tan \theta = -1$

$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ or $135^\circ$

However, since we are dealing with the absolute value (magnitude) of the dot product, the equation $|\cos \theta| = \sin \theta$ restricts $\theta$ to the first and second quadrants where $\sin \theta$ is positive. Both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ satisfy the equality $|\cos \theta| = \sin \theta$. Specifically, for $\theta = \frac{\pi}{4}$, $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. For $\theta = \frac{3\pi}{4}$, $|\cos \frac{3\pi}{4}| = |-\frac{1}{\sqrt{2}}| = \frac{1}{\sqrt{2}}$ and $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.

Visual Representation of the Vector Relationship

Final Solution: $\theta = \frac{\pi}{4}$

Step 1: Assigning Position Vectors

Let the vertices of the triangle be $A$, $B$, and $C$. We represent these points by their position vectors with respect to the origin:

• $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}$
• $\vec{c} = 3\hat{i} - 4\hat{j} - 4\hat{k}$

Step 2: Calculating Side Vectors

To determine the sides of the triangle, we calculate the displacement vectors $\vec{AB}$, $\vec{BC}$, and $\vec{CA}$ [By the vector definition of displacement, $\vec{PQ} = \vec{pos}(Q) - \vec{pos}(P)$]:

$\vec{AB} = \vec{b} - \vec{a} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$

$\vec{BC} = \vec{c} - \vec{b} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$

$\vec{CA} = \vec{a} - \vec{c} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$

Step 3: Calculating Magnitudes (Squared)

We calculate the squared magnitude of each side to apply the Converse of Pythagoras Theorem [Magnitude $\|\vec{r}\| = \sqrt{x^2 + y^2 + z^2}$]:

Step 4: Verification of Right-Angled Triangle Criterion

For a triangle to be right-angled, the square of the longest side must equal the sum of the squares of the other two sides [Pythagorean Theorem].

Comparing the values: $41 = 6 + 35$.

Since $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$, the triangle satisfies the Pythagorean condition.

Visual Representation of the Geometry

Final Solution: Since $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$, the vectors form the vertices of a right-angled triangle, specifically right-angled at vertex $C$.

Step 1: Definition and Verification of Unit Vectors

A vector $\vec{v}$ is defined as a unit vector if its magnitude $|\vec{v}|$ is equal to $1$ [Per the definition of normalized vectors]. The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $\sqrt{x^2 + y^2 + z^2}$.

Let the three vectors be:

• $\vec{a} = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}$
• $\vec{b} = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}$
• $\vec{c} = \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) = \frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} - \frac{3}{7}\hat{k}$

Calculating magnitudes:

• $|\vec{a}| = \sqrt{(\frac{2}{7})^2 + (\frac{3}{7})^2 + (\frac{6}{7})^2} = \sqrt{\frac{4+9+36}{49}} = \sqrt{\frac{49}{49}} = 1$
• $|\vec{b}| = \sqrt{(\frac{3}{7})^2 + (-\frac{6}{7})^2 + (\frac{2}{7})^2} = \sqrt{\frac{9+36+4}{49}} = \sqrt{\frac{49}{49}} = 1$
• $|\vec{c}| = \sqrt{(\frac{6}{7})^2 + (\frac{2}{7})^2 + (-\frac{3}{7})^2} = \sqrt{\frac{36+4+9}{49}} = \sqrt{\frac{49}{49}} = 1$

Since $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$, all three are unit vectors.

Step 2: Verification of Mutual Perpendicularity

Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product $\vec{u} \cdot \vec{v} = 0$ [By the Orthogonality Criterion].

Step 3: Analytical Conclusion

The dot products of the vectors taken in pairs are zero, implying they are mutually orthogonal. Combined with the unit magnitude confirmed in Step 1, these vectors form an orthonormal basis in $\mathbb{R}^3$.

Final Solution: The vectors are proven to be unit vectors (magnitude = 1) and mutually perpendicular (dot product = 0).

Step 1: Theoretical Foundation of Magnitude

The magnitude of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is defined by the Euclidean norm [Per the definition of the magnitude of a vector in 3D space]:

$|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$

To construct two different vectors with the same magnitude, we must select components $(x, y, z)$ such that the sum of the squares of the components remains constant while at least one component differs between the two vectors.

Step 2: Defining Vector $\vec{a}$

Let us define vector $\vec{a}$ by assigning arbitrary values to its components. For simplicity, let:

$\vec{a} = 1\hat{i} + 2\hat{j} + 2\hat{k}$

Calculating the magnitude $|\vec{a}|$:

$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ units

Step 3: Defining Vector $\vec{b}$ (Permutation Method)

To ensure $\vec{b} \neq \vec{a}$, we can permute the components of vector $\vec{a}$. Let:

$\vec{b} = 2\hat{i} + 1\hat{j} + 2\hat{k}$

Calculating the magnitude $|\vec{b}|$:

$|\vec{b}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ units

[Justification: Since the addition of squares is commutative, swapping components preserves the resultant magnitude.]

Visual Representation of Vectors in $\mathbb{R}^3$

Step 4: Comparative Analysis

Since $\vec{a} \neq \vec{b}$ (as the ordered triplets are distinct) and $|\vec{a}| = |\vec{b}| = 3$, these satisfy the problem criteria.

Final Solution: Two such vectors are $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$, both having a magnitude of $3$ units.

Given Vectors and Algebraic Expansion

We are required to prove the vector identity: $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$.

Let $\vec{a}$ and $\vec{b}$ be two arbitrary vectors in $\mathbb{R}^3$. We apply the distributive property of the cross product over vector addition and subtraction [Per the Distributive Law of Cross Product over Addition: $\vec{A} \times (\vec{B} + \vec{C}) = (\vec{A} \times \vec{B}) + (\vec{A} \times \vec{C})$].

Step 1: Expanding the Cross Product

Expanding the left-hand side (LHS):

$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times (\vec{a} + \vec{b}) - \vec{b} \times (\vec{a} + \vec{b})$

Applying the distributive property again to both terms:

$= (\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b})$

Step 2: Applying Vector Identity Properties

We utilize two fundamental properties of the cross product:

• Property 1 (Self-Cross Product): The cross product of any vector with itself is the zero vector, i.e., $\vec{v} \times \vec{v} = \vec{0}$ [Because $\sin(0^\circ) = 0$]. Thus, $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$.
• Property 2 (Anti-Commutative Property): The cross product is anti-commutative, i.e., $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$.

Step 3: Simplification

Substituting these properties into the expanded expression:

$= \vec{0} + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{a}) - \vec{0}$

Substituting $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:

$= (\vec{a} \times \vec{b}) - (-(\vec{a} \times \vec{b}))$

$= (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b})$

$= 2(\vec{a} \times \vec{b})$

Visual Representation of Vector Cross Product

Final Solution: $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$, which confirms the original identity.

Step 1: Establishing Position Vectors

Given the points $A(1, -2, -8)$, $B(5, 0, -2)$, and $C(11, 3, 7)$, we represent them as position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ relative to the origin $O(0, 0, 0)$:

• $\vec{OA} = \hat{i} - 2\hat{j} - 8\hat{k}$
• $\vec{OB} = 5\hat{i} + 0\hat{j} - 2\hat{k}$
• $\vec{OC} = 11\hat{i} + 3\hat{j} + 7\hat{k}$

Step 2: Calculating Direction Vectors

To prove collinearity, we determine the vectors $\vec{AB}$ and $\vec{BC}$ [By the definition of collinear vectors, two vectors are collinear if they are scalar multiples of each other].

$\vec{AB} = \vec{OB} - \vec{OA} = (5-1)\hat{i} + (0 - (-2))\hat{j} + (-2 - (-8))\hat{k} = 4\hat{i} + 2\hat{j} + 6\hat{k}$

$\vec{BC} = \vec{OC} - \vec{OB} = (11-5)\hat{i} + (3-0)\hat{j} + (7 - (-2))\hat{k} = 6\hat{i} + 3\hat{j} + 9\hat{k}$

Step 3: Proving Collinearity

We analyze the relationship between $\vec{AB}$ and $\vec{BC}$:

$\vec{BC} = 6\hat{i} + 3\hat{j} + 9\hat{k} = \frac{3}{2}(4\hat{i} + 2\hat{j} + 6\hat{k}) = \frac{3}{2}\vec{AB}$

Since $\vec{BC} = k\vec{AB}$ where $k = 1.5$, the vectors $\vec{AB}$ and $\vec{BC}$ are parallel. Furthermore, they share the common point $B$. Therefore, $A, B, C$ are collinear.

Step 4: Determining the Ratio of Division

Let point $B$ divide $AC$ in the ratio $k:1$. Using the Section Formula in vector form [Per the Section Formula for internal division, $\vec{b} = \frac{m\vec{c} + n\vec{a}}{m+n}$]:

$\vec{b} = \frac{k\vec{c} + 1\vec{a}}{k+1}$

Substituting the components for the $x$-coordinate:

$5 = \frac{k(11) + 1(1)}{k+1}$

$5(k+1) = 11k + 1$

$5k + 5 = 11k + 1 \implies 6k = 4 \implies k = \frac{4}{6} = \frac{2}{3}$

Step 5: Verification of the Ratio

Applying the ratio to the $y$-coordinates to ensure consistency:

$y = \frac{2(3) + 3(-2)}{2+3} = \frac{6-6}{5} = 0$. (Matches $B_y = 0$).

Applying the ratio to the $z$-coordinates:

$z = \frac{2(7) + 3(-8)}{2+3} = \frac{14-24}{5} = \frac{-10}{5} = -2$. (Matches $B_z = -2$).

Final Solution: The points A, B, and C are collinear, and point B divides the line segment AC in the ratio 2:3 internally.

Step 1: Theoretical Definition of Coinitial Vectors

In vector algebra, two or more vectors are classified as coinitial if they originate from the same initial point [Definition: A set of vectors $\{ \vec{a}, \vec{b}, \vec{c}, ... \}$ are coinitial if there exists a point $P$ such that $P$ is the tail of all vectors in the set].

Step 2: Analysis of the Geometric Figure

Given a square $ABCD$ with vertices mapped in a 2D Cartesian plane, we identify the vectors emanating from each vertex. Based on standard conventions for a square $ABCD$ labeled in counter-clockwise order:

Step 3: Identification of Sets of Coinitial Vectors

By inspecting the vertices where two vectors share a common starting point:

Step 4: Logical Conclusion

Since the problem asks to identify the coinitial vectors in the square, we provide the sets emanating from the vertices. These pairs satisfy the requirement of sharing a common origin [Per the properties of vectors in a geometric polygon].

Final Solution: The pairs of coinitial vectors are $(\vec{AB}, \vec{AD})$, $(\vec{BA}, \vec{BC})$, $(\vec{CB}, \vec{CD})$, and $(\vec{DA}, \vec{DC})$.

Step 1: Theoretical Foundation of Direction in Vectors

A vector in 3D space is defined by its components $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$. The direction of a vector is uniquely determined by its unit vector, denoted as $\hat{a}$. The unit vector is calculated as: $$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{x}{\sqrt{x^2+y^2+z^2}}\hat{i} + \frac{y}{\sqrt{x^2+y^2+z^2}}\hat{j} + \frac{z}{\sqrt{x^2+y^2+z^2}}\hat{k}$$ [Per the Definition of Directional Vectors].

Step 2: Condition for Same Direction

Two vectors $\vec{a}$ and $\vec{b}$ have the same direction if and only if there exists a positive scalar $k > 0$ such that $\vec{b} = k\vec{a}$.
If $k > 0$, the unit vectors are identical: $\hat{b} = \frac{k\vec{a}}{|k\vec{a}|} = \frac{k\vec{a}}{k|\vec{a}|} = \hat{a}$. [By the property of Scalar Multiplication of Vectors].

Step 3: Constructing Two Different Vectors

Let us choose a reference vector $\vec{a} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

• Vector 1 ($\vec{A}$): Set $k=1$. Thus, $\vec{A} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.
• Vector 2 ($\vec{B}$): Set $k=2$. Thus, $\vec{B} = 2(1\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$.

Since $\vec{B} = 2\vec{A}$ and $2 > 0$, the vectors $\vec{A}$ and $\vec{B}$ are different in magnitude but share the same direction.

Visual Representation

Summary Table of Vectors

Final Solution: Two such vectors are $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + 4\hat{j} + 6\hat{k}$, which share the same direction as they are positive scalar multiples of each other.