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: 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: 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: Establishing Given Vectors and Variables

Let the position vectors of points $P$ and $Q$ be represented by $\vec{p}$ and $\vec{q}$ respectively:

• $\vec{p} = 2\vec{a} + \vec{b}$
• $\vec{q} = \vec{a} - 3\vec{b}$

The ratio of division is given as $m:n = 1:2$. Since the division is external, we utilize the external division formula for position vectors [By Section Formula for External Division]:

$\vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n}$

Step 2: Calculating the Position Vector of Point R

Substitute the values of $m, n, \vec{p},$ and $\vec{q}$ into the formula:

$\vec{r} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2}$

Perform the algebraic expansion [By Distributive Property]:

$\vec{r} = \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1}$

Combine like terms:

$\vec{r} = \frac{-3\vec{a} - 5\vec{b}}{-1}$

$\vec{r} = 3\vec{a} + 5\vec{b}$

Step 3: Proving P is the Midpoint of RQ

For $P$ to be the midpoint of the line segment $RQ$, the position vector of $P$ must satisfy the midpoint condition: $\vec{p} = \frac{\vec{r} + \vec{q}}{2}$ [By Midpoint Theorem].

Calculate the average of vectors $\vec{r}$ and $\vec{q}$:

$\frac{\vec{r} + \vec{q}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2}$

Simplify the numerator:

$\frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b}$

Since this result equals $\vec{p}$, $P$ is confirmed to be the midpoint of segment $RQ$.

Visual Representation of Points

Final Solution: The position vector of point R is $3\vec{a} + 5\vec{b}$. Since $\frac{\vec{r} + \vec{q}}{2} = \vec{p}$, it is verified that P is the midpoint of segment RQ.

Step 1: Formal Definition of Given Parameters

Let the magnitudes of vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ be equal to $k$, where $k > 0$. Thus:

$|\vec{a}| = |\vec{b}| = |\vec{c}| = k$ [Given magnitude equality]

Since the vectors are mutually perpendicular, the dot product of any pair of these vectors is zero:

$\vec{a} \cdot \vec{b} = 0$, $\vec{b} \cdot \vec{c} = 0$, $\vec{c} \cdot \vec{a} = 0$ [Definition of orthogonal vectors]

Step 2: Determining the Magnitude of the Resultant Vector

Let $\vec{r} = \vec{a} + \vec{b} + \vec{c}$. The magnitude of $\vec{r}$ is calculated using the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$:

$|\vec{r}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})$

$|\vec{r}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$

Substituting $|\vec{v}|^2 = |\vec{v}|^2$ and the zero dot products:

$|\vec{r}|^2 = k^2 + k^2 + k^2 + 2(0 + 0 + 0) = 3k^2$

$|\vec{r}| = \sqrt{3k^2} = k\sqrt{3}$

Step 3: Calculating Inclination Angles

Let $\theta_1, \theta_2, \theta_3$ be the angles that vector $\vec{r}$ makes with $\vec{a}, \vec{b}$, and $\vec{c}$ respectively. By the definition of the dot product $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos \theta$:

$\cos \theta_1 = \frac{\vec{r} \cdot \vec{a}}{|\vec{r}| |\vec{a}|} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{r}| k}$

Expanding the dot product in the numerator:

$\vec{r} \cdot \vec{a} = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a} = k^2 + 0 + 0 = k^2$

Substituting back:

$\cos \theta_1 = \frac{k^2}{(k\sqrt{3})(k)} = \frac{1}{\sqrt{3}}$

Step 4: Symmetry Analysis

By symmetric logic applied to $\vec{b}$ and $\vec{c}$:

$\cos \theta_2 = \frac{\vec{r} \cdot \vec{b}}{|\vec{r}| |\vec{b}|} = \frac{k^2}{(k\sqrt{3})(k)} = \frac{1}{\sqrt{3}}$

$\cos \theta_3 = \frac{\vec{r} \cdot \vec{c}}{|\vec{r}| |\vec{c}|} = \frac{k^2}{(k\sqrt{3})(k)} = \frac{1}{\sqrt{3}}$

Since $\cos \theta_1 = \cos \theta_2 = \cos \theta_3 = \frac{1}{\sqrt{3}}$, it follows that $\theta_1 = \theta_2 = \theta_3 = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

Final Solution: Since the cosines of the angles between the resultant vector and each component vector are identical ($\frac{1}{\sqrt{3}}$), the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b},$ and $\vec{c}$.

Step 1: Defining the Given Vectors

Let the given vectors be defined as follows:

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

Step 2: Calculating the Sum Vector

Let $\vec{s}$ be the sum of vectors $\vec{b}$ and $\vec{c}$ [Vector Addition Law].

$\vec{s} = \vec{b} + \vec{c}$

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

$\vec{s} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}$

Step 3: Determining the Unit Vector along $\vec{s}$

The unit vector $\hat{s}$ is defined as $\frac{\vec{s}}{|\vec{s}|}$ [Definition of a Unit Vector].

$|\vec{s}| = \sqrt{(2 + \lambda)^2 + (6)^2 + (-2)^2}$

$|\vec{s}| = \sqrt{4 + 4\lambda + \lambda^2 + 36 + 4} = \sqrt{\lambda^2 + 4\lambda + 44}$

$\hat{s} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}}$

Step 4: Formulating the Scalar Product Equation

The problem states that the scalar product of $\vec{a}$ and $\hat{s}$ is equal to $1$.

$\vec{a} \cdot \hat{s} = 1$

$(\hat{i} + \hat{j} + \hat{k}) \cdot \left[ \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}} \right] = 1$

Step 5: Algebraic Manipulation and Solving for $\lambda$

Perform the dot product of the components [Properties of the Scalar Product]:

$\frac{1(2 + \lambda) + 1(6) + 1(-2)}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1$

$\frac{2 + \lambda + 6 - 2}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1$

$\frac{\lambda + 6}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1$

$\lambda + 6 = \sqrt{\lambda^2 + 4\lambda + 44}$

Square both sides to eliminate the radical:

$(\lambda + 6)^2 = \lambda^2 + 4\lambda + 44$

$\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44$

Subtract $\lambda^2$ from both sides and simplify:

$12\lambda - 4\lambda = 44 - 36$

$8\lambda = 8$

$\lambda = 1$

Final Solution: \lambda = 1

Step 1: Theoretical Foundation of a Unit Vector in 2D

In the Cartesian plane, any unit vector $\hat{r}$ can be expressed in terms of its angle $\theta$ with the positive $x$-axis using trigonometric components. Given that the vector lies in the $XY$-plane, its $z$-component is $0$.

The vector components are defined as:

• $x = \cos(\theta)$
• $y = \sin(\theta)$

Step 2: Substitution of Given Parameters

Given $\theta = 30^{\circ}$ (or $\frac{\pi}{6}$ radians), we substitute the value into the coordinate expression:
$\hat{r} = \cos(30^{\circ})\hat{i} + \sin(30^{\circ})\hat{j}$

Step 3: Evaluation of Trigonometric Ratios

Using standard trigonometric values:

• $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$
• $\sin(30^{\circ}) = \frac{1}{2}$

Step 4: Construction of the Final Vector

Substituting the values into the component form:
$\hat{r} = \left( \frac{\sqrt{3}}{2} \right)\hat{i} + \left( \frac{1}{2} \right)\hat{j}$

Step 5: Visual Representation

Verification of Magnitude

To ensure this is a unit vector, we calculate its magnitude $|\hat{r}|$:
$|\hat{r}| = \sqrt{(\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2}$
$|\hat{r}| = \sqrt{\frac{3}{4} + \frac{1}{4}}$
$|\hat{r}| = \sqrt{1} = 1$ [By the Pythagorean Theorem for vector magnitude].

Final Solution: The unit vector is $\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$

Step 1: Defining the Given Vector Components

The vectors are provided in Cartesian coordinate form:

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

Step 2: Performing Linear Combination Calculation

Let the resultant vector be $\vec{v} = 2\vec{a} - \vec{b} + 3\vec{c}$. We substitute the components into the expression [By the Property of Scalar Multiplication and Vector Addition]:

$\vec{v} = 2(\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + 3(\hat{i} - 2\hat{j} + \hat{k})$

Distributing the scalar factors:

$\vec{v} = (2\hat{i} + 2\hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k})$

Grouping the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$:

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

$\vec{v} = 3\hat{i} - 3\hat{j} + 2\hat{k}$

Step 3: Calculating the Magnitude of $\vec{v}$

The magnitude $|\vec{v}|$ is determined using the Euclidean norm formula $|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$ [By the Definition of Magnitude in $\mathbb{R}^3$]:

$|\vec{v}| = \sqrt{(3)^2 + (-3)^2 + (2)^2}$

$|\vec{v}| = \sqrt{9 + 9 + 4} = \sqrt{22}$

Step 4: Deriving the Unit Vector

A unit vector $\hat{u}$ in the direction of vector $\vec{v}$ is defined as $\hat{u} = \frac{\vec{v}}{|\vec{v}|}$ [By the Definition of Normalized Vectors]:

$\hat{u} = \frac{3\hat{i} - 3\hat{j} + 2\hat{k}}{\sqrt{22}}$

$\hat{u} = \frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}$

Final Solution: The unit vector parallel to $2\vec{a} - \vec{b} + 3\vec{c}$ is $\frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}$.

Step 1: Establishing the Position Vectors

To determine the vector joining two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in a three-dimensional Cartesian coordinate system, we first define the position vectors $\vec{OP}$ and $\vec{OQ}$ with respect to the origin $O(0, 0, 0)$ [Per the definition of position vectors in $\mathbb{R}^3$]:

$\vec{OP} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$

$\vec{OQ} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$

Step 2: Deriving the Vector $\vec{PQ}$

By the Triangle Law of Vector Addition, for the triangle $\triangle OPQ$, the vector joining $P$ to $Q$ is given by the difference of the position vectors [Ref: Vector displacement principle]:

$\vec{PQ} = \vec{OQ} - \vec{OP}$

Substituting the coordinate expressions:

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

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

Step 3: Identifying Scalar Components

The scalar components of a vector $\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}$ are the coefficients $a, b,$ and $c$. Thus, for vector $\vec{PQ}$:

Step 4: Calculating the Magnitude

The magnitude of a vector $\vec{PQ} = a\hat{i} + b\hat{j} + c\hat{k}$ is defined as the Euclidean norm $||\vec{PQ}|| = \sqrt{a^2 + b^2 + c^2}$ [By the Pythagorean Theorem extended to 3D space].

$||\vec{PQ}|| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Final Solution: The scalar components are $(x_2 - x_1), (y_2 - y_1),$ and $(z_2 - z_1)$. The magnitude is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

Step 1: Evaluation of Fundamental Triple Products

To evaluate the expression $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$, we utilize the properties of the cross product of unit vectors in a right-handed Cartesian coordinate system. By definition of the right-hand rule for orthonormal basis vectors:

• $\hat{i} \times \hat{j} = \hat{k}$
• $\hat{j} \times \hat{k} = \hat{i}$
• $\hat{k} \times \hat{i} = \hat{j}$

[Per the cyclic property of orthonormal cross products: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, and $\hat{k} \times \hat{i} = \hat{j}$].

Step 2: Analysis of Anti-Cyclic Cross Products

We must also account for the anti-commutative property of the cross product ($ \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) $):

• $\hat{i} \times \hat{k} = -(\hat{k} \times \hat{i}) = -\hat{j}$

Step 3: Substitution and Scalar Dot Product Calculation

We substitute these values into the given expression $E = \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$:

$E = \hat{i} \cdot (\hat{i}) + \hat{j} \cdot (-\hat{j}) + \hat{k} \cdot (\hat{k})$

[By the definition of the scalar dot product where $\hat{n} \cdot \hat{n} = |\hat{n}|^2 = 1$]:

$E = (1) + (-1) + (1)$

Step 4: Geometric Visualization

Step 5: Final Arithmetic Summation

$E = 1 - 1 + 1$

$E = 1$

Final Solution: 1

Step 1: Determining the Direction of Vector $\vec{d}$

Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, the vector $\vec{d}$ must be parallel to the cross product of $\vec{a}$ and $\vec{b}$. Let $\vec{n} = \vec{a} \times \vec{b}$. [Per the property that the cross product of two vectors produces a vector orthogonal to the plane containing them].

Given: $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$ $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix}$

Expanding the determinant: $\vec{n} = \hat{i}(4 \cdot 7 - 2 \cdot -2) - \hat{j}(1 \cdot 7 - 2 \cdot 3) + \hat{k}(1 \cdot -2 - 4 \cdot 3)$ $\vec{n} = \hat{i}(28 + 4) - \hat{j}(7 - 6) + \hat{k}(-2 - 12)$ $\vec{n} = 32\hat{i} - \hat{j} - 14\hat{k}$

Step 2: Expressing $\vec{d}$ in Scalar Form

Since $\vec{d}$ is parallel to $\vec{n}$, we can define $\vec{d} = \lambda \vec{n}$, where $\lambda$ is a scalar constant. [By definition of collinear vectors].

$\vec{d} = \lambda(32\hat{i} - \hat{j} - 14\hat{k})$

Step 3: Solving for $\lambda$ using the Scalar Product Constraint

We are given the condition $\vec{c} \cdot \vec{d} = 15$. $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$

$(2\hat{i} - \hat{j} + 4\hat{k}) \cdot \lambda(32\hat{i} - \hat{j} - 14\hat{k}) = 15$ $\lambda [ (2)(32) + (-1)(-1) + (4)(-14) ] = 15$ [By the distributive property of dot product over addition]

$\lambda [ 64 + 1 - 56 ] = 15$ $\lambda [ 9 ] = 15$ $\lambda = \frac{15}{9} = \frac{5}{3}$

Step 4: Final Calculation of Vector $\vec{d}$

Substitute $\lambda$ back into the expression for $\vec{d}$:

$\vec{d} = \frac{5}{3}(32\hat{i} - \hat{j} - 14\hat{k})$ $\vec{d} = \frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}$

Final Solution: $\vec{d} = \frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}$

Step 1: Establishing the Relation for Direction Cosines

Let the vector $\vec{r}$ be equally inclined to the coordinate axes $OX$, $OY$, and $OZ$. Let the angles made by the vector with these axes be $\alpha$, $\beta$, and $\gamma$ respectively.

Given the condition of equal inclination:

$\alpha = \beta = \gamma$

The direction cosines of a vector are defined as $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. [Per the definition of direction cosines in 3D space].

Since $\alpha = \beta = \gamma$, it follows that:

$l = m = n$

Step 2: Applying the Fundamental Identity

For any vector in three-dimensional space, the sum of the squares of its direction cosines is always unity [Fundamental identity for direction cosines]:

$l^2 + m^2 + n^2 = 1$

Substituting $l = m = n$ into the identity:

$l^2 + l^2 + l^2 = 1$

$3l^2 = 1$

Step 3: Solving for the Direction Cosines

Solving the algebraic equation derived in Step 2:

$l^2 = \frac{1}{3}$

$l = \pm \frac{1}{\sqrt{3}}$

Since $l = m = n$, the values for $m$ and $n$ are identical to $l$.

Therefore, the direction cosines are $\left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \right)$.

Visual Representation of a Unit Vector along the Diagonal

Analytical Conclusion

The calculation confirms that there are two possible sets of direction cosines depending on the direction of the vector along the line equally inclined to the axes. The result holds as $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.

Final Solution: The direction cosines are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.

Step 1: Expanding the Scalar Product

Given the expression $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$, we apply the distributive property of the dot product over vector addition [Distributive Law of Dot Product: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$]:

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

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

Step 2: Applying Algebraic Identities for Vectors

Utilizing the property that $\vec{u} \cdot \vec{u} = |\vec{u}|^2$ and the commutativity of the dot product [Commutative Law: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$], we simplify the expression:

$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$

Step 3: Equating to the Given Condition

The problem states the condition $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$. Substituting our expansion into this equation:

$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2$

Subtracting $|\vec{a}|^2 + |\vec{b}|^2$ from both sides, we obtain:

$2(\vec{a} \cdot \vec{b}) = 0$

$\vec{a} \cdot \vec{b} = 0$

Step 4: Interpretation of Orthogonality

By 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. Since $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$, the condition $\vec{a} \cdot \vec{b} = 0$ implies $\cos(\theta) = 0$, meaning $\theta = 90^\circ$ or $\frac{\pi}{2}$ radians.

Step 5: Logical Conclusion

The sequence of operations confirms that the equality holds if and only if the scalar product of the vectors is zero, which is the necessary and sufficient condition for two non-zero vectors to be perpendicular.

Final Solution: The equality holds if and only if $\vec{a} \perp \vec{b}$, as implied by the vanishing of the scalar product term $2(\vec{a} \cdot \vec{b}) = 0$.

Step 1: Theoretical Foundation

A vector $\vec{a}$ is defined as a unit vector if and only if its magnitude (norm) is equal to unity ($1$). Mathematically, for a vector $\vec{v}$, $\vec{v}$ is a unit vector if $|\vec{v}| = 1$ [Definition of Unit Vector].

Step 2: Defining the Given Vector

Let the given vector be $\vec{v} = x(\hat{i} + \hat{j} + \hat{k})$. Distributing the scalar $x$ across the components, we get:

$\vec{v} = x\hat{i} + x\hat{j} + x\hat{k}$

Step 3: Calculating the Magnitude

The magnitude of a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$ [Magnitude of a Vector in 3D Space].

Substituting the components $a=x, b=x, c=x$:

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

$|\vec{v}| = \sqrt{3x^2}$

$|\vec{v}| = |x|\sqrt{3}$

Step 4: Solving for $x$

Since the vector is a unit vector, we set the magnitude equal to $1$:

$|x|\sqrt{3} = 1$

$|x| = \frac{1}{\sqrt{3}}$

Removing the absolute value, we obtain the two possible values for $x$:

$x = \pm \frac{1}{\sqrt{3}}$

Visual Representation of Vector Components

Step 5: Verification of Result

Final Solution: $x = \pm \frac{1}{\sqrt{3}}$

Step 1: Calculating the Resultant Vector

To find the resultant vector $\vec{r}$, we perform the vector addition of $\vec{a}$ and $\vec{b}$ by summing their respective components along the $\hat{i}$, $\hat{j}$, and $\hat{k}$ axes [By the Parallelogram Law of Vector Addition].

Given:

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

Summation:

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

$\vec{r} = 3\hat{i} + 1\hat{j} + 0\hat{k}$

$\vec{r} = 3\hat{i} + \hat{j}$

Step 2: Determining the Magnitude of the Resultant Vector

The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is defined as $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.

$|\vec{r}| = \sqrt{(3)^2 + (1)^2 + (0)^2}$

$|\vec{r}| = \sqrt{9 + 1 + 0} = \sqrt{10}$

Step 3: Finding the Unit Vector in the Direction of $\vec{r}$

The unit vector $\hat{r}$ in the direction of $\vec{r}$ is calculated by dividing the vector by its magnitude [By definition of normalized vectors]:

$\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{3\hat{i} + \hat{j}}{\sqrt{10}}$

$\hat{r} = \frac{3}{\sqrt{10}}\hat{i} + \frac{1}{\sqrt{10}}\hat{j}$

Step 4: Scaling to a Magnitude of 5 Units

To obtain a vector $\vec{v}$ of magnitude $k=5$ parallel to $\vec{r}$, we multiply the unit vector $\hat{r}$ by the desired magnitude:

$\vec{v} = 5 \cdot \hat{r} = 5 \left( \frac{3}{\sqrt{10}}\hat{i} + \frac{1}{\sqrt{10}}\hat{j} \right)$

Simplifying the coefficients:

$\vec{v} = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j}$

Rationalizing the denominator ($\frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}$):

$\vec{v} = \frac{15\sqrt{10}}{10}\hat{i} + \frac{5\sqrt{10}}{10}\hat{j} = \frac{3\sqrt{10}}{2}\hat{i} + \frac{\sqrt{10}}{2}\hat{j}$

Final Solution: The vector is $\frac{3\sqrt{10}}{2}\hat{i} + \frac{\sqrt{10}}{2}\hat{j}$ (or $\pm \frac{15}{\sqrt{10}}\hat{i} \pm \frac{5}{\sqrt{10}}\hat{j}$ depending on direction orientation).

Step 1: Establishing the Coordinate System

To analyze the displacement, we define a Cartesian coordinate system where the origin $O(0, 0)$ represents the initial point of departure. Let the positive $x$-axis represent the East direction and the positive $y$-axis represent the North direction.

• The unit vector along the East is $\hat{i}$.
• The unit vector along the North is $\hat{j}$.
• The unit vector along the West is $-\hat{i}$.

Step 2: Vector Representation of Displacements

Vector $\vec{d_1}$ (First displacement): The girl walks 4 km towards the West.

$\vec{d_1} = -4\hat{i} + 0\hat{j}$ [Per Cartesian mapping of cardinal directions]

Vector $\vec{d_2}$ (Second displacement): The girl walks 3 km at an angle of $30^{\circ}$ East of North.

In this orientation, "North" corresponds to the positive $y$-axis ($90^{\circ}$). Moving $30^{\circ}$ East of North shifts the angle towards the positive $x$-axis. Thus, the angle with the positive $y$-axis is $30^{\circ}$, making the angle with the positive $x$-axis $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.

Using trigonometric resolution: $\vec{d_2} = 3(\cos 60^{\circ})\hat{i} + 3(\sin 60^{\circ})\hat{j}$

$\vec{d_2} = 3(\frac{1}{2})\hat{i} + 3(\frac{\sqrt{3}}{2})\hat{j} = 1.5\hat{i} + 1.5\sqrt{3}\hat{j}$

Step 3: Calculation of Net Displacement

The total displacement $\vec{R}$ is the vector sum of $\vec{d_1}$ and $\vec{d_2}$ [By Triangle Law of Vector Addition].

$\vec{R} = \vec{d_1} + \vec{d_2}$

$\vec{R} = (-4\hat{i} + 0\hat{j}) + (1.5\hat{i} + 1.5\sqrt{3}\hat{j})$

$\vec{R} = (-4 + 1.5)\hat{i} + (1.5\sqrt{3})\hat{j}$

$\vec{R} = -2.5\hat{i} + 1.5\sqrt{3}\hat{j}$

Step 4: Magnitude and Direction Analysis

The magnitude of displacement $|\vec{R}|$ is given by $\sqrt{x^2 + y^2}$.

$|\vec{R}| = \sqrt{(-2.5)^2 + (1.5\sqrt{3})^2}$

$|\vec{R}| = \sqrt{6.25 + (2.25 \times 3)} = \sqrt{6.25 + 6.75} = \sqrt{13}$

$|\vec{R}| \approx 3.605 \text{ km}$

Final Solution: The girl's displacement from the initial point is $\sqrt{13} \text{ km}$ at an angle of $\tan^{-1}\left(\frac{1.5\sqrt{3}}{-2.5}\right)$ relative to the West axis.

Given Variables & Initial Setup

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. By definition of unit vectors:

• $|\vec{a}| = 1$ [Definition of unit vector]
• $|\vec{b}| = 1$ [Definition of unit vector]
• The angle between $\vec{a}$ and $\vec{b}$ is $\theta$. Thus, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = \cos \theta$ [By the definition of dot product].

Step 1: Establishing the Condition for the Resultant

We are given that $\vec{c} = \vec{a} + \vec{b}$ is also a unit vector. Therefore:

$|\vec{a} + \vec{b}| = 1$

Squaring both sides to eliminate the square root typically found in the magnitude formula:

$|\vec{a} + \vec{b}|^2 = 1^2$

$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1$ [By the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$]

Step 2: Algebraic Expansion and Substitution

Expanding the dot product using the distributive property:

$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = 1$

$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 1$ [Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ by commutativity of dot product]

Substitute the known magnitudes ($|\vec{a}| = 1, |\vec{b}| = 1$) and the dot product ($\vec{a} \cdot \vec{b} = \cos \theta$):

$(1)^2 + 2 \cos \theta + (1)^2 = 1$

Step 3: Solving for $\theta$

$1 + 2 \cos \theta + 1 = 1$

$2 + 2 \cos \theta = 1$

$2 \cos \theta = 1 - 2$

$2 \cos \theta = -1$

$\cos \theta = -\frac{1}{2}$

Since $\theta$ is the angle between two vectors, $0 \le \theta \le \pi$. The value of $\theta$ for which $\cos \theta = -\frac{1}{2}$ is:

$\theta = \frac{2\pi}{3}$ or $120^\circ$

Geometric Representation

Final Solution: $\theta = \frac{2\pi}{3}$

Step 1: Analyzing the Proposition

The question asks whether the magnitude of the sum of two vectors, $\vec{a} = \vec{b} + \vec{c}$, is necessarily equal to the sum of the magnitudes of the individual vectors, $|\vec{a}| = |\vec{b}| + |\vec{c}|$. [Per the Triangle Inequality for vectors, $|\vec{b} + \vec{c}| \leq |\vec{b}| + |\vec{c}|$.]

Step 2: Geometrical Interpretation

Geometrically, $\vec{a}$, $\vec{b}$, and $\vec{c}$ form the sides of a triangle (or a degenerate triangle). The condition $|\vec{b} + \vec{c}| = |\vec{b}| + |\vec{c}|$ holds if and only if vectors $\vec{b}$ and $\vec{c}$ are collinear and point in the same direction.

Step 3: Counter-Example Construction

To disprove the general statement, we choose non-collinear vectors $\vec{b}$ and $\vec{c}$.

Let $\vec{b} = \hat{i}$ and $\vec{c} = \hat{j}$.

Then $\vec{a} = \vec{b} + \vec{c} = \hat{i} + \hat{j}$.

Step 4: Magnitude Calculations

Calculate $|\vec{a}|$, $|\vec{b}|$, and $|\vec{c}|$:

• $|\vec{b}| = \sqrt{1^2 + 0^2} = 1$
• $|\vec{c}| = \sqrt{0^2 + 1^2} = 1$
• $|\vec{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}$

Step 5: Verification of Inequality

Substitute these values into the expression $|\vec{b}| + |\vec{c}|$:

$|\vec{b}| + |\vec{c}| = 1 + 1 = 2$

Since $\sqrt{2} \approx 1.414$ and $1.414 \neq 2$, it is evident that:

$|\vec{a}| \neq |\vec{b}| + |\vec{c}|$

Step 6: Logical Justification

The equality $|\vec{a}| = |\vec{b}| + |\vec{c}|$ only holds when the angle $\theta$ between $\vec{b}$ and $\vec{c}$ is $0^\circ$. Given the definition of vector addition in $\mathbb{R}^3$, the magnitude is defined as:

$|\vec{b} + \vec{c}| = \sqrt{|\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos\theta}$

For the equation $|\vec{a}| = |\vec{b}| + |\vec{c}|$ to be true, we require $\cos\theta = 1$, which implies $\theta = 0$. Since this is not true for all vectors, the statement is false in the general case.

Final Solution: No, it is not true in general. It is only true when $\vec{b}$ and $\vec{c}$ are collinear and in the same direction.