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

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.

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: 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.

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.

Step 1: Coordinate Identification & Vector Representation

Given the initial point $A(x_1, y_1) = (2, 1)$ and the terminal point $B(x_2, y_2) = (-5, 7)$, the vector $\vec{AB}$ is defined as the displacement vector from $A$ to $B$. [Per the definition of a position vector in $\mathbb{R}^2$].

The vector $\vec{AB}$ is calculated as:

Substituting the given values:

Step 2: Determining Scalar Components

The scalar components of a vector $\vec{r} = a\hat{i} + b\hat{j}$ are the coefficients $a$ and $b$. These represent the signed magnitudes of the vector along the coordinate axes.

• Scalar component along the $x$-axis ($a$): $-7$
• Scalar component along the $y$-axis ($b$): $6$

Step 3: Determining Vector Components

The vector components are the individual components of the vector including their respective unit vectors. [By the Principle of Vector Decomposition].

• Vector component along the $x$-axis: $-7\hat{i}$
• Vector component along the $y$-axis: $6\hat{j}$

Step 4: Geometric Visualization

Summary of Data

Final Solution: The scalar components are $-7$ and $6$, and the vector components are $-7\hat{i}$ and $6\hat{j}$.

Given Variables & Initial Setup

Let the points be $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in a three-dimensional Cartesian coordinate system. The given coordinates are:

• $P(2, 3, 4) \implies \vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
• $Q(4, 1, -2) \implies \vec{q} = 4\hat{i} + 1\hat{j} - 2\hat{k}$

We are required to find the position vector $\vec{r}$ of the midpoint $R$ of the line segment $PQ$. [By the Midpoint Formula in Vector Algebra, the position vector of the midpoint of a line segment connecting two points with position vectors $\vec{a}$ and $\vec{b}$ is given by $\frac{\vec{a} + \vec{b}}{2}$].

Step 1: Applying the Midpoint Formula

The position vector $\vec{r}$ of the midpoint $R$ is calculated as:

$\vec{r} = \frac{\vec{p} + \vec{q}}{2}$

Step 2: Vector Addition

Substitute the given vectors into the equation:

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

Group the components corresponding to the unit vectors $\hat{i}, \hat{j},$ and $\hat{k}$:

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

$\vec{r} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2}$

Step 3: Scalar Multiplication

Divide each component by the scalar $2$:

$\vec{r} = \left(\frac{6}{2}\right)\hat{i} + \left(\frac{4}{2}\right)\hat{j} + \left(\frac{2}{2}\right)\hat{k}$

$\vec{r} = 3\hat{i} + 2\hat{j} + 1\hat{k}$

Visual Representation

Final Solution: The position vector of the midpoint is $3\hat{i} + 2\hat{j} + \hat{k}$.

Step 1: Theoretical Foundation of Equality of Vectors

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 defined as equal if and only if their corresponding components are identical [Definition of Equality of Vectors]. Specifically:
$a_1 = b_1$, $a_2 = b_2$, and $a_3 = b_3$.

Step 2: Mapping Given Vectors to Components

Given the two vectors:

• Vector $\vec{A} = 2\hat{i} + 3\hat{j}$
• Vector $\vec{B} = x\hat{i} + y\hat{j}$

We decompose these vectors into their scalar components:

Step 3: Algebraic Equivalence and Solution

By equating the components of $\vec{A}$ and $\vec{B}$ [Equality of Corresponding Components]:

For the $\hat{i}$ components: $x = 2$

For the $\hat{j}$ components: $y = 3$

Step 4: Vector Visualization

Final Solution: $x = 2$ and $y = 3$

Step 1: Establishing the Vector Component Representation

Given the vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, we identify the scalar components along the Cartesian axes:

• $x = 1$
• $y = 2$
• $z = 3$

[Per the definition of a vector in 3D space $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$]

Step 2: Calculating the Magnitude of the Vector

The magnitude $|\vec{a}|$ is defined as the square root of the sum of the squares of its components:

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

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

$|\vec{a}| = \sqrt{1 + 4 + 9}$

$|\vec{a}| = \sqrt{14}$

[By the Euclidean Norm formula for vectors in $\mathbb{R}^3$]

Step 3: Determining Direction Cosines

The direction cosines $(l, m, n)$ of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ are given by the ratios of the scalar components to the magnitude of the vector:

$l = \frac{x}{|\vec{a}|}, \quad m = \frac{y}{|\vec{a}|}, \quad n = \frac{z}{|\vec{a}|}$

Substituting the calculated values:

• $l = \frac{1}{\sqrt{14}}$
• $m = \frac{2}{\sqrt{14}}$
• $n = \frac{3}{\sqrt{14}}$

Visual Representation of Vector Components

Step 4: Verification of Identity

For any direction cosines, the identity $l^2 + m^2 + n^2 = 1$ must hold true:

$(\frac{1}{\sqrt{14}})^2 + (\frac{2}{\sqrt{14}})^2 + (\frac{3}{\sqrt{14}})^2 = \frac{1}{14} + \frac{4}{14} + \frac{9}{14} = \frac{14}{14} = 1$

[By the property of Direction Cosines $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$]

Final Solution: The direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

Step 1: Formal Definition of Collinearity

Two vectors $\vec{a}$ and $\vec{b}$ are said to be collinear if and only if there exists a non-zero scalar $\lambda \in \mathbb{R}$ such that $\vec{b} = \lambda \vec{a}$ [Per the definition of parallel/collinear vectors in vector space]. If such a scalar exists, the vectors possess the same or opposite direction and are proportional component-wise.

Step 2: Assigning Vector Variables

Let the given vectors be defined as:

• $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$
• $\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}$

Step 3: Component-wise Scalar Comparison

To determine if $\vec{b} = \lambda \vec{a}$, we equate the components:

Step 4: Vector Synthesis and Logical Deduction

Since the ratio of the corresponding components is constant for all three dimensions ($x, y, z$), we can express $\vec{b}$ as follows:

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

$\vec{b} = -2\vec{a}$

By the scalar multiplication property, since $\vec{b} = \lambda \vec{a}$ where $\lambda = -2$, the vectors $\vec{a}$ and $\vec{b}$ are parallel. Because they share a line of action in vector space, they are defined as collinear [By the geometric property of parallel vectors].

Final Solution: Since $\vec{b} = -2\vec{a}$, the vectors are scalar multiples of each other, satisfying the condition for collinearity. Hence, the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} - 8\hat{k}$ are collinear.

Step 1: Define the Given Vector

Let the given vector be $\vec{a}$. $$\vec{a} = 5\hat{i} - 1\hat{j} + 2\hat{k}$$ The components of the vector are $x = 5$, $y = -1$, and $z = 2$.

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

The magnitude of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by the formula $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$ [By the definition of the Euclidean norm in $\mathbb{R}^3$].

Substituting the given values: $$|\vec{a}| = \sqrt{(5)^2 + (-1)^2 + (2)^2}$$ $$|\vec{a}| = \sqrt{25 + 1 + 4}$$ $$|\vec{a}| = \sqrt{30}$$

Step 3: Determine the Unit Vector in the Direction of $\vec{a}$

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

$$\hat{a} = \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}}$$ $$\hat{a} = \frac{5}{\sqrt{30}}\hat{i} - \frac{1}{\sqrt{30}}\hat{j} + \frac{2}{\sqrt{30}}\hat{k}$$

Step 4: Scale the Unit Vector to the Required Magnitude

To find a vector $\vec{v}$ with magnitude $8$ units in the direction of $\vec{a}$, we multiply the unit vector $\hat{a}$ by the scalar $8$ [Property of scalar multiplication of vectors]:

$$\vec{v} = 8 \cdot \hat{a}$$ $$\vec{v} = 8 \left( \frac{5}{\sqrt{30}}\hat{i} - \frac{1}{\sqrt{30}}\hat{j} + \frac{2}{\sqrt{30}}\hat{k} \right)$$ $$\vec{v} = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$$

Step 5: Rationalization (Optional for simplified form)

Multiplying the numerator and denominator by $\sqrt{30}$:

$$\vec{v} = \frac{40\sqrt{30}}{30}\hat{i} - \frac{8\sqrt{30}}{30}\hat{j} + \frac{16\sqrt{30}}{30}\hat{k}$$ $$\vec{v} = \frac{4\sqrt{30}}{3}\hat{i} - \frac{4\sqrt{30}}{15}\hat{j} + \frac{8\sqrt{30}}{15}\hat{k}$$

Final Solution: The vector is $\frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$ units.

Step 1: Define the Given Vectors

The vectors are provided in component form as follows:

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

Step 2: Calculate the Resultant Vector $\vec{c} = \vec{a} + \vec{b}$

To find the sum of two vectors, we add their respective components [Per the principle of Vector Addition]:

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

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

$\vec{c} = \hat{i} + \hat{k}$

Step 3: Determine the Magnitude of the Resultant Vector

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

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

$|\vec{c}| = \sqrt{1 + 0 + 1} = \sqrt{2}$

Step 4: Compute the Unit Vector $\hat{c}$

The unit vector in the direction of any vector $\vec{v}$ is given by the formula $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$ [Per the definition of normalized vectors].

$\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$

$\hat{c} = \frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$

Visual Representation of the Resultant Vector

Final Solution: The unit vector in the direction of $\vec{a} + \vec{b}$ is $\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$.

Step 1: Theoretical Foundation of Vector Magnitude

The magnitude (or length) of a vector $\vec{v}$ in three-dimensional space, defined in Cartesian components as $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, is determined by the Euclidean norm. This is derived from the Pythagorean theorem extended to three dimensions. The formula for the magnitude, denoted as $|\vec{v}|$, is given by:

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

[Justification: Per the definition of the norm of a vector in Euclidean space $\mathbb{R}^3$]

Step 2: Identifying Scalar Components

Given the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, we identify the scalar components by comparing it to the standard form $x\hat{i} + y\hat{j} + z\hat{k}$:

• $x = 1$
• $y = 1$
• $z = 1$

Step 3: Calculating the Magnitude

Substitute the identified components into the magnitude formula:

$|\vec{a}| = \sqrt{(1)^2 + (1)^2 + (1)^2}$

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

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

Visual Representation of the Vector Magnitude

Step 4: Conclusion

The calculation demonstrates that the length of the vector in 3D space is the square root of the sum of the squares of its directional components.

Final Solution: $|\vec{a}| = \sqrt{3}$ units

Step 1: Given Variables & Theoretical Foundation

Let the position vectors of points $P$ and $Q$ be represented by $\vec{a}$ and $\vec{b}$ respectively. The point $R$ divides the line segment $PQ$ externally in the ratio $m:n$.

• $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$
• $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$
• Ratio $m : n = 2 : 1$

[Per the Section Formula for external division, the position vector $\vec{r}$ of a point $R$ dividing $PQ$ externally in the ratio $m:n$ is given by: $\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$]

Step 2: Substitution into the External Division Formula

Substitute the given vectors and scalar values into the formula:

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

Step 3: Algebraic Simplification

Perform the scalar multiplication and vector subtraction:

Numerator = $(2(-\hat{i}) + 2(\hat{j}) + 2(\hat{k})) - (\hat{i} + 2\hat{j} - \hat{k})$

Numerator = $(-2\hat{i} + 2\hat{j} + 2\hat{k}) - \hat{i} - 2\hat{j} + \hat{k}$

Group the components by unit vectors:

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

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

Visual Representation of Vector Positioning

Step 4: Final Vector Consolidation

The vector $\vec{r}$ simplifies to:

$\vec{r} = -3\hat{i} + 3\hat{k}$

Final Solution: The position vector of point R is $-3\hat{i} + 3\hat{k}$

Step 1: Identification of Vector Components

The vector $\vec{b}$ is expressed in terms of its rectangular components in a 3D Cartesian coordinate system:

$\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$

Given the vector $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$, we identify the scalar components as:

• $x = 2$
• $y = -7$
• $z = -3$

Step 2: Mathematical Formulation of Magnitude

The magnitude of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, denoted by $|\vec{a}|$, is defined as the square root of the sum of the squares of its components [Per the definition of the Euclidean norm in $\mathbb{R}^3$]:

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

Step 3: Execution of Calculation

Substituting the identified values into the magnitude formula:

$|\vec{b}| = \sqrt{(2)^2 + (-7)^2 + (-3)^2}$

$|\vec{b}| = \sqrt{4 + 49 + 9}$

$|\vec{b}| = \sqrt{62}$

Step 4: Vector Visualization

Analytical Conclusion

The magnitude represents the scalar length of the vector in Euclidean space. Since $62$ is a square-free integer, the radical form is the most precise representation.

Final Solution: $|\vec{b}| = \sqrt{62}$ units

Step 1: Identifying Given Variables

Let the position vector of point $P$ be $\vec{p}$ and the position vector of point $Q$ be $\vec{q}$. We are given:

• $\vec{p} = 1\hat{i} + 2\hat{j} - 1\hat{k}$
• $\vec{q} = -1\hat{i} + 1\hat{j} + 1\hat{k}$
• The ratio $m : n = 2 : 1$

Step 2: Applying the Section Formula for Internal Division

The position vector $\vec{r}$ of a point $R$ dividing the line segment $PQ$ internally in the ratio $m : n$ is given by the formula:

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

[Per the Section Formula in Vector Algebra]

Step 3: Algebraic Substitution and Calculation

Substitute the given vectors $\vec{p}$ and $\vec{q}$ and the scalar ratio values $m=2$ and $n=1$ into the equation:

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

Distribute the scalars across the vector components:

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

Group the like components (unit vectors $\hat{i}, \hat{j}, \hat{k}$):

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

$$\vec{r} = \frac{-1\hat{i} + 4\hat{j} + 1\hat{k}}{3}$$

Step 4: Final Vector Representation

Distribute the denominator 3 across each component:

$$\vec{r} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$$

Final Solution: The position vector of point R is $-\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$

Step 1: Define the Given Vector

Let the given vector be $\vec{a}$. We define the components as follows:

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

The components of the vector along the $x, y,$ and $z$ axes are $a_x = 1, a_y = 1,$ and $a_z = 1$, respectively. [By Definition of Cartesian Components of a Vector].

Step 2: Calculate the Magnitude of the Vector

The magnitude of the vector $\vec{a}$, denoted as $|\vec{a}|$, is calculated using the Euclidean norm formula:

$|\vec{a}| = \sqrt{(a_x)^2 + (a_y)^2 + (a_z)^2}$

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

Step 3: Determine the Direction Cosines

If a vector makes angles $\alpha, \beta, \gamma$ with the $OX, OY,$ and $OZ$ axes, its direction cosines are given by $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$. These are defined as:

$\cos \alpha = \frac{a_x}{|\vec{a}|} = \frac{1}{\sqrt{3}}$

$\cos \beta = \frac{a_y}{|\vec{a}|} = \frac{1}{\sqrt{3}}$

$\cos \gamma = \frac{a_z}{|\vec{a}|} = \frac{1}{\sqrt{3}}$

[By Definition of Direction Cosines for a Vector in 3D Space].

Step 4: Analyze Equality of Inclination

Since $\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$, and considering the domain of angles $\alpha, \beta, \gamma \in [0, \pi]$, the values of the angles must be equal:

$\alpha = \beta = \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Because the angles made with the coordinate axes are identical, the vector is equally inclined to the axes $OX, OY,$ and $OZ$.

Visual Representation

Final Solution: Since the direction cosines $\cos \alpha, \cos \beta, \cos \gamma$ are all equal to $\frac{1}{\sqrt{3}}$, the angles $\alpha, \beta, \gamma$ are equal; thus, the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

Step 1: Identifying the Given Vector Components

To find the sum of three vectors, we represent each vector in its standard Cartesian form $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, where $x, y, z$ are the scalar components along the $X, Y,$ and $Z$ axes respectively.

• $\vec{a} = 1\hat{i} - 2\hat{j} + 1\hat{k}$
• $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$
• $\vec{c} = 1\hat{i} - 6\hat{j} - 7\hat{k}$

Step 2: Applying the Addition of Vectors Law

According to the triangle law and the commutative property of vector addition [Per the Algebraic Properties of Vectors], the sum of vectors $\vec{r} = \vec{a} + \vec{b} + \vec{c}$ is obtained by summing their respective components:

$\vec{r} = (x_a + x_b + x_c)\hat{i} + (y_a + y_b + y_c)\hat{j} + (z_a + z_b + z_c)\hat{k}$

Step 3: Execution of Algebraic Summation

Step 4: Vector Visualization

Step 5: Final Compilation

By substituting the results from the table into the vector equation:

$\vec{r} = (0)\hat{i} + (-4)\hat{j} + (-1)\hat{k}$

$\vec{r} = -4\hat{j} - \hat{k}$

Final Solution: The sum of the vectors is $-4\hat{j} - \hat{k}$.

Step 1: Determining the Vector Component Form

To find the vector $\vec{AB}$ directed from point $A(x_1, y_1, z_1)$ to $B(x_2, y_2, z_2)$, we subtract the coordinates of $A$ from $B$: $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$ [By the definition of a position vector between two points].

Given $A(1, 2, -3)$ and $B(-1, -2, 1)$:

• $\Delta x = -1 - 1 = -2$
• $\Delta y = -2 - 2 = -4$
• $\Delta z = 1 - (-3) = 4$

Thus, $\vec{AB} = -2\hat{i} - 4\hat{j} + 4\hat{k}$.

Step 2: Calculating the Magnitude of Vector $\vec{AB}$

The magnitude $|\vec{AB}|$ is given by the square root of the sum of the squares of the components: $|\vec{AB}| = \sqrt{a^2 + b^2 + c^2}$ [Per the magnitude formula for a 3D vector].

$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2}$

$|\vec{AB}| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$ units.

Step 3: Calculating Direction Cosines

The direction cosines $(l, m, n)$ are defined as the ratio of each component to the magnitude of the vector: $l = \frac{a}{|\vec{r}|}, m = \frac{b}{|\vec{r}|}, n = \frac{c}{|\vec{r}|}$

• $l = \frac{-2}{6} = -\frac{1}{3}$
• $m = \frac{-4}{6} = -\frac{2}{3}$
• $n = \frac{4}{6} = \frac{2}{3}$

Geometric Representation

Step 4: Verification of Consistency

Check the identity $l^2 + m^2 + n^2 = 1$ [Fundamental property of direction cosines]:

$(-\frac{1}{3})^2 + (-\frac{2}{3})^2 + (\frac{2}{3})^2 = \frac{1}{9} + \frac{4}{9} + \frac{4}{9} = \frac{9}{9} = 1$.

The calculation is verified as accurate.

Final Solution: The direction cosines of vector $\vec{AB}$ are $\left( -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right)$.

Step 1: Identification of the Vector

The given vector is $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$. In component form, this can be written as $\vec{a} = (1, 1, 2)$.

Step 2: Calculation of the Magnitude

The magnitude of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is defined by the formula: $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$ [By the definition of the Euclidean norm in $\mathbb{R}^3$].

Substituting the given components:

Step 3: Determining the Unit Vector

A unit vector $\hat{a}$ in the direction of vector $\vec{a}$ is defined as the vector divided by its own magnitude: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ [Per the definition of normalized vectors].

Substituting the values derived in Step 1 and Step 2:

Geometric Representation

Step 4: Verification (Optional)

To verify that $\hat{a}$ is indeed a unit vector, we calculate its magnitude: $|\hat{a}| = \sqrt{\left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{2}{\sqrt{6}}\right)^2}$ $|\hat{a}| = \sqrt{\frac{1}{6} + \frac{1}{6} + \frac{4}{6}} = \sqrt{\frac{6}{6}} = 1$ Since the magnitude is unity, the vector is correctly normalized.

Final Solution: The unit vector in the direction of $\vec{a}$ is $\frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$.