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: 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: Analyzing the First Condition ($\vec{a} \cdot \vec{a} = 0$)

Given the equation $\vec{a} \cdot \vec{a} = 0$, we apply the definition of the dot product of a vector with itself, which is equivalent to the square of its magnitude:

$\vec{a} \cdot \vec{a} = |\vec{a}|^2$ [By the definition of Scalar Product: $\vec{A} \cdot \vec{A} = |\vec{A}||\vec{A}|\cos(0^\circ) = |\vec{A}|^2$]

Setting this equal to the given value: $|\vec{a}|^2 = 0$ $\implies |\vec{a}| = 0$

This implies that $\vec{a}$ is the zero vector, denoted as $\vec{a} = \vec{0}$.

Step 2: Analyzing the Second Condition ($\vec{a} \cdot \vec{b} = 0$)

We are given $\vec{a} \cdot \vec{b} = 0$. Substituting the result from Step 1 ($\vec{a} = \vec{0}$):

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

By the property of the scalar product with a zero vector, the dot product of the zero vector with any arbitrary vector $\vec{b}$ is always zero, regardless of the components or the direction of $\vec{b}$.

$\vec{0} \cdot \vec{b} = 0 \cdot |\vec{b}| \cdot \cos(\theta) = 0$ [Per the Zero Vector Property in Vector Algebra]

Step 3: Determining the Status of Vector $\vec{b}$

Since the condition $\vec{a} \cdot \vec{b} = 0$ is satisfied for any vector $\vec{b}$ when $\vec{a} = \vec{0}$, we conclude that $\vec{b}$ is not constrained by the provided equations.

Conclusion

Because $\vec{a}$ is the zero vector, the product $\vec{a} \cdot \vec{b}$ will result in $0$ for any vector $\vec{b}$ in the vector space. Thus, no specific constraints are imposed on $\vec{b}$.

Final Solution: The vector $\vec{b}$ can be any arbitrary vector in the given vector space.

Step 1: Theoretical Foundation

Two vectors are perpendicular if and only if their scalar (dot) product is equal to zero. Let the two given vectors be defined as follows:

Let $\vec{u} = |\vec{a}|\vec{b} + |\vec{b}|\vec{a}$

Let $\vec{v} = |\vec{a}|\vec{b} - |\vec{b}|\vec{a}$

To prove $\vec{u} \perp \vec{v}$, we must demonstrate that $\vec{u} \cdot \vec{v} = 0$ [By Definition of Orthogonal Vectors].

Step 2: Performing the Scalar Product

We calculate the dot product $\vec{u} \cdot \vec{v}$ using the distributive property of the dot product over vector addition: $\vec{A} \cdot (\vec{B} - \vec{C}) = \vec{A} \cdot \vec{B} - \vec{A} \cdot \vec{C}$.

$\vec{u} \cdot \vec{v} = (|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} - |\vec{b}|\vec{a})$

Expanding the product [By Distributive Law]:

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

Step 3: Algebraic Simplification

Utilizing the properties of scalar multiplication with vectors ($k\vec{A} \cdot m\vec{B} = km(\vec{A} \cdot \vec{B})$) and the property $\vec{A} \cdot \vec{A} = |\vec{A}|^2$:

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

Since the dot product is commutative ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$) [By Commutative Property of Dot Product]:

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

Step 4: Cancellation and Final Verification

Observing the terms:

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

$= 0 + 0$

$= 0$

Final Solution: Since the scalar product of $(|\vec{a}|\vec{b} + |\vec{b}|\vec{a})$ and $(|\vec{a}|\vec{b} - |\vec{b}|\vec{a})$ is 0, the two vectors are confirmed to be perpendicular.

Step 1: Identifying Given Parameters

Based on the problem statement, we identify the following scalar and vector quantities:

• Magnitude of vector $\vec{a}$: $|\vec{a}| = \sqrt{3}$
• Magnitude of vector $\vec{b}$: $|\vec{b}| = 2$
• Scalar product (dot product): $\vec{a} \cdot \vec{b} = \sqrt{6}$

Step 2: Theoretical Foundation

The relationship between the dot product of two vectors, their magnitudes, and the angle $\theta$ between them is defined by the formula [Per the Geometric Definition of the Scalar Product]:

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

Step 3: Algebraic Substitution and Manipulation

Rearranging the equation to solve for $\cos \theta$:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

Substituting the known values into the expression:

$\cos \theta = \frac{\sqrt{6}}{\sqrt{3} \cdot 2}$

Step 4: Simplifying the Expression

Using the property of surds where $\frac{\sqrt{m}}{\sqrt{n}} = \sqrt{\frac{m}{n}}$:

$\cos \theta = \frac{\sqrt{6}}{2\sqrt{3}}$

$\cos \theta = \frac{1}{2} \cdot \sqrt{\frac{6}{3}}$

$\cos \theta = \frac{1}{2} \cdot \sqrt{2}$

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

Step 5: Determining the Angle

Given $\cos \theta = \frac{1}{\sqrt{2}}$, we find the principal value of $\theta$ in the interval $[0, \pi]$:

$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

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

Vector Representation

Final Solution: The angle between the two vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ radians or $45^\circ$.

Step 1: Formal Definition and Given Variables

Let the two vectors be $\vec{a}$ and $\vec{b}$. We are given the following conditions:

• The magnitudes of the vectors are equal: $|\vec{a}| = |\vec{b}| = x$ [Assumption of scalar magnitude $x$].
• The angle between the vectors, $\theta = 60^{\circ}$.
• The scalar product (dot product) of the vectors, $\vec{a} \cdot \vec{b} = \frac{1}{2}$.

Step 2: Applying the Scalar Product Definition

The definition of the scalar product for any two vectors $\vec{a}$ and $\vec{b}$ is given by:
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$ [By the definition of the dot product].

Step 3: Algebraic Substitution and Solving for x

Substituting the known values into the equation derived in Step 2:

$\frac{1}{2} = (x)(x) \cos(60^{\circ})$

Since $\cos(60^{\circ}) = \frac{1}{2}$ [Per trigonometric identity], we have:

$\frac{1}{2} = x^2 \left( \frac{1}{2} \right)$

Multiplying both sides by $2$ to isolate $x^2$:

$1 = x^2$

$x = \sqrt{1} = 1$ [Since magnitude must be a non-negative scalar, $x > 0$].

Step 4: Geometric Visualization

Step 5: Final Verification

Final Solution: The magnitude of both vectors $\vec{a}$ and $\vec{b}$ is $1$.

Step 1: Application of the Distributive Law of Scalar Product

To evaluate the dot product of two linear combinations of vectors, we apply the distributive property of the scalar (dot) product over vector addition, defined as $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$.

Given the expression: $(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$

Expanding the product term-by-term:

$= 3\vec{a} \cdot (2\vec{a} + 7\vec{b}) - 5\vec{b} \cdot (2\vec{a} + 7\vec{b})$

$= (3\vec{a} \cdot 2\vec{a}) + (3\vec{a} \cdot 7\vec{b}) - (5\vec{b} \cdot 2\vec{a}) - (5\vec{b} \cdot 7\vec{b})$

Step 2: Applying Scalar Multiples and Commutative Property

We utilize the following properties of the dot product:

• Scalar Multiplication: $(k\vec{u}) \cdot (l\vec{v}) = kl(\vec{u} \cdot \vec{v})$ [Scalar Property]
• Commutative Property: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ [Dot Product Property]
• Self-Dot Product: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ [Geometric Definition]

Simplifying the expansion:

$= 6(\vec{a} \cdot \vec{a}) + 21(\vec{a} \cdot \vec{b}) - 10(\vec{b} \cdot \vec{a}) - 35(\vec{b} \cdot \vec{b})$

$= 6|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) - 10(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2$ [Applying Commutative Property $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}$]

Step 3: Final Algebraic Simplification

Grouping the common dot product terms $(\vec{a} \cdot \vec{b})$:

$= 6|\vec{a}|^2 + (21 - 10)(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2$

$= 6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2$

Vector Representation Visualization

Final Solution: $6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2$

Step 1: Identifying the Given Conditions

We are provided with the following parameters:

• Let $\vec{a}$ be a non-zero vector such that its magnitude is $|\vec{a}| = a$ (where $a \neq 0$).
• Let $\lambda$ be a non-zero scalar ($\lambda \neq 0$).
• The resultant vector $\vec{b} = \lambda\vec{a}$ is defined as a unit vector.

Step 2: Theoretical Foundation of a Unit Vector

By definition, a vector is a unit vector if and only if its magnitude is equal to unity ($1$). [Per the definition of normalized vectors in Vector Algebra].

Therefore, we must satisfy the condition:

$|\lambda\vec{a}| = 1$

Step 3: Applying Scalar Multiplication Properties

Using the property of the norm of a scalar product, where $|k\vec{v}| = |k| \cdot |\vec{v}|$ for any scalar $k$ and vector $\vec{v}$: [Property: Magnitude of Scalar Multiplication]

$|\lambda| \cdot |\vec{a}| = 1$

Step 4: Algebraic Manipulation for $\lambda$

Substitute the given magnitude $|\vec{a}| = a$ into the equation:

$|\lambda| \cdot a = 1$

To isolate $\lambda$, divide both sides by the magnitude $a$ (given $a \neq 0$):

$|\lambda| = \frac{1}{a}$

Step 5: Geometric and Visual Representation

Step 6: Interpretation of the Result

Since $|\lambda| = \frac{1}{a}$, this implies that the scalar $\lambda$ can take two possible values based on the directionality, though the magnitude is fixed as the reciprocal of the magnitude of vector $\vec{a}$.

Final Solution: $\lambda = \pm\frac{1}{a}$ or $|\lambda| = \frac{1}{a}$

Step 1: Expanding the Scalar Product Expression

Given the scalar product equation: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8$.

Applying the distributive property of the dot product [Distributive Law: $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$]:

$\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 8$

Since the dot product is commutative [Commutative Law: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$], the terms $-\vec{a} \cdot \vec{b}$ and $+\vec{b} \cdot \vec{a}$ cancel out. Utilizing the identity $\vec{x} \cdot \vec{x} = |\vec{x}|^2$ [Definition of Dot Product], we obtain:

$|\vec{a}|^2 - |\vec{b}|^2 = 8$ --- (Equation 1)

Step 2: Substituting the Given Relation

We are given the relation: $|\vec{a}| = 8|\vec{b}|$. Substitute this expression into Equation 1:

$(8|\vec{b}|)^2 - |\vec{b}|^2 = 8$

$64|\vec{b}|^2 - |\vec{b}|^2 = 8$

$63|\vec{b}|^2 = 8$

Step 3: Calculating Magnitudes

Solving for $|\vec{b}|$:

$|\vec{b}|^2 = \frac{8}{63}$

$|\vec{b}| = \sqrt{\frac{8}{63}} = \sqrt{\frac{4 \cdot 2}{9 \cdot 7}} = \frac{2\sqrt{2}}{3\sqrt{7}}$

To rationalize the denominator: $|\vec{b}| = \frac{2\sqrt{14}}{21}$.

Now, solve for $|\vec{a}|$ using $|\vec{a}| = 8|\vec{b}|$:

$|\vec{a}| = 8 \cdot \left( \frac{2\sqrt{14}}{21} \right) = \frac{16\sqrt{14}}{21}$.

Visual Representation of Vector Components

Final Solution: $|\vec{a}| = \frac{16\sqrt{14}}{21}$ and $|\vec{b}| = \frac{2\sqrt{14}}{21}$

Step 1: Identifying Given Vectors

Let the two given vectors be represented as $\vec{a}$ and $\vec{b}$:

• $\vec{a} = 1\hat{i} + 3\hat{j} + 7\hat{k}$
• $\vec{b} = 7\hat{i} - 1\hat{j} + 8\hat{k}$

Step 2: Defining the Projection Formula

The scalar projection of a vector $\vec{a}$ on a vector $\vec{b}$ is defined as the dot product of $\vec{a}$ and the unit vector in the direction of $\vec{b}$. [Per the definition of Scalar Projection]:

$\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

Step 3: Calculating the Dot Product ($\vec{a} \cdot \vec{b}$)**