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.

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).

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: Setting up the Cross Product

Given the cross product of two vectors is the zero vector, $\vec{A} \times \vec{B} = \vec{0}$, we define:
$\vec{A} = 2\hat{i} + 6\hat{j} + 27\hat{k}$
$\vec{B} = \hat{i} + \lambda\hat{j} + \mu\hat{k}$

The cross product is computed using the determinant of a $3 \times 3$ matrix [Per the definition of the vector product in Euclidean space]:

$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \vec{0}$

Step 2: Evaluating the Determinant

Expanding the determinant along the first row:

$\hat{i}(6\mu - 27\lambda) - \hat{j}(2\mu - 27) + \hat{k}(2\lambda - 6) = 0\hat{i} + 0\hat{j} + 0\hat{k}$

Step 3: Solving the System of Equations

Equating the components of the resultant vector to zero [By the Principle of Equality of Vectors]:

• Component of $\hat{k}$: $2\lambda - 6 = 0 \implies 2\lambda = 6 \implies \lambda = 3$
• Component of $\hat{j}$: $-(2\mu - 27) = 0 \implies 2\mu = 27 \implies \mu = \frac{27}{2} = 13.5$
• Verification via $\hat{i}$ component: Substitute $\lambda = 3$ and $\mu = 13.5$ into $6\mu - 27\lambda$:
  $6(13.5) - 27(3) = 81 - 81 = 0$

Step 4: Vector Alignment Visualization

Since the cross product of two non-zero vectors is zero if and only if the vectors are collinear [Per the properties of the cross product], we can also verify the ratio of components:
$\frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu}$
$\frac{2}{1} = \frac{6}{\lambda} \implies \lambda = 3$
$\frac{2}{1} = \frac{27}{\mu} \implies \mu = 13.5$

Final Solution: \lambda = 3, \mu = 13.5

Step 1: Setting up the Cross Product Matrix

To find the cross product $\vec{a} \times \vec{b}$, we utilize the determinant form of the cross product for vectors in three-dimensional Cartesian space [Per the definition of the vector cross product]:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$

Given $\vec{a} = 1\hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$, we substitute the components:

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

Step 2: Expanding the Determinant

Expanding along the first row [By Laplace expansion]:

$\vec{a} \times \vec{b} = \hat{i}[(-7)(2) - (7)(-2)] - \hat{j}[(1)(2) - (7)(3)] + \hat{k}[(1)(-2) - (-7)(3)]$

$\vec{a} \times \vec{b} = \hat{i}[-14 + 14] - \hat{j}[2 - 21] + \hat{k}[-2 + 21]$

$\vec{a} \times \vec{b} = 0\hat{i} - (-19)\hat{j} + 19\hat{k}$

$\vec{a} \times \vec{b} = 19\hat{j} + 19\hat{k}$

Step 3: Calculating the Magnitude

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

$|\vec{a} \times \vec{b}| = \sqrt{(0)^2 + (19)^2 + (19)^2}$

$|\vec{a} \times \vec{b}| = \sqrt{0 + 361 + 361}$

$|\vec{a} \times \vec{b}| = \sqrt{722}$

Step 4: Simplifying the Radical

Factoring $722$:

$722 = 2 \times 361 = 2 \times 19^2$

$|\vec{a} \times \vec{b}| = \sqrt{19^2 \times 2} = 19\sqrt{2}$

Final Solution: $|\vec{a} \times \vec{b}| = 19\sqrt{2}$

Step 1: Calculating Vector Sums and Differences

Given the vectors $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$, we first define the resultant vectors $\vec{c} = \vec{a} + \vec{b}$ and $\vec{d} = \vec{a} - \vec{b}$.

Summation: $\vec{c} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k}$

Difference: $\vec{d} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k}$

Step 2: Finding the Perpendicular Vector via Cross Product

A vector perpendicular to both $\vec{c}$ and $\vec{d}$ is given by their cross product $\vec{n} = \vec{c} \times \vec{d}$. [By the Property of Cross Product: $\vec{A} \times \vec{B}$ is orthogonal to both $\vec{A}$ and $\vec{B}$].

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$

Expanding along the first row: $\vec{n} = \hat{i}(4 \cdot 4 - 0 \cdot 0) - \hat{j}(4 \cdot 4 - 0 \cdot 2) + \hat{k}(4 \cdot 0 - 4 \cdot 2)$

$\vec{n} = 16\hat{i} - 16\hat{j} - 8\hat{k}$

Step 3: Calculating the Magnitude and Unit Vector

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

Magnitude $|\vec{n}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$.

Thus, $\hat{n} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24}$.

Simplifying coefficients: $\hat{n} = \pm \left( \frac{16}{24}\hat{i} - \frac{16}{24}\hat{j} - \frac{8}{24}\hat{k} \right) = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right)$.

Visual Representation of Orthogonal Basis

Final Solution: The unit vector perpendicular to both vectors is $\pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right)$.

Step 1: Establishing the Magnitude Equation

The magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is given by the formula:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta)$

where $\theta$ is the angle between the two vectors, such that $0 \leq \theta \leq \pi$ [Per the definition of the Cross Product Magnitude].

Step 2: Substituting Given Values

Given the following values:

• $|\vec{a}| = 3$
• $|\vec{b}| = \frac{\sqrt{2}}{3}$
• $|\vec{a} \times \vec{b}| = 1$ (as it is defined as a unit vector)

Substitute these values into the cross product magnitude formula:

$1 = (3) \times \left(\frac{\sqrt{2}}{3}\right) \times \sin(\theta)$

Step 3: Algebraic Solving for $\theta$

Simplify the right-hand side of the equation:

$1 = 3 \times \frac{\sqrt{2}}{3} \times \sin(\theta)$

$1 = \sqrt{2} \sin(\theta)$

Isolating the trigonometric function:

$\sin(\theta) = \frac{1}{\sqrt{2}}$

[By the Inverse Trigonometric Property, where $\sin^{-1}(x) = \theta$ for $x \in [-1, 1]$]

Step 4: Determining the Angle

Since $\sin(\theta) = \frac{1}{\sqrt{2}}$, we evaluate the principle value of $\theta$ within the range $[0, \pi]$:

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

$\theta = \frac{\pi}{4}$

Visual Representation of Vector Cross Product

Final Solution: The angle between the vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$.

Step 1: Identification of Position Vectors

The given position vectors of the vertices of the rectangle $ABCD$ are:

• $\vec{a} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$
• $\vec{b} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$
• $\vec{c} = \hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$
• $\vec{d} = -\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$

Step 2: Calculating Adjacent Side Vectors

To find the area of the rectangle, we determine the vectors representing two adjacent sides, $\vec{AB}$ and $\vec{AD}$.

$\vec{AB} = \vec{b} - \vec{a} = (\hat{i} - (-\hat{i})) + (\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k}$

$\vec{AD} = \vec{d} - \vec{a} = (-\hat{i} - (-\hat{i})) + (-\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} = 0\hat{i} - 1\hat{j} + 0\hat{k}$

[By the definition of a displacement vector between two points in 3D space, $\vec{PQ} = \vec{OQ} - \vec{OP}$]

Step 3: Calculating Magnitudes of Sides

The lengths of the adjacent sides are the magnitudes of the displacement vectors:

• $|\vec{AB}| = \sqrt{2^2 + 0^2 + 0^2} = \sqrt{4} = 2$ units
• $|\vec{AD}| = \sqrt{0^2 + (-1)^2 + 0^2} = \sqrt{1} = 1$ unit

Step 4: Calculation of Area

The area of a rectangle is the product of the magnitudes of its adjacent side vectors:

$\text{Area} = |\vec{AB}| \times |\vec{AD}|$

$\text{Area} = 2 \times 1 = 2$ square units

Visual Representation

Final Solution: The area of the rectangle is 2 square units.

Step 1: Establishing Vector Components

Given the vectors in their component form:

• $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$
• $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$
• $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$

Step 2: Evaluating the Left-Hand Side (LHS)

First, calculate the sum of vectors $\vec{b}$ and $\vec{c}$ [Per the definition of Vector Addition]:

$\vec{b} + \vec{c} = (b_1 + c_1)\hat{i} + (b_2 + c_2)\hat{j} + (b_3 + c_3)\hat{k}$

Now, compute the cross product $\vec{a} \times (\vec{b} + \vec{c})$ using the determinant expansion [Per the definition of the Cross Product in Cartesian components]:

$\vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ (b_1+c_1) & (b_2+c_2) & (b_3+c_3) \end{vmatrix}$

Expanding along the first row:

$= \hat{i}[a_2(b_3+c_3) - a_3(b_2+c_2)] - \hat{j}[a_1(b_3+c_3) - a_3(b_1+c_1)] + \hat{k}[a_1(b_2+c_2) - a_2(b_1+c_1)]$

$= \hat{i}[a_2b_3 + a_2c_3 - a_3b_2 - a_3c_2] - \hat{j}[a_1b_3 + a_1c_3 - a_3b_1 - a_3c_1] + \hat{k}[a_1b_2 + a_1c_2 - a_2b_1 - a_2c_1]$

Step 3: Evaluating the Right-Hand Side (RHS)

Calculate $\vec{a} \times \vec{b}$ and $\vec{a} \times \vec{c}$ individually:

$\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} + (a_3b_1 - a_1b_3)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$

$\vec{a} \times \vec{c} = (a_2c_3 - a_3c_2)\hat{i} + (a_3c_1 - a_1c_3)\hat{j} + (a_1c_2 - a_2c_1)\hat{k}$

Adding the two resultant vectors:

$\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = [(a_2b_3 - a_3b_2) + (a_2c_3 - a_3c_2)]\hat{i} + [(a_3b_1 - a_1b_3) + (a_3c_1 - a_1c_3)]\hat{j} + [(a_1b_2 - a_2b_1) + (a_1c_2 - a_2c_1)]\hat{k}$

Step 4: Comparative Analysis

By regrouping the terms in the RHS expression, we obtain:

$= [(a_2b_3 + a_2c_3) - (a_3b_2 + a_3c_2)]\hat{i} + [(a_3b_1 + a_3c_1) - (a_1b_3 + a_1c_3)]\hat{j} + [(a_1b_2 + a_1c_2) - (a_2b_1 + a_2c_1)]\hat{k}$

Comparing this term-for-term with the expanded LHS in Step 2, we observe that the coefficients of $\hat{i}, \hat{j},$ and $\hat{k}$ are identical.

Final Solution: Since the components of $\vec{a} \times (\vec{b} + \vec{c})$ and $\vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ are identical, it is proven that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.

Theoretical Foundation & Definition

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin(\theta) \hat{n}$, where $\theta$ is the angle between the two vectors and $\hat{n}$ is a unit vector perpendicular to both. Given the condition $\vec{a} \times \vec{b} = \vec{0}$, we examine the mathematical requirements:

• $|\vec{a}| = 0$ (The zero vector case)
• $|\vec{b}| = 0$ (The zero vector case)
• $\sin(\theta) = 0 \implies \theta = 0^\circ$ or $180^\circ$ (The collinear vector case)

Analyzing the Converse Statement

The converse states: "If $\vec{a} \times \vec{b} = \vec{0}$, then either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$." To test the validity of this converse, we must determine if there exist non-zero vectors $\vec{a}$ and $\vec{b}$ such that their cross product is still the zero vector.

Counter-Example Formulation

Consider two non-zero vectors that are parallel (collinear). Let:
$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
$\vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k}$

Note that $\vec{b} = 2\vec{a}$, which confirms the vectors are linearly dependent and parallel. Thus, the angle $\theta$ between them is $0$.

Calculating the Cross Product

Using the determinant method for the cross product:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{vmatrix}$
$= \hat{i}(3 \times 8 - 4 \times 6) - \hat{j}(2 \times 8 - 4 \times 4) + \hat{k}(2 \times 6 - 3 \times 4)$
$= \hat{i}(24 - 24) - \hat{j}(16 - 16) + \hat{k}(12 - 12)$
$= 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$

Geometric Representation

Logical Conclusion

Because we have identified vectors $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$ such that $\vec{a} \times \vec{b} = \vec{0}$ (due to them being collinear), the condition "either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$" is not strictly necessary for the product to be zero. Therefore, the converse is false.

Final Solution: The converse is not true. If two non-zero vectors are parallel, their cross product is the zero vector, as demonstrated by $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k}$.

Step 1: Identifying Direction Cosines

Let the unit vector $\vec{a}$ be defined in terms of its components as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$. By definition of a unit vector, the components $a_1, a_2, a_3$ correspond to the direction cosines $\cos \alpha, \cos \beta, \cos \gamma$, where $\alpha, \beta, \gamma$ are the angles made by the vector with the axes $x, y, z$ respectively.

Given:

• $\alpha = \frac{\pi}{3}$
• $\beta = \frac{\pi}{4}$
• $\gamma = \theta$ (where $\theta$ is an acute angle)

Step 2: Applying the Direction Cosine Identity

For any unit vector, the sum of the squares of its direction cosines is equal to unity: $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$ [By the property of direction cosines in 3D Euclidean space]

Substituting the known values: $$\cos^2\left(\frac{\pi}{3}\right) + \cos^2\left(\frac{\pi}{4}\right) + \cos^2 \theta = 1$$ $$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1$$ $$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$$ $$\frac{3}{4} + \cos^2 \theta = 1$$ $$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$$

Step 3: Solving for $\theta$ and Components

Since $\cos^2 \theta = \frac{1}{4}$, we have $\cos \theta = \pm \frac{1}{2}$. Given the constraint that $\theta$ is an acute angle ($0 \leq \theta \leq \frac{\pi}{2}$), $\cos \theta$ must be positive. $$\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$$

The components of the vector $\vec{a}$ are given by $(\cos \alpha, \cos \beta, \cos \gamma)$:

• $a_1 = \cos(\frac{\pi}{3}) = \frac{1}{2}$
• $a_2 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
• $a_3 = \cos(\frac{\pi}{3}) = \frac{1}{2}$

Visual Representation of Directional Angles

Final Solution: The angle $\theta = \frac{\pi}{3}$ and the components of $\vec{a}$ are $\left( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \right)$.

Step 1: Coordinate Representation and Vector Formation

Given the vertices of the triangle in $\mathbb{R}^3$: $A = (1, 1, 2)$, $B = (2, 3, 5)$, and $C = (1, 5, 5)$.

To determine the area of triangle $ABC$, we represent two sides originating from a common vertex as vectors. We select vertex $A$ as the reference point:

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

$\vec{AC} = (1 - 1)\hat{i} + (5 - 1)\hat{j} + (5 - 2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$

Step 2: Theoretical Foundation for Area Calculation

[Per the definition of the cross product in vector algebra], the area of a triangle defined by two vectors $\vec{AB}$ and $\vec{AC}$ is given by the formula:

$\text{Area} = \frac{1}{2} | \vec{AB} \times \vec{AC} |$

Step 3: Computation of the Cross Product

The cross product $\vec{AB} \times \vec{AC}$ is calculated using the determinant form:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix}$

Expanding along the first row: $= \hat{i}((2)(3) - (3)(4)) - \hat{j}((1)(3) - (3)(0)) + \hat{k}((1)(4) - (2)(0))$ $= \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0)$ $= -6\hat{i} - 3\hat{j} + 4\hat{k}$

Step 4: Magnitude Calculation

The magnitude of the resulting vector $| \vec{AB} \times \vec{AC} |$ is: $| -6\hat{i} - 3\hat{j} + 4\hat{k} | = \sqrt{(-6)^2 + (-3)^2 + (4)^2}$ $= \sqrt{36 + 9 + 16} = \sqrt{61}$

Step 5: Final Area Determination

Substituting the magnitude back into the area formula: $\text{Area} = \frac{1}{2} \sqrt{61}$ square units.

Final Solution: The area of the triangle is $\frac{\sqrt{61}}{2}$ square units.