Step 1: Identifying the Direction Vector of the Line

A line parallel to the $x$-axis maintains a constant direction vector. The unit vector along the $x$-axis is defined as $\vec{i} = (1, 0, 0)$. Therefore, the direction vector $\vec{b}$ of the required line is:

$\vec{b} = a\hat{i} + 0\hat{j} + 0\hat{k} = (1, 0, 0)$ [Per the definition of Cartesian coordinate axes].

Step 2: Identifying the Position Vector of the Given Point

The line is specified to pass through the origin. The coordinates of the origin are $(0, 0, 0)$. Thus, the position vector $\vec{a}$ of the point on the line is:

$\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = (0, 0, 0)$.

Step 3: Formulating the Vector Equation

The vector equation of a line passing through a point $\vec{a}$ and parallel to a vector $\vec{b}$ is given by the formula:

$\vec{r} = \vec{a} + \lambda\vec{b}$ [By the standard vector form of a line equation].

Substituting the values from Step 1 and Step 2:

$\vec{r} = (0, 0, 0) + \lambda(1, 0, 0)$

$\vec{r} = \lambda(1, 0, 0)$, where $\lambda$ is a scalar parameter.

Step 4: Deriving the Cartesian Equations

Let the coordinates of any point on the line be $(x, y, z)$. The vector equation $\vec{r} = (x, y, z)$ implies:

$(x, y, z) = (\lambda, 0, 0)$

By comparing the corresponding components:

• $x = \lambda$
• $y = 0$
• $z = 0$

Since $\lambda$ can be any real number, the constraints on the coordinates of the line are $y = 0$ and $z = 0$.

Visual Representation

Final Solution: The equation of the line is $y = 0$ and $z = 0$.

Step 1: Identification of Direction Vectors

For a line represented in the symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, the direction ratios of the line are given by the vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$.

• For the first line, $L_1: \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$, the direction vector is $\vec{v}_1 = -3\hat{i} + 2k\hat{j} + 2\hat{k}$.
• For the second line, $L_2: \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$, the direction vector is $\vec{v}_2 = 3k\hat{i} + 1\hat{j} - 5\hat{k}$.

Step 2: Applying the Orthogonality Condition

Two lines with direction vectors $\vec{v}_1$ and $\vec{v}_2$ are perpendicular if and only if their scalar (dot) product is zero [Per the condition for perpendicular vectors, $\vec{v}_1 \cdot \vec{v}_2 = 0$].

The dot product is defined as: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Step 3: Algebraic Derivation

Substituting the components into the equation:

$(-3)(3k) + (2k)(1) + (2)(-5) = 0$

Expanding the terms:

$-9k + 2k - 10 = 0$

Combining the $k$ terms:

$-7k - 10 = 0$

Solving for $k$:

$-7k = 10$

$k = -\frac{10}{7}$

Visual Representation of Perpendicular Vectors

Verification of Results

Verification Calculation: $(-3)(-30/7) + (-20/7)(1) + (2)(-5) = \frac{90}{7} - \frac{20}{7} - 10 = \frac{70}{7} - 10 = 10 - 10 = 0$. The condition is satisfied.

Final Solution: k = -\frac{10}{7}

Step 1: Identifying Direction Ratios

Let the two lines be $L_1$ and $L_2$. The direction ratios of line $L_1$ are given by the vector $\vec{v_1} = \langle a, b, c \rangle$. The direction ratios of line $L_2$ are given by the vector $\vec{v_2} = \langle b-c, c-a, a-b \rangle$.

Step 2: Theoretical Formulation of the Angle Between Two Lines

The angle $\theta$ between two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ is defined by the formula:
$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$ [By the definition of the dot product of two vectors, $\vec{v_1} \cdot \vec{v_2} = |\vec{v_1}| |\vec{v_2}| \cos \theta$].

Step 3: Calculating the Numerator (Dot Product)

The dot product $\vec{v_1} \cdot \vec{v_2}$ is calculated as:
$\text{Numerator} = a(b-c) + b(c-a) + c(a-b)$
$= ab - ac + bc - ba + ca - cb$
$= (ab - ba) + (bc - cb) + (ca - ac)$
$= 0 + 0 + 0 = 0$
[By the Commutative Property of multiplication and the Additive Inverse Property].

Step 4: Analyzing the Geometric Implications

Since the numerator of the expression for $\cos \theta$ is $0$, we have:
$\cos \theta = \frac{0}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} = 0$
For $\cos \theta = 0$, the angle $\theta$ must satisfy $\cos \theta = \cos(90^\circ)$.
Therefore, $\theta = \frac{\pi}{2}$ radians or $90^\circ$. [Condition for orthogonality: two vectors are perpendicular if and only if their dot product is zero].

Visual Representation of Orthogonality

Final Solution: The angle between the two lines is 90° or $\frac{\pi}{2}$ radians.

Step 1: Identifying Direction Vectors of Given Lines

A line in 3D space represented in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ has a direction vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

• For the first line: $L_1: \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$, the direction vector is $\vec{v_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$.
• For the second line: $L_2: \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$, the direction vector is $\vec{v_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.

Step 2: Determining the Direction of the Required Line

Since the required line is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d}$ must be parallel to the cross product of $\vec{v_1}$ and $\vec{v_2}$ [Per the definition of the cross product of two vectors].

$\vec{d} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$

Expanding the determinant:

$\vec{d} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$

$\vec{d} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$

$\vec{d} = 24\hat{i} + 36\hat{j} + 72\hat{k}$

Step 3: Simplifying the Direction Vector

We can simplify the direction vector by dividing by the common scalar factor $12$ (since the direction of a line remains the same for any scalar multiple of its direction vector):

$\vec{d'} = 2\hat{i} + 3\hat{j} + 6\hat{k}$

Step 4: Formulating the Vector Equation of the Line

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{d}$ is given by $\vec{r} = \vec{a} + \lambda\vec{d}$ [Standard form of a line in vector space].

• The given point is $(1, 2, -4)$, so $\vec{a} = 1\hat{i} + 2\hat{j} - 4\hat{k}$.
• The direction vector is $\vec{d'} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.

Substituting these values:

$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

Visual Representation of Direction Vectors

Final Solution: The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$, where $\lambda$ is a scalar parameter.

Step 1: Identification of Line Parameters

The given lines are in the form $\vec{r} = \vec{a} + \lambda\vec{b}$ and $\vec{r} = \vec{c} + \mu\vec{d}$. From the provided equations:

• Line 1: $\vec{a_1} = 6\hat{i} + 2\hat{j} + 2\hat{k}$, $\vec{b_1} = \hat{i} - 2\hat{j} + 2\hat{k}$
• Line 2: $\vec{a_2} = -4\hat{i} + 0\hat{j} - \hat{k}$, $\vec{b_2} = 3\hat{i} - 2\hat{j} - 2\hat{k}$

Step 2: Theoretical Foundation for Shortest Distance

The shortest distance $d$ between two skew lines is given by the formula [Per the vector projection theorem]:

$d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$

Step 3: Calculating $(\vec{a_2} - \vec{a_1})$ and $(\vec{b_1} \times \vec{b_2})$

Calculate the vector connecting the points on the lines:

$\vec{a_2} - \vec{a_1} = (-4 - 6)\hat{i} + (0 - 2)\hat{j} + (-1 - 2)\hat{k} = -10\hat{i} - 2\hat{j} - 3\hat{k}$

Calculate the cross product of the direction vectors:

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix}$

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

$= \hat{i}(4 + 4) - \hat{j}(-2 - 6) + \hat{k}(-2 + 6) = 8\hat{i} + 8\hat{j} + 4\hat{k}$

Step 4: Scalar Triple Product and Magnitude Calculation

Calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(-10\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (8\hat{i} + 8\hat{j} + 4\hat{k}) = (-10)(8) + (-2)(8) + (-3)(4) = -80 - 16 - 12 = -108$

Calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$

Step 5: Final Shortest Distance Evaluation

$d = \left| \frac{-108}{12} \right| = |-9| = 9$

Final Solution: The shortest distance between the two lines is 9 units.