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

Component	Line 1 Vector ($\vec{v}_1$)	Line 2 Vector ($\vec{v}_2$)
x	$-3$	$3(-10/7) = -30/7$
y	$2(-10/7) = -20/7$	$1$
z	$2$	$-5$

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: Identification of Direction Vectors

The equation of a line in vector form is given by $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{b}$ represents the direction vector of the line. Given the two lines:

• Line 1: $\vec{r} = (2\hat{i} - 5\hat{j} + \hat{k}) + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$
• Line 2: $\vec{r} = (7\hat{i} - 6\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$

We extract the direction vectors $\vec{b}_1$ and $\vec{b}_2$:

$\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}$

$\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$

Step 2: Theoretical Formulation for the Angle Between Lines

The angle $\theta$ between two lines with direction vectors $\vec{b}_1$ and $\vec{b}_2$ is defined by the formula [Per the definition of the dot product]:

$\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$

Step 3: Calculating Dot Product and Magnitudes

Calculation of the Dot Product ($\vec{b}_1 \cdot \vec{b}_2$):

$\vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19$

Calculation of Magnitudes:

$|\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

$|\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Step 4: Determination of the Angle

Substituting the computed values into the cosine formula:

$\cos \theta = \frac{19}{(7)(3)} = \frac{19}{21}$

Therefore, the angle $\theta$ is given by:

$\theta = \cos^{-1}\left(\frac{19}{21}\right)$

Final Solution: $\theta = \cos^{-1}\left(\frac{19}{21}\right)$

Step 1: Identifying Parametric Vector Forms

The standard vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$. We must rewrite the given equations to extract the position vector $\vec{a}$ (a point on the line) and the direction vector $\vec{b}$.

For the first line: $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k})$ Thus, $\vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}$.

For the second line: $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$ $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})$ Thus, $\vec{a_2} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}$.

Step 2: Calculating Necessary Vectors

To find the shortest distance ($d$), we utilize the formula $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$.

Difference in position vectors: $(\vec{a_2} - \vec{a_1}) = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 0\hat{i} + 1\hat{j} - 4\hat{k} = \hat{j} - 4\hat{k}$.

Cross product of direction vectors ($\vec{b_1} \times \vec{b_2}$): $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix}$ $= \hat{i}(-2 - (-4)) - \hat{j}(2 - (-2)) + \hat{k}(-2 - 1)$ $= 2\hat{i} - 4\hat{j} - 3\hat{k}$.

Step 3: Calculating Magnitude and Dot Product

Magnitude of the cross product: $|\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.

Dot product of $(\vec{a_2} - \vec{a_1})$ and $(\vec{b_1} \times \vec{b_2})$: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0)(2) + (1)(-4) + (-4)(-3) = 0 - 4 + 12 = 8$.

Step 4: Final Distance Computation

Substituting the values into the shortest distance formula: $d = \left| \frac{8}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}}$.

Final Solution: The shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units.

Step 1: Theoretical Foundation

The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by the general form:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$

The vector equation of a line passing through a position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is defined as:

$$\vec{r} = \vec{a} + \lambda\vec{b}$$

[By the fundamental definition of vector equations in 3D Euclidean space]

Step 2: Identifying the Position Vector of the Point

Given the Cartesian equation:

$$\frac{x - 5}{3} = \frac{y - (-4)}{7} = \frac{z - 6}{2}$$

Comparing this to the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, we extract the coordinates of a fixed point on the line:

• $x_1 = 5$
• $y_1 = -4$
• $z_1 = 6$

The position vector $\vec{a}$ of the point $(5, -4, 6)$ is:

$$\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$$

Step 3: Identifying the Direction Vector

The denominators in the Cartesian form represent the direction ratios $(a, b, c)$ of the line. From the given equation, these values are:

• $a = 3$
• $b = 7$
• $c = 2$

The vector $\vec{b}$ parallel to the line is constructed from these direction ratios:

$$\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$$

Step 4: Constructing the Vector Equation

Substituting $\vec{a}$ and $\vec{b}$ into the standard vector equation form $\vec{r} = \vec{a} + \lambda\vec{b}$:

$$\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$$

Where $\lambda$ is a scalar parameter [Per the definition of the parametric form of a line in $\mathbb{R}^3$].

Visual Representation of the Line Vector

Final Solution: The vector form of the line is $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$, where $\lambda$ is a scalar.

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: Establishing Theoretical Foundation

The Cartesian equation of a line passing through a fixed point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula [Per the standard form of Cartesian equations of a line in 3D space]:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Step 2: Identifying Given Parameters

• Fixed Point ($P$): The line passes through $(-2, 4, -5)$. Thus, $x_1 = -2$, $y_1 = 4$, and $z_1 = -5$.
• Direction Vectors: Two lines are parallel if their direction ratios are proportional [By the condition of parallelism for lines in 3D geometry]. The given line is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.
• The direction ratios of the given line are $(3, 5, 6)$. Therefore, the direction ratios $(a, b, c)$ of our required line are also $(3, 5, 6)$.

Step 3: Deriving the Cartesian Equation

Substitute the coordinates of the point $(-2, 4, -5)$ and the direction ratios $(3, 5, 6)$ into the standard form equation:

$\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6}$

Simplifying the signs within the numerators:

$\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}$

Step 4: Geometric Visualization

Step 5: Verification of Analytical Constraints

The direction ratios $(3, 5, 6)$ satisfy the vector requirements for parallelism, and substituting $(-2, 4, -5)$ into the final expression yields $\frac{0}{3} = \frac{0}{5} = \frac{0}{6}$, which is a valid satisfaction of the point-slope form for a line [By the property of identity in Cartesian geometry].

Final Solution: The Cartesian equation of the line is $\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}$.

Step 1: Identifying Vertices and Defining Sides

Let the vertices of the triangle be denoted as follows:

• $A = (3, 5, -4)$
• $B = (-1, 1, 2)$
• $C = (-5, -5, -2)$

The sides of the triangle are the line segments $AB$, $BC$, and $CA$. The direction cosines of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $\frac{x_2-x_1}{d}, \frac{y_2-y_1}{d}, \frac{z_2-z_1}{d}$, where $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ [Per the distance formula in 3D space].

Step 2: Calculating Direction Cosines for Side AB

Vector $\vec{AB} = (-1-3, 1-5, 2-(-4)) = (-4, -4, 6)$.

Length $d_{AB} = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}$.

Direction Cosines ($l_1, m_1, n_1$):

• $l_1 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$
• $m_1 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$
• $n_1 = \frac{6}{2\sqrt{17}} = \frac{3}{\sqrt{17}}$

Step 3: Calculating Direction Cosines for Side BC

Vector $\vec{BC} = (-5-(-1), -5-1, -2-2) = (-4, -6, -4)$.

Length $d_{BC} = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}$.

Direction Cosines ($l_2, m_2, n_2$):

• $l_2 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$
• $m_2 = \frac{-6}{2\sqrt{17}} = -\frac{3}{\sqrt{17}}$
• $n_2 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

Step 4: Calculating Direction Cosines for Side CA

Vector $\vec{CA} = (3-(-5), 5-(-5), -4-(-2)) = (8, 10, -2)$.

Length $d_{CA} = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}$.

Direction Cosines ($l_3, m_3, n_3$):

• $l_3 = \frac{8}{2\sqrt{42}} = \frac{4}{\sqrt{42}}$
• $m_3 = \frac{10}{2\sqrt{42}} = \frac{5}{\sqrt{42}}$
• $n_3 = \frac{-2}{2\sqrt{42}} = -\frac{1}{\sqrt{42}}$

Step 5: Summary Table of Results

Side	Direction Cosines $(l, m, n)$
$AB$	$(-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}})$
$BC$	$(-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}})$
$CA$	$(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}})$

Final Solution: The direction cosines are as listed in the table above for sides AB, BC, and CA respectively.

Step 1: Identification of Given Vectors

To determine the equation of a line in space, we require a point through which the line passes and a direction vector parallel to the line. Let $\vec{a}$ be the position vector of the given point and $\vec{b}$ be the direction vector.

• Position vector of the point: $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$
• Direction vector: $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$

[Per the definition of the vector equation of a line passing through a point $\vec{a}$ and parallel to a vector $\vec{b}$]

Step 2: Deriving the Vector Equation

The vector equation of a line passing through point $A(\vec{a})$ and parallel to vector $\vec{b}$ is defined as:

$\vec{r} = \vec{a} + \lambda \vec{b}$

Where $\lambda$ is a scalar parameter ($ \lambda \in \mathbb{R} $). Substituting the given values:

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

Step 3: Deriving the Cartesian Equation

Let the coordinates of a general point on the line be $(x, y, z)$. Thus, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Let the given point $\vec{a}$ have coordinates $(x_1, y_1, z_1) = (2, -1, 4)$ and the direction vector $\vec{b}$ have direction ratios $(a, b, c) = (1, 2, -1)$.

The Cartesian form of the line is given by [Ref: Symmetric form of a line equation]:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Substituting the values $(x_1, y_1, z_1) = (2, -1, 4)$ and $(a, b, c) = (1, 2, -1)$:

$\frac{x - 2}{1} = \frac{y - (-1)}{2} = \frac{z - 4}{-1}$

Simplifying the denominator for the $y$-term:

$\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$

Visual Representation of the Line Direction

Step 4: Summary of Equations

Form	Equation
Vector Form	$\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k})$
Cartesian Form	$\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$

Final Solution: The vector equation is $\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k})$ and the Cartesian equation is $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$.

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: Theoretical Foundation of Direction Cosines

In three-dimensional coordinate geometry, if a line makes angles $\alpha$, $\beta$, and $\gamma$ with the positive directions of the $x$, $y$, and $z$-axes respectively, the direction cosines of the line are defined as:

• $l = \cos \alpha$
• $m = \cos \beta$
• $n = \cos \gamma$

[Definition of Direction Cosines in Cartesian Space]

Step 2: Substitution of Given Angles

Given the angles:

• $\alpha = 90^\circ$
• $\beta = 135^\circ$
• $\gamma = 45^\circ$

Step 3: Calculating Trigonometric Values

Applying the cosine function to each angle:

$l = \cos(90^\circ) = 0$

$m = \cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ) = -\frac{1}{\sqrt{2}}$

$n = \cos(45^\circ) = \frac{1}{\sqrt{2}}$

[Trigonometric Identity: $\cos(180^\circ - \theta) = -\cos(\theta)$]

Step 4: Verification via Direction Cosine Property

The sum of the squares of the direction cosines must equal unity: $l^2 + m^2 + n^2 = 1$.

$0^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$

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

Step 5: Visual Representation

Final Solution: The direction cosines of the line are $0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Step 1: Standardizing the Direction Ratios of Line 1

The given equation of the first line is: $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$

To identify the direction ratios $(a_1, b_1, c_1)$, the variables $x, y, z$ must have a coefficient of $1$ and be positive. We rearrange as follows:

• $\frac{-(x-1)}{3} = \frac{x-1}{-3}$
• $\frac{7(y-2)}{2p} = \frac{y-2}{2p/7}$
• $\frac{z-3}{2}$

Thus, the direction ratios of Line 1 are: $a_1 = -3, b_1 = \frac{2p}{7}, c_1 = 2$ [By the property of standard form of line equations $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$]

Step 2: Standardizing the Direction Ratios of Line 2

The given equation of the second line is: $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$

Rearranging to ensure the coefficients of $x, y, z$ are $+1$:

• $\frac{-7(x-1)}{3p} = \frac{x-1}{-3p/7}$
• $\frac{y-5}{1}$
• $\frac{-(z-6)}{5} = \frac{z-6}{-5}$

Thus, the direction ratios of Line 2 are: $a_2 = -\frac{3p}{7}, b_2 = 1, c_2 = -5$

Step 3: Applying the Condition for Perpendicularity

Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are at right angles if and only if their dot product is zero [Per the condition of orthogonality]: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

Substituting the values derived in Steps 1 and 2:

$(-3) \left( -\frac{3p}{7} \right) + \left( \frac{2p}{7} \right)(1) + (2)(-5) = 0$

Step 4: Algebraic Solution for $p$

Multiply the terms:

$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$

Combine the fractions:

$\frac{11p}{7} = 10$

Solve for $p$:

$11p = 70$ $p = \frac{70}{11}$

Final Solution: $p = \frac{70}{11}$

Step 1: Identification of Direction Vectors

The given lines are in the Cartesian form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. The direction ratios $(a, b, c)$ of a line determine its orientation in $3D$ space.

• For the first line: $L_1: \frac{x}{2} = \frac{y}{2} = \frac{z}{1}$. The direction ratios are $\vec{b_1} = (2, 2, 1)$.
• For the second line: $L_2: \frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$. The direction ratios are $\vec{b_2} = (4, 1, 8)$.

Step 2: Theoretical Formulation for the Angle Between Two Lines

The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the cosine formula [By the Definition of the Scalar Product of Vectors]:

$\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$

Step 3: Calculating Magnitudes and Scalar Product

First, calculate the scalar product (dot product) of vectors $\vec{b_1}$ and $\vec{b_2}$:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$

Next, calculate the magnitudes of the direction vectors $|\vec{b_1}|$ and $|\vec{b_2}|$:

$|\vec{b_1}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

$|\vec{b_2}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$

Step 4: Final Substitution and Calculation

Substitute the derived values into the cosine formula:

$\cos \theta = \frac{18}{3 \times 9}$

$\cos \theta = \frac{18}{27} = \frac{2}{3}$

Therefore, $\theta = \cos^{-1}\left(\frac{2}{3}\right)$

Visual Representation of Direction Vectors

Final Solution: The angle between the two lines is $\cos^{-1}\left(\frac{2}{3}\right)$.

Step 1: Establishing the Direction Vector for the First Line ($L_1$)

To determine the orientation of a line in three-dimensional space passing through two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, we calculate the direction ratios $(a, b, c)$ using the vector displacement between the points: $a = x_2 - x_1$, $b = y_2 - y_1$, and $c = z_2 - z_1$.

Given points for Line 1: $P(4, 7, 8)$ and $Q(2, 3, 4)$.

• $a_1 = 2 - 4 = -2$
• $b_1 = 3 - 7 = -4$
• $c_1 = 4 - 8 = -4$

Thus, the direction vector for $L_1$ is $\vec{v_1} = (-2, -4, -4)$ [By the definition of direction ratios of a line passing through two points].

Step 2: Establishing the Direction Vector for the Second Line ($L_2$)

Given points for Line 2: $R(-1, -2, 1)$ and $S(1, 2, 5)$.

• $a_2 = 1 - (-1) = 2$
• $b_2 = 2 - (-2) = 4$
• $c_2 = 5 - 1 = 4$

Thus, the direction vector for $L_2$ is $\vec{v_2} = (2, 4, 4)$ [By the definition of direction ratios of a line passing through two points].

Step 3: Verification of Parallelism

Two lines are considered parallel if and only if their direction ratios are proportional, satisfying the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k$, where $k$ is a scalar constant [Per the condition for parallelism in 3D geometry].

Applying the ratio test:

Component	Ratio Calculation	Result
x-ratio	$-2 / 2$	$-1$
y-ratio	$-4 / 4$	$-1$
z-ratio	$-4 / 4$	$-1$

Since $\frac{-2}{2} = \frac{-4}{4} = \frac{-4}{4} = -1$, the ratios are equal.

Step 4: Geometric Visualization

Final Solution: Since the direction ratios of the line through $(4, 7, 8)$ and $(2, 3, 4)$ are proportional to the direction ratios of the line through $(-1, -2, 1)$ and $(1, 2, 5)$ with a constant of proportionality $k = -1$, the two lines are parallel.

Step 1: Identification of Direction Vectors

The equation of a line in 3D space is given by the vector form $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{b}$ represents the direction vector of the line. For the two given lines:

• Line 1: $\vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k})$
• Line 2: $\vec{r} = (2\hat{i} - \hat{j} - 56\hat{k}) + \mu(3\hat{i} - 5\hat{j} - 4\hat{k})$

From the above equations, the direction vectors $\vec{b_1}$ and $\vec{b_2}$ are:

$\vec{b_1} = 1\hat{i} - 1\hat{j} - 2\hat{k}$

$\vec{b_2} = 3\hat{i} - 5\hat{j} - 4\hat{k}$

Step 2: Theoretical Framework for Angle Calculation

The acute angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by the formula [Per the definition of the dot product]:

$\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$

Step 3: Scalar Product and Magnitudes

Calculating the dot product ($\vec{b_1} \cdot \vec{b_2}$):

$\vec{b_1} \cdot \vec{b_2} = (1)(3) + (-1)(-5) + (-2)(-4)$

$\vec{b_1} \cdot \vec{b_2} = 3 + 5 + 8 = 16$

Calculating the magnitude of $\vec{b_1}$ ($|\vec{b_1}|$):

$|\vec{b_1}| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

Calculating the magnitude of $\vec{b_2}$ ($|\vec{b_2}|$):

$|\vec{b_2}| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$

Step 4: Computing the Angle

Substitute the calculated values into the formula derived in Step 2:

$\cos \theta = \frac{16}{\sqrt{6} \cdot 5\sqrt{2}}$

$\cos \theta = \frac{16}{5\sqrt{12}} = \frac{16}{5 \cdot 2\sqrt{3}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$

To rationalize the denominator:

$\cos \theta = \frac{8\sqrt{3}}{15}$

Therefore, $\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$

Visual Representation of Direction Vectors

Final Solution: The angle between the two lines is $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$

Step 1: Identifying the Vector Equation Parameters

To determine the equation of a line in three-dimensional space, we utilize the vector form of a line equation, defined as:

$\vec{r} = \vec{a} + \lambda\vec{b}$

Where:

• $\vec{a}$ is the position vector of a point through which the line passes.
• $\vec{b}$ is the vector parallel to the line.
• $\lambda$ is a scalar parameter.

[Per the standard vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$]

Step 2: Assigning Values to Given Variables

Given the point $(1, 2, 3)$, we define the position vector $\vec{a}$ as:

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

Given the parallel vector $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$, we substitute these values into the general formula.

Step 3: Constructing the Vector Equation

Substituting $\vec{a}$ and $\vec{b}$ into $\vec{r} = \vec{a} + \lambda\vec{b}$:

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

Step 4: Deriving the Cartesian Equation (Optional Analysis)

If the line passes through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$, the Cartesian form is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

Using $(x_1, y_1, z_1) = (1, 2, 3)$ and $(a, b, c) = (3, 2, -2)$:

$\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}$

Final Solution: The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$ and the Cartesian equation is $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}$.

Step 1: Identification of Direction Vectors

For any line given in the Cartesian form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, the vector parallel to the line is $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$.

Given the first line: $\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$

The direction vector is $\vec{a_1} = 2\hat{i} + 5\hat{j} - 3\hat{k}$

Given the second line: $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$

The direction vector is $\vec{a_2} = -1\hat{i} + 8\hat{j} + 4\hat{k}$

Step 2: Theoretical Formulation for the Angle Between Two Lines

The angle $\theta$ between two lines with direction vectors $\vec{a_1}$ and $\vec{a_2}$ is defined by the cosine formula derived from the dot product property $\vec{a_1} \cdot \vec{a_2} = |\vec{a_1}| |\vec{a_2}| \cos \theta$ [Per the definition of the Scalar Product]:

$\cos \theta = \frac{|\vec{a_1} \cdot \vec{a_2}|}{|\vec{a_1}| |\vec{a_2}|}$

Step 3: Calculating the Dot Product and Magnitudes

Part A: Scalar Product $\vec{a_1} \cdot \vec{a_2}$

$\vec{a_1} \cdot \vec{a_2} = (2)(-1) + (5)(8) + (-3)(4)$

$\vec{a_1} \cdot \vec{a_2} = -2 + 40 - 12 = 26$

Part B: Magnitude of $\vec{a_1}$

$|\vec{a_1}| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$

Part C: Magnitude of $\vec{a_2}$

$|\vec{a_2}| = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$

Step 4: Computation of $\theta$

Substituting the values into the cosine formula:

$\cos \theta = \frac{|26|}{\sqrt{38} \cdot 9} = \frac{26}{9\sqrt{38}}$

Therefore, $\theta = \cos^{-1} \left( \frac{26}{9\sqrt{38}} \right)$

Visual Representation of Directional Vectors

Final Solution: The angle $\theta$ between the two lines is $\cos^{-1} \left( \frac{26}{9\sqrt{38}} \right)$

Step 1: Identification of Line Parameters

The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$. For two lines:

• Line 1: $\vec{r} = \vec{a_1} + \lambda\vec{b_1} \implies \vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = \hat{i} - 3\hat{j} + 2\hat{k}$
• Line 2: $\vec{r} = \vec{a_2} + \mu\vec{b_2} \implies \vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$ and $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$

Step 2: Vector Differences and Cross Product

The shortest distance $d$ between two skew lines is given by the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ [Per the formula for shortest distance between skew lines].

First, calculate $(\vec{a_2} - \vec{a_1})$:

$\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$ using the determinant method:

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ $= \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6))$ $= -9\hat{i} + 3\hat{j} + 9\hat{k}$

Step 3: Calculating Magnitudes and Scalar Triple Product

Calculate the magnitude of the cross product:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-9)^2 + (3)^2 + (9)^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = \sqrt{9 \times 19} = 3\sqrt{19}$

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

$(3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = (3)(-9) + (3)(3) + (3)(9) = -27 + 9 + 27 = 9$

Step 4: Final Distance Calculation

Substitute into the distance formula:

$d = \frac{|9|}{3\sqrt{19}} = \frac{3}{\sqrt{19}}$

Rationalizing the denominator:

$d = \frac{3\sqrt{19}}{19}$

Final Solution: The shortest distance between the two lines is $\frac{3\sqrt{19}}{19}$ units.

Step 1: Identifying Direction Cosine Vectors

Given the direction cosines of three lines $L_1$, $L_2$, and $L_3$, we define their corresponding unit direction vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ in $\mathbb{R}^3$:

• $\vec{a} = \langle \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} \rangle$
• $\vec{b} = \langle \frac{4}{13}, \frac{12}{13}, \frac{3}{13} \rangle$
• $\vec{c} = \langle \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \rangle$

Step 2: Theoretical Condition for Mutual Perpendicularity

Two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular if and only if the dot product of their direction vectors is zero [Per the definition of the dot product in Euclidean space]:

$l_1l_2 + m_1m_2 + n_1n_2 = 0$

Step 3: Calculating Dot Products

Pair	Dot Product Calculation	Result
$\vec{a} \cdot \vec{b}$	$(\frac{12}{13} \cdot \frac{4}{13}) + (\frac{-3}{13} \cdot \frac{12}{13}) + (\frac{-4}{13} \cdot \frac{3}{13})$	$\frac{48 - 36 - 12}{169} = \frac{0}{169} = 0$
$\vec{b} \cdot \vec{c}$	$(\frac{4}{13} \cdot \frac{3}{13}) + (\frac{12}{13} \cdot \frac{-4}{13}) + (\frac{3}{13} \cdot \frac{12}{13})$	$\frac{12 - 48 + 36}{169} = \frac{0}{169} = 0$
$\vec{a} \cdot \vec{c}$	$(\frac{12}{13} \cdot \frac{3}{13}) + (\frac{-3}{13} \cdot \frac{-4}{13}) + (\frac{-4}{13} \cdot \frac{12}{13})$	$\frac{36 + 12 - 48}{169} = \frac{0}{169} = 0$

Step 4: Geometric Visualization

Step 5: Conclusion of Orthogonality

Since the dot products of the direction vectors taken pairwise ($\vec{a} \cdot \vec{b} = 0$, $\vec{b} \cdot \vec{c} = 0$, and $\vec{a} \cdot \vec{c} = 0$) are all equal to zero, the lines are mutually perpendicular [By the property of orthogonal vectors].

Final Solution: Since the sum of products of corresponding direction cosines for every pair of lines is zero, the three lines are mutually perpendicular.