Step 1: Identifying the Theoretical Framework

The direction cosines of a line in three-dimensional space are defined as the cosines of the angles ($\alpha, \beta, \gamma$) that the line makes with the positive directions of the $x$, $y$, and $z$-axes, respectively. These are denoted by $l, m,$ and $n$.

According to the definition:

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

[Per the definition of direction cosines in $\mathbb{R}^3$]

Step 2: Assigning Given Angular Values

From the problem statement, the angles are given as:

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

Step 3: Calculating Direction Cosines

We calculate each direction cosine individually using trigonometric evaluations:

For $l$ (x-axis):
$l = \cos(90^{\circ}) = 0$

For $m$ (y-axis):
$m = \cos(135^{\circ}) = \cos(180^{\circ} - 45^{\circ})$
Using the trigonometric identity $\cos(180^{\circ} - \theta) = -\cos(\theta)$ [By the Supplementary Angle Theorem]:
$m = -\cos(45^{\circ}) = -\frac{1}{\sqrt{2}}$

For $n$ (z-axis):
$n = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}$

Step 4: Verification of Identity

The direction cosines must satisfy the fundamental identity $l^2 + m^2 + n^2 = 1$.
$(0)^2 + (-\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$.
[The identity holds, confirming the values are correct.]

Step 5: Geometric Visualization

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

Step 1: Theoretical Foundation of the Line Equation

A straight line in 3D space can be expressed in vector form as $\vec{r} = \vec{a} + \lambda \vec{b}$, where:

• $\vec{a}$ is the position vector of a fixed point on the line: $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.
• $\vec{b}$ is the vector parallel to the line, defined by direction ratios $(a, b, c)$: $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.
• $\lambda$ is a scalar parameter.

The cartesian form is given by: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. [Reference: Standard Cartesian-to-Vector Conversion Theorem]

Step 2: Identifying Given Parameters

The given cartesian equation is:

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

By mapping this to the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, we extract the components:

Component	Value
Point $(x_1, y_1, z_1)$	$(5, -4, 6)$
Direction Ratios $(a, b, c)$	$(3, 7, 2)$

Step 3: Constructing the Vector Form

Using the identified coordinates and direction ratios, we construct the vectors:

• Position vector $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$
• Parallel vector $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$

Substituting these into the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$ yields:

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

Step 4: Geometric Visualization

Final Solution: The vector equation 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: Identification of Direction Vectors

The given equations of the lines are in the standard symmetric form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where the vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ represents the direction of the line.

For Line 1: $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$
The direction vector $\vec{b_1} = 2\hat{i} + 2\hat{j} + 1\hat{k}$.

For Line 2: $\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$
The direction vector $\vec{b_2} = 4\hat{i} + 1\hat{j} + 8\hat{k}$.

Step 2: Theoretical Foundation for Angle Calculation

The angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by the cosine formula derived from the scalar (dot) product definition [Per the definition of the dot product: $\vec{b_1} \cdot \vec{b_2} = |\vec{b_1}| |\vec{b_2}| \cos \theta$]:

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

Step 3: Calculating Components

1. Dot Product ($\vec{b_1} \cdot \vec{b_2}$):

$\vec{b_1} \cdot \vec{b_2} = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$

2. Magnitude of $\vec{b_1}$:

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

3. Magnitude of $\vec{b_2}$:

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

Step 4: Solving for $\theta$

Substituting the calculated values into the formula:

$\cos \theta = \frac{18}{(3)(9)} = \frac{18}{27}$

Simplifying the fraction: $\cos \theta = \frac{2}{3}$

Therefore, the angle $\theta$ is defined as: $\theta = \cos^{-1}\left(\frac{2}{3}\right)$

Visual Representation of Direction Vectors

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

Step 1: Standardizing the Line Equations

To determine the direction ratios of the lines, the equations must be expressed in the standard Cartesian form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

For Line 1:

$\frac{-(x-1)}{3} = \frac{7(y-2)}{2p} = \frac{z-3}{2}$
$\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$
The direction ratios $\vec{b_1} = (a_1, b_1, c_1)$ are $\langle -3, \frac{2p}{7}, 2 \rangle$. [Per standard form conversion rule].

For Line 2:

$\frac{-7(x-1)}{3p} = \frac{y-5}{1} = \frac{-(z-6)}{5}$
$\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$
The direction ratios $\vec{b_2} = (a_2, b_2, c_2)$ are $\langle -\frac{3p}{7}, 1, -5 \rangle$.

Step 2: Applying the Condition for Perpendicularity

Two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ are perpendicular if and only if their dot product is zero [By the condition of orthogonality: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$].

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

Step 3: Algebraic Simplification and Solving for $p$

Multiply the terms to form the linear equation:

$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$
$\frac{11p}{7} = 10$
$11p = 70$
$p = \frac{70}{11}$

Step 4: Geometric Visualization

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

Step 1: Identifying Direction Vectors

The equation of a line in vector form is given by $\vec{r} = \vec{a} + \lambda\vec{b}$. For the given pair of 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: $\vec{r} = (7\hat{i} - 6\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$

The direction vectors are defined as $\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{b}_2 = 1\hat{i} + 2\hat{j} + 2\hat{k}$ [Per the standard vector representation of a line].

Step 2: Theoretical Formulation of the Angle

The angle $\theta$ between two lines with direction vectors $\vec{b}_1$ and $\vec{b}_2$ is given by the cosine formula [By 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

First, calculate the scalar dot product $\vec{b}_1 \cdot \vec{b}_2$:

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

Next, calculate the magnitudes $|\vec{b}_1|$ and $|\vec{b}_2|$:

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

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

Step 4: Final Trigonometric Evaluation

Substitute the derived values into the cosine formula:

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

Therefore, the angle $\theta$ is defined as:

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

Final Solution: The angle between the given pair of lines is $\cos^{-1}\left(\frac{19}{21}\right)$ radians.

Step 1: Identifying Direction Ratios of the Given Lines

The standard form of a line in 3D space is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $\langle a, b, c \rangle$ represents the direction ratios of the line.

Given Line 1 ($L_1$): $$\frac{x - 5}{7} = \frac{y + 2}{-5} = \frac{z - 0}{1}$$ The direction ratios of $L_1$ are $\vec{b_1} = \langle 7, -5, 1 \rangle$.

Given Line 2 ($L_2$): $$\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{3}$$ The direction ratios of $L_2$ are $\vec{b_2} = \langle 1, 2, 3 \rangle$.

Step 2: Theoretical Criterion for Perpendicularity

Two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ are perpendicular if and only if the dot product of their direction vectors is equal to zero [Per the condition of orthogonality of vectors in $\mathbb{R}^3$]:

$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$

Step 3: Calculating the Dot Product

We compute the sum of the products of the corresponding direction ratios:

Component	Calculation	Product
x-direction	$7 \times 1$	$7$
y-direction	$-5 \times 2$	$-10$
z-direction	$1 \times 3$	$3$

Applying the formula:

$$\text{Sum} = (7)(1) + (-5)(2) + (1)(3)$$ $$\text{Sum} = 7 - 10 + 3$$ $$\text{Sum} = -3 + 3 = 0$$

Step 4: Geometric Visualization

Step 5: Conclusion

Since the dot product of the direction vectors $\vec{b_1} \cdot \vec{b_2} = 0$, the vectors are orthogonal [By the definition of the scalar product]. Consequently, the lines $L_1$ and $L_2$ are perpendicular to each other.

Final Solution: The lines are perpendicular because the dot product of their direction ratios $(7)(1) + (-5)(2) + (1)(3) = 0$.

Step 1: Identifying Given Parameters

The direction ratios of a line are given as:

• $a = -18$
• $b = 12$
• $c = -4$

[Per the definition of direction ratios, any triplet $(a, b, c)$ proportional to the direction cosines $(l, m, n)$ represents the orientation of the line.]

Step 2: Calculating the Magnitude of the Direction Ratios

To find the direction cosines, we must first determine the magnitude of the vector representing the direction ratios, denoted by $r = \sqrt{a^2 + b^2 + c^2}$.

$r = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$

$r = \sqrt{324 + 144 + 16}$

$r = \sqrt{484}$

$r = 22$

Step 3: Deriving Direction Cosines

The direction cosines $(l, m, n)$ are calculated by dividing each direction ratio by the magnitude $r$:

$l = \frac{a}{r} = \frac{-18}{22} = -\frac{9}{11}$

$m = \frac{b}{r} = \frac{12}{22} = \frac{6}{11}$

$n = \frac{c}{r} = \frac{-4}{22} = -\frac{2}{11}$

[By the relation $l^2 + m^2 + n^2 = 1$, we verify: $(-\frac{9}{11})^2 + (\frac{6}{11})^2 + (-\frac{2}{11})^2 = \frac{81 + 36 + 4}{121} = \frac{121}{121} = 1$.]

Step 4: Consideration of Alternative Direction Cosines

Since the direction cosines are defined by the vector divided by its magnitude, taking the negative of the magnitude also yields a valid set of direction cosines for the same line (representing the opposite orientation):

$(l, m, n) = (\frac{9}{11}, -\frac{6}{11}, \frac{2}{11})$

Visual Representation

Final Solution: The direction cosines are $\left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$ or $\left( \frac{9}{11}, -\frac{6}{11}, \frac{2}{11} \right)$.

Step 1: Identification of Line Parameters

The given lines are in the vector form $\vec{r} = \vec{a} + \lambda\vec{b}$. Comparing the given equations:

• Line 1: $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$
• Line 2: $\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}$ and $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$

Step 2: Calculating Vector Differences and Cross Product

The shortest distance $d$ between two skew lines is given by the formula: $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ [By the Shortest Distance Formula for Skew Lines].

First, calculate $\vec{a}_2 - \vec{a}_1$:
$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$

Next, calculate the cross product $\vec{b}_1 \times \vec{b}_2$ using the determinant method:

Expanding the determinant:

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} + 0\hat{j} + 3\hat{k}$

Step 3: Calculating Magnitude and Dot Product

Calculate the magnitude $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

Calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 0\hat{j} + 3\hat{k}) = (1)(-3) + (-3)(0) + (-2)(3) = -3 + 0 - 6 = -9$

Step 4: Final Distance Computation

Substitute the values into the distance formula:

$d = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}$

Rationalizing the denominator:

$d = \frac{3\sqrt{2}}{2}$

Final Solution: The shortest distance between the lines is $\frac{3\sqrt{2}}{2}$ units.

Step 1: Identifying 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. For the given pair of 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 equations, we extract the direction vectors $\vec{b_1}$ and $\vec{b_2}$:

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

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

Step 2: Theoretical Formulation for Angle Between Lines

The angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by the cosine formula [By Definition of Scalar Product]:

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

Step 3: Calculating Dot Product and Magnitudes

Scalar Product ($\vec{b_1} \cdot \vec{b_2}$):

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

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

Magnitudes ($|\vec{b_1}|$ and $|\vec{b_2}|$):

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

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

Step 4: Computing $\cos \theta$ and Determining $\theta$

Substituting the values into the formula:

$\cos \theta = \frac{16}{\sqrt{6} \cdot \sqrt{50}} = \frac{16}{\sqrt{300}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$

To rationalize the denominator:

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

Therefore, the angle $\theta$ is defined as:

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

Final Solution: $\theta = \cos^{-1}\left( \frac{8\sqrt{3}}{15} \right)$

Step 1: Establishing the Geometric Relationship

Let the line make an angle $\alpha$ with each of the $x$, $y$, and $z$ axes. Consequently, the direction angles are $\alpha = \beta = \gamma = \alpha$.

The direction cosines of a line are defined as $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. [Per the definition of direction cosines in 3D Euclidean space].

Step 2: Applying the Fundamental Property of Direction Cosines

For any line, the sum of the squares of its direction cosines is unity [By the fundamental identity for direction cosines]:

$l^2 + m^2 + n^2 = 1$

Substituting the condition of equal angles ($\cos \alpha = \cos \beta = \cos \gamma$):

$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$

$3\cos^2 \alpha = 1$

Step 3: Algebraic Derivation of Values

Solving for $\cos \alpha$:

$\cos^2 \alpha = \frac{1}{3}$

$\cos \alpha = \pm \frac{1}{\sqrt{3}}$

Since $\cos \alpha = \cos \beta = \cos \gamma$, the direction cosines are:

$(l, m, n) = \left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \right)$

Step 4: Interpretation of Results

The set of direction cosines identifies the orientation of the line in space. Given the symmetry of the problem relative to the axes, two cases emerge based on the direction of the vector:

Orientation	Direction Cosines $(l, m, n)$
Case 1	$\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$
Case 2	$\left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right)$

Final Solution: The direction cosines are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.

Step 1: Coordinate Definition and Vector Representation

Let the vertices of the triangle be defined as:

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

The direction cosines of a line segment connecting two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are proportional to the direction ratios $(a, b, c) = (x_2-x_1, y_2-y_1, z_2-z_1)$. If $L$ is the length of the segment, where $L = \sqrt{a^2 + b^2 + c^2}$, the direction cosines $(l, m, n)$ are given by $\left( \frac{a}{L}, \frac{b}{L}, \frac{c}{L} \right)$ [Per the definition of direction cosines in $\mathbb{R}^3$].

Step 2: Analysis of Side AB

Direction ratios for $\vec{AB}$: $a_1 = -1 - 3 = -4$, $b_1 = 1 - 5 = -4$, $c_1 = 2 - (-4) = 6$.

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

Direction cosines $(l_1, m_1, n_1) = \left( \frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} \right) = \mathbf{\left( -\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \right)}$.

Step 3: Analysis of Side BC

Direction ratios for $\vec{BC}$: $a_2 = -5 - (-1) = -4$, $b_2 = -5 - 1 = -6$, $c_2 = -2 - 2 = -4$.

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

Direction cosines $(l_2, m_2, n_2) = \left( \frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}} \right) = \mathbf{\left( -\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}} \right)}$.

Step 4: Analysis of Side CA

Direction ratios for $\vec{CA}$: $a_3 = 3 - (-5) = 8$, $b_3 = 5 - (-5) = 10$, $c_3 = -4 - (-2) = -2$.

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

Direction cosines $(l_3, m_3, n_3) = \left( \frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}} \right) = \mathbf{\left( \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}} \right)}$.

Summary Table of Direction Cosines

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

Final Solution: The direction cosines for the sides are $AB: (-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}})$, $BC: (-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}})$, and $CA: (\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}})$.

Step 1: Theoretical Foundation for Perpendicularity

Two lines in three-dimensional space with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are mutually perpendicular if and only if the sum of the products of their corresponding direction cosines is equal to zero. This is derived from the dot product property of direction vectors where $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$. Since $|\vec{a}| = 1$ and $|\vec{b}| = 1$ for direction cosines, $\vec{a} \cdot \vec{b} = 0$ implies $\cos(90^\circ) = 0$ [Per the condition of orthogonality].

The condition is defined as: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

Step 2: Defining the Given Direction Cosines

Let the three lines be denoted as $L_1, L_2,$ and $L_3$ with direction cosines as follows:

• $L_1: (\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13})$
• $L_2: (\frac{4}{13}, \frac{12}{13}, \frac{3}{13})$
• $L_3: (\frac{3}{13}, \frac{-4}{13}, \frac{12}{13})$

Step 3: Verification of Orthogonality (Pairwise Analysis)

Pair	Calculation ($l_i l_j + m_i m_j + n_i n_j$)	Result
$L_1 \perp L_2$	$(\frac{12}{13} \cdot \frac{4}{13}) + (\frac{-3}{13} \cdot \frac{12}{13}) + (\frac{-4}{13} \cdot \frac{3}{13})$	$\frac{48 - 36 - 12}{169} = \frac{0}{169} = 0$
$L_2 \perp L_3$	$(\frac{4}{13} \cdot \frac{3}{13}) + (\frac{12}{13} \cdot \frac{-4}{13}) + (\frac{3}{13} \cdot \frac{12}{13})$	$\frac{12 - 48 + 36}{169} = \frac{0}{169} = 0$
$L_1 \perp L_3$	$(\frac{12}{13} \cdot \frac{3}{13}) + (\frac{-3}{13} \cdot \frac{-4}{13}) + (\frac{-4}{13} \cdot \frac{12}{13})$	$\frac{36 + 12 - 48}{169} = \frac{0}{169} = 0$

Step 4: Visualizing the Orthogonal Basis

Step 5: Logical Synthesis

Since the sum of the products of the direction cosines for each pair $(L_1, L_2)$, $(L_2, L_3)$, and $(L_1, L_3)$ is zero, it confirms that the angle $\theta$ between every pair satisfies $\cos(\theta) = 0$, which implies $\theta = 90^\circ$.

Final Solution: The three lines are mutually perpendicular as they satisfy the condition $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ for all pairs.

Step 1: Calculating Direction Ratios for Line 1 ($L_1$)

Let $L_1$ be the line passing through points $A(1, -1, 2)$ and $B(3, 4, -2)$. The direction ratios $(a_1, b_1, c_1)$ of a line passing through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

$a_1 = 3 - 1 = 2$

$b_1 = 4 - (-1) = 5$

$c_1 = -2 - 2 = -4$

Thus, the direction vector for $L_1$ is $\vec{v_1} = 2\hat{i} + 5\hat{j} - 4\hat{k}$. [By the definition of direction vectors in 3D space].

Step 2: Calculating Direction Ratios for Line 2 ($L_2$)

Let $L_2$ be the line passing through points $C(0, 3, 2)$ and $D(3, 5, 6)$. Applying the same formula:

$a_2 = 3 - 0 = 3$

$b_2 = 5 - 3 = 2$

$c_2 = 6 - 2 = 4$

Thus, the direction vector for $L_2$ is $\vec{v_2} = 3\hat{i} + 2\hat{j} + 4\hat{k}$.

Step 3: Verification of Perpendicularity Condition

Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if and only if the dot product of their direction vectors is zero: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$. [Per the condition of perpendicularity for vectors].

Substituting the values calculated:

$(2)(3) + (5)(2) + (-4)(4)$

$= 6 + 10 - 16$

$= 16 - 16$

$= 0$

Step 4: Conclusion

Since the sum of the products of the corresponding direction ratios is equal to zero, the vectors are orthogonal. Therefore, line $AB$ is perpendicular to line $CD$.

Final Solution: The lines are proved to be perpendicular as $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Step 1: Identifying Given Parameters

To determine the equation of a line in 3D space, we require a point through which the line passes and a vector indicating its direction.

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

[Per the standard definition of a line in vector form, $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\lambda$ is a scalar.]

Step 2: Deriving the Vector Equation

Substituting the given vectors $\vec{a}$ and $\vec{b}$ into the standard formula:

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

Where $\vec{r}$ represents the position vector of any arbitrary point $(x, y, z)$ on the line and $\lambda \in \mathbb{R}$.

Step 3: Deriving the Cartesian Equation

Let the point $A$ be represented by coordinates $(x_1, y_1, z_1) = (2, -1, 4)$.

Let the direction ratios of the vector $\vec{b}$ be $(a, b, c) = (1, 2, -1)$.

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

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

Substituting the values:

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

Simplifying the expression:

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

Visual Representation of the Directional Vector

Analytical Summary

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: Identification of Given Parameters

The problem requires the determination of the Cartesian equation of a line passing through a fixed point $A(x_1, y_1, z_1)$ and parallel to a given vector $\vec{b}$.

• Point $A$ through which the line passes: $(-2, 4, -5)$. Thus, $x_1 = -2$, $y_1 = 4$, $z_1 = -5$.
• The given line equation is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.
• The direction ratios of the given line are $(a, b, c) = (3, 5, 6)$. [Per the standard form of a line equation $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$].

Step 2: Establishing the Parallel Condition

Since the required line is parallel to the given line, their direction ratios must be proportional. In this specific case, the required line can share the exact same direction ratios as the given line, $\vec{v} = 3\hat{i} + 5\hat{j} + 6\hat{k}$.

[Justification: Two lines are parallel if and only if their direction vectors are scalar multiples of each other. By choosing the same direction vector $(3, 5, 6)$, we satisfy the condition of parallelism.]

Step 3: Formulation of the Cartesian Equation

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

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

Substituting the identified values:

• $x_1 = -2 \implies x - (-2) = x + 2$
• $y_1 = 4 \implies y - 4$
• $z_1 = -5 \implies z - (-5) = z + 5$
• $a = 3, b = 5, c = 6$

Step 4: Geometrical Representation

Step 5: Final Equation Assembly

Combining the variables into the standard Cartesian form:

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

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

Step 1: Establishing the Theoretical Framework

To prove that three points $A$, $B$, and $C$ are collinear, we must demonstrate that they lie on the same straight line. In three-dimensional coordinate geometry, three points are collinear if the direction ratios of the line segments joining them are proportional. Alternatively, one can show that the distance between the two furthest points is equal to the sum of the distances between the adjacent segments [By the Triangle Inequality Theorem, if $d(A, B) + d(B, C) = d(A, C)$, the points are collinear].

Step 2: Calculating Direction Ratios

Let the given points be:

• $A = (2, 3, 4)$
• $B = (-1, -2, 1)$
• $C = (5, 8, 7)$

The direction ratios $(a, b, c)$ of a line segment connecting $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

For vector $\vec{AB}$:

$a_1 = -1 - 2 = -3$

$b_1 = -2 - 3 = -5$

$c_1 = 1 - 4 = -3$

Direction ratios of $AB = (-3, -5, -3)$.

For vector $\vec{BC}$:

$a_2 = 5 - (-1) = 6$

$b_2 = 8 - (-2) = 10$

$c_2 = 7 - 1 = 6$

Direction ratios of $BC = (6, 10, 6)$.

Step 3: Verification of Proportionality

Two lines are parallel (or collinear if they share a common point) if their direction ratios are proportional [Condition of Collinearity: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$].

Taking the ratio of components of $\vec{AB}$ and $\vec{BC}$:

$\frac{a_1}{a_2} = \frac{-3}{6} = -\frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-5}{10} = -\frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-3}{6} = -\frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$, the direction ratios are proportional. Because the line segments $AB$ and $BC$ share a common point $B$, the points $A$, $B$, and $C$ must lie on the same straight line.

Step 4: Spatial Representation

Final Solution: Since the direction ratios of the segments $AB$ and $BC$ are proportional and the segments share the common point $B$, the points $(2, 3, 4)$, $(-1, -2, 1)$, and $(5, 8, 7)$ are proved to be collinear.

Step 1: Identifying Parameters from Symmetric Equations

The standard symmetric form of a line is $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$. Comparing this to the given lines:

• Line 1: $\frac{x-(-1)}{7} = \frac{y-(-1)}{-6} = \frac{z-(-1)}{1}$
• Line 2: $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$

We extract the point vectors $\vec{a_1}, \vec{a_2}$ and direction vectors $\vec{b_1}, \vec{b_2}$:

• $\vec{a_1} = -\hat{i} - \hat{j} - \hat{k}$
• $\vec{b_1} = 7\hat{i} - 6\hat{j} + \hat{k}$
• $\vec{a_2} = 3\hat{i} + 5\hat{j} + 7\hat{k}$
• $\vec{b_2} = \hat{i} - 2\hat{j} + \hat{k}$

Step 2: Calculating Vector Components

Calculate the difference vector $(\vec{a_2} - \vec{a_1})$:

$\vec{a_2} - \vec{a_1} = (3 - (-1))\hat{i} + (5 - (-1))\hat{j} + (7 - (-1))\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k}$ [Vector Subtraction Axiom]

Calculate the cross product $(\vec{b_1} \times \vec{b_2})$:

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}$ [Determinant Expansion]

$= \hat{i}(-6 + 2) - \hat{j}(7 - 1) + \hat{k}(-14 + 6) = -4\hat{i} - 6\hat{j} - 8\hat{k}$

Step 3: Calculating Magnitude of Cross Product

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2}$ [Magnitude Formula]

$= \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$

Step 4: Applying Shortest Distance Formula

The shortest distance $d$ between two skew lines is defined by $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ [Per Shortest Distance Theorem for Skew Lines].

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

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

$= -16 - 36 - 64 = -116$

Substitute into the formula:

$d = \left| \frac{-116}{2\sqrt{29}} \right| = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}}$

Rationalize the denominator:

$d = \frac{58\sqrt{29}}{29} = 2\sqrt{29}$

Visual Representation of Geometry

Final Solution: The shortest distance between the two lines is $2\sqrt{29}$ units.

Step 1: Identification of Direction Vectors

The equation of a line in symmetric form is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ represents the direction vector of the line. [Per the Cartesian representation of lines in $\mathbb{R}^3$].

For the given pair of lines:

• Line 1: $\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$
  Direction vector $\vec{a_1} = 2\hat{i} + 5\hat{j} - 3\hat{k}$
• Line 2: $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$
  Direction vector $\vec{a_2} = -1\hat{i} + 8\hat{j} + 4\hat{k}$

Step 2: Theoretical Formulation for Angle Between Two Lines

The acute angle $\theta$ between two lines with direction vectors $\vec{a_1}$ and $\vec{a_2}$ is defined by the dot product formula: [By the definition of the scalar product in Euclidean space]:

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

Step 3: Execution of Scalar Computations

Part A: Calculate the dot 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: Calculate the magnitudes $|\vec{a_1}|$ and $|\vec{a_2}|$

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

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

Step 4: Substitution and Final Angle Calculation

Substituting the values into the cosine formula:

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

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

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

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

Visual Representation of Direction Vectors

Final Solution: \theta = \cos^{-1} \left( \frac{26}{9\sqrt{38}} \right)

Step 1: Identification of Parameters

The shortest distance $d$ between two skew lines given by $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is calculated using 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 the shortest distance between skew lines].

From the given equations:

• $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$
• $\vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}$
• $\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}$
• $\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$

Step 2: Vector Subtraction and Cross Product

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 Magnitude and Dot Product

Calculate the magnitude $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + (3)^2 + (9)^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$

Calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

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

Step 4: Final Distance Calculation

Substitute the values into the formula:

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

Final Solution: The shortest distance is $\frac{3}{\sqrt{19}}$ units (or rationalized as $\frac{3\sqrt{19}}{19}$ units).

Step 1: Conversion to Standard Vector Form

The standard vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$. We rearrange the given equations to isolate the position vector $\vec{a}$ and the direction vector $\vec{b}$.

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

Separating terms with and without $t$:

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

Here, $\vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}$.

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

Separating terms with and without $s$:

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

Here, $\vec{a_2} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}$.

Step 2: Calculating Necessary Vectors

The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ is given by the formula:

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

Vector $(\vec{a_2} - \vec{a_1})$:

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

Vector Cross Product $(\vec{b_1} \times \vec{b_2})$:

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

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

$= 2\hat{i} - 4\hat{j} - 3\hat{k}$

Step 3: Calculating Magnitudes and Dot Product

Magnitude of $(\vec{b_1} \times \vec{b_2})$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$

Dot Product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k})$

$= (0)(2) + (1)(-4) + (-4)(-3) = 0 - 4 + 12 = 8$

Step 4: Computing Final Distance

$d = \frac{|8|}{\sqrt{29}} = \frac{8}{\sqrt{29}}$

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