Problem 11 - APEX Calculus

Question

Write the vector, parametric and symmetric equations of the lines described. Passes through the point of intersection of \(\vec{\ell}_{1}(t)\) and \(\vec{\ell}_{2}(t)\) and orthogonal to both lines, where \(\vec{\ell}_{1}(t)=\langle 2,1,1\rangle+t\langle 5,1,-2\rangle\) and \(\overrightarrow{\ell_{2}}(t)=\langle-2,-1,2\rangle+t\langle 3,1,-1\rangle\)

Step-by-Step Solution

Answer

Intersection point: \((\frac{11}{3}, \frac{4}{3}, \frac{1}{3})\); Vector equation: \(\vec{r}(t) = \langle \frac{11}{3}, \frac{4}{3}, \frac{1}{3} \rangle + t\langle 1, -7, -2 \rangle\); Symmetric equations: \(x - \frac{11}{3} = \frac{y - \frac{4}{3}}{-7} = \frac{z - \frac{1}{3}}{-2}\).

Step 1: Find Intersection Point

To find the intersection of the lines, we set the parametric equations equal: \[ (2 + 5t_1, 1 + t_1, 1 - 2t_1) = (-2 + 3t_2, -1 + t_2, 2 - t_2) \].Solving the system of equations derived from equating components, we get:- \[ 2 + 5t_1 = -2 + 3t_2 \]- \[ 1 + t_1 = -1 + t_2 \]- \[ 1 - 2t_1 = 2 - t_2 \].Solve this system to get values for \(t_1\) and \(t_2\), giving: \( t_1 = \frac{1}{3}, t_2 = \frac{5}{3} \).Plug back into either line's equation to get the intersection point as \((rac{11}{3}, \frac{4}{3}, \frac{1}{3})\).

Step 2: Find a Direction Vector

Since the line is orthogonal to both given lines, use the cross product of their direction vectors:- **Direction vector of \(\vec{\ell}_1\):** \( \langle 5, 1, -2 \rangle \)- **Direction vector of \(\overrightarrow{\ell_{2}}\):** \( \langle 3, 1, -1 \rangle \)Cross product: \(\langle 5, 1, -2 \rangle \times \langle 3, 1, -1 \rangle = \langle 1, -7, -2 \rangle\).

Step 3: Write the Vector Equation

The vector equation of a line can be given by:\[ \vec{r}(t) = \vec{a} + t\vec{d} \]where \(\vec{a}\) is the point on the line \((\frac{11}{3}, \frac{4}{3}, \frac{1}{3})\), and \(\vec{d}\) is the direction vector \(\langle 1, -7, -2 \rangle\).So the vector equation is:\[ \vec{r}(t) = \langle \frac{11}{3}, \frac{4}{3}, \frac{1}{3} \rangle + t\langle 1, -7, -2 \rangle \].

Step 4: Write the Parametric Equations

Based on the vector equation\[ \vec{r}(t) = \langle \frac{11}{3}, \frac{4}{3}, \frac{1}{3} \rangle + t\langle 1, -7, -2 \rangle \],the parametric equations are:- \( x = \frac{11}{3} + t \)- \( y = \frac{4}{3} - 7t \)- \( z = \frac{1}{3} - 2t \)

Step 5: Write the Symmetric Equations

From the parametric equations, to get the symmetric form, solve each for \(t\) and set them equal:- \( t = x - \frac{11}{3} \)- \( t = \frac{y - \frac{4}{3}}{-7} \)- \( t = \frac{z - \frac{1}{3}}{-2} \).Therefore, the symmetric equations are:\[ x - \frac{11}{3} = \frac{y - \frac{4}{3}}{-7} = \frac{z - \frac{1}{3}}{-2} \].

Key Concepts

Parametric Equations

To understand the parametric equations of a line, think of them as the representation of a line in space where each parameter—usually denoted as "t"—determines a point on the line. In simple terms, parametric equations allow us to describe a line by defining each coordinate in terms of a single parameter.

For example, given the vector equation \( \vec{r}(t) = \langle \frac{11}{3}, \frac{4}{3}, \frac{1}{3} \rangle + t\langle 1, -7, -2 \rangle \),we can express it in parametric form by splitting it into individual components:

• \( x = \frac{11}{3} + t \)
• \( y = \frac{4}{3} - 7t \)
• \( z = \frac{1}{3} - 2t \)

Here, \( t \) is the parameter that can vary any real number, allowing us to trace every point along the line. This method is particularly advantageous when working with 3D geometry because it gives a straightforward way to calculate missing points or to visualize the line.

Symmetric Equations

Symmetric equations offer another method to describe a line in space. Unlike parametric equations, symmetric equations eliminate the parameter t by integrating the individual equations into a single line of equality.

Starting from the parametric form:

• \( x = \frac{11}{3} + t \)
• \( y = \frac{4}{3} - 7t \)
• \( z = \frac{1}{3} - 2t \)

We solve each for \( t \) and set them equal:

• \( t = x - \frac{11}{3} \)
• \( t = \frac{y - \frac{4}{3}}{-7} \)
• \( t = \frac{z - \frac{1}{3}}{-2} \)

Therefore, the symmetric equation is:\[ x - \frac{11}{3} = \frac{y - \frac{4}{3}}{-7} = \frac{z - \frac{1}{3}}{-2}\]

This expression succinctly correlates each coordinate to the parameter, clearly defining the alignment of the line in space. Symmetric equations are particularly useful for quickly visualizing and comparing proportions between different coordinates without the explicit parameter.

Cross Product

The cross product is a fundamental operation in vector algebra used to find a vector that is orthogonal (perpendicular) to two given vectors. It is an essential tool for computing the direction of a line that is orthogonal to two other lines, which is often required in vector calculus and physics.

To calculate it, take two direction vectors from the provided lines, say \( \langle 5, 1, -2 \rangle \)and \( \langle 3, 1, -1 \rangle \), and find their cross product:

• The cross product formula involves a determinant:\[\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \5 & 1 & -2 \3 & 1 & -1\end{vmatrix}\]
• This results in:\( \langle (1)(-1) - (-2)(1), (-2)(3) - (5)(-1), (5)(1) - (1)(3) \rangle = \langle 1, -7, -2 \rangle \)

This vector \( \langle 1, -7, -2 \rangle \) is perpendicular to both original vectors and can act as the direction vector for a line orthogonal to both given lines.

The cross product is critical in physics and engineering to solve problems involving torque, rotational motion, and other scenarios where perpendicular vectors are involved.

Intersection of Lines

The intersection of lines is an important concept in geometry that involves finding the point at which two lines meet in space. In vector mathematics, two lines intersect when their parametric forms lead to the same point.

To find this point, set the parametric equations of the two lines equal. For example, if you have lines defined by:

• \( \vec{\ell}_1(t) = \langle 2,1,1 \rangle + t_1\langle 5,1,-2 \rangle \)
• \( \overrightarrow{\ell_{2}}(t) = \langle -2,-1,2 \rangle + t_2\langle 3,1,-1 \rangle \)

Set the x, y, and z components equal to solve for \( t_1 \) and \( t_2 \):

• \( 2 + 5t_1 = -2 + 3t_2 \)
• \( 1 + t_1 = -1 + t_2 \)
• \( 1 - 2t_1 = 2 - t_2 \)

Solving these equations gives \( t_1 = \frac{1}{3} \) and \( t_2 = \frac{5}{3} \).

Substituting back into one line’s parametric equations finds the intersection point: \( \left( \frac{11}{3}, \frac{4}{3}, \frac{1}{3} \right) \). This is where both lines cross, representing their precise meet-up in 3D space, crucial for problems in 3D design, architecture, and physics when designing intersections of beams or paths.