Problem 12 - APEX Calculus

Question

Write the vector, parametric and symmetric equations of the lines described. Passes through the point of intersection of \(\ell_{1}(t)\) and \(\ell_{2}(t)\) and orthogonal to both lines, where \(\ell_{1}=\left\\{\begin{array}{l}x=t \\\ y=-2+2 t \\ z=1+t\end{array}\right.\) and \(\ell_{2}=\left\\{\begin{array}{l}x=2+t \\ y=2-t \\ z=3+2 t\end{array}\right.\)

Step-by-Step Solution

Answer

Intersection at (2, 2, 3), direction <5, -1, -3>. Equations: vector \(\mathbf{r} = (2,2,3) + t(5,-1,-3)\), parametric (x=2+5t, y=2-t, z=3-3t), symmetric \((x-2)/5 = (y-2)/-1 = (z-3)/-3\).

Step 1: Find Intersection of Lines

First, we need to find the point where the two lines \( \ell_1 \) and \( \ell_2 \) intersect. By equating the parametric equations of both lines, we'll solve for the parameter values where both lines meet. Set \( x_1(t_1) = x_2(t_2) \), \( y_1(t_1) = y_2(t_2) \), and \( z_1(t_1) = z_2(t_2) \): \[ t = 2 + t_2 \] \[ -2 + 2t = 2 - t_2 \] \[ 1 + t = 3 + 2t_2 \]From the first equation, \( t_2 = t - 2 \). Substitute \( t_2 \) in the second and third equations to obtain \( t \):\[ -2 + 2t = 2 - (t - 2) \implies 3t = 6 \implies t = 2 \]Substitute \( t = 2 \) into \( t_2 = t - 2 \) gives \( t_2 = 0 \).Substitute \( t = 2 \) into \( \ell_1 \) to get the intersection point \((2, 2, 3)\).

Step 2: Find the Direction Vector Orthogonal to Both Lines

Next, we need to find a vector orthogonal to both lines. This can be found by taking the cross product of their direction vectors. For \( \ell_1 \), the direction vector is \<1, 2, 1\>. For \( \ell_2 \), the direction vector is \<1, -1, 2\>. Compute the cross product:\[ \begin{vmatrix} i & j & k \ 1 & 2 & 1 \ 1 & -1 & 2 \end{vmatrix} = i(4 + 1) - j(2 - 1) + k(-1 - 2) = 5i - j + (-3)k \]The orthogonal vector is \<5, -1, -3\>.

Step 3: Write the Vector Equation

Now, we can write the vector equation of the line passing through the intersection point \((2, 2, 3)\) with direction \<5, -1, -3\>:\[ \mathbf{r} = \begin{pmatrix} 2 \ 2 \ 3 \end{pmatrix} + t\begin{pmatrix} 5 \ -1 \ -3 \end{pmatrix} \]

Step 4: Write the Parametric Equations

From the vector equation, derive the parametric equations. Each component gives:\[ x = 2 + 5t \] \[ y = 2 - t \] \[ z = 3 - 3t \]

Step 5: Write the Symmetric Equations

Eliminate the parameter \( t \) from the parametric equations to get the symmetric form:\[ \frac{x-2}{5} = \frac{y-2}{-1} = \frac{z-3}{-3} \]

Key Concepts

Parametric Equations

Parametric equations are a way to express geometric figures, such as lines or curves, using a set of parameters. These parameters allow us to describe the coordinates of points along the line as they vary. The key feature is observing how the parameter changes the position of a point in space. In the case of a line, parametric equations can typically be written in the form:

• For x: \( x = x_0 + at \)
• For y: \( y = y_0 + bt \)
• For z: \( z = z_0 + ct \)

Here, \( (x_0, y_0, z_0) \) is a fixed point on the line, and \( (a, b, c) \) is the direction vector that indicates the direction in which the line extends. Understanding this set-up helps bridge the connection to writing a line's equation given any line problem in vector calculus and algebra.

Symmetric Equations

Symmetric equations are another powerful way to describe lines in three-dimensional space. They are derived by eliminating the parameter from parametric equations. This form is particularly nice because it provides a clear condition that points must satisfy to lie on the line.
The general symmetric equation form obtained from parametric equations is:
\[\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\]This representation is neat and compact, presenting conditions directly required for any point \((x, y, z)\) to be on the line. They serve as a quick check to see if a point lies on the line by inserting the values into the appropriate fractions and verifying equality.
When working with symmetric equations, it's important to note that a zero is not permitted in a denominator as it would make the equation undefined, so care must be taken when transitioning from parametric forms where any of \(a\), \(b\), or \(c\) are zero.

Intersection of Lines

Finding the intersection of lines involves solving for points that satisfy the equations of two lines simultaneously. Two lines in space can intersect at a single point if they are not parallel and not skew (i.e., non-parallel and non-intersecting).
To find this intersection point, set the parametric equations of the two lines equal to each other and solve for the parameters. This involves setting:

• \( x_1 = x_2 \) for the x-components,
• \( y_1 = y_2 \) for the y-components,
• \( z_1 = z_2 \) for the z-components.

Subsequently, solve these simultaneous equations to find the parameter values, which can then be used to determine the actual point of intersection of the two lines in space. This process highlights the relationship and point-sharing between intersecting lines.

Cross Product

The cross product is a mathematical operation that can be used to find a vector perpendicular to two given vectors in three-dimensional space. It is especially useful in line problems where one needs a direction vector orthogonal to two others, such as when determining a plane or a new line.
The formula for computing the cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = \mathbf{i}(a_2b_3 - a_3b_2) - \mathbf{j}(a_1b_3 - a_3b_1) + \mathbf{k}(a_1b_2 - a_2b_1)\]This results in a new vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). The magnitude of this vector also represents the area of the parallelogram formed by the two vectors. The cross product is an essential tool for understanding orientation and space in vector calculus.