Problem 8 - Essential Calculus Early Transcendentals

Question

\(6-10=\) Find parametric equations and symmetric equations for the line. The line through \((2,1,0)\) and perpendicular to both \(\mathbf{i}+\mathbf{j}\) and \(\mathbf{j}+\mathbf{k}\)

Step-by-Step Solution

Answer

Parametric: \(x = 2 + t, \ y = 1 - t, \ z = t\). Symmetric: \(\frac{x-2}{1} = \frac{y-1}{-1} = \frac{z}{1}\).

Step 1: Identify Direction Vector

To find the line perpendicular to both \( \mathbf{i} + \mathbf{j} \) and \( \mathbf{j} + \mathbf{k} \), calculate the cross product of these vectors: \((1, 1, 0) \times (0, 1, 1)\).

Step 2: Calculate Cross Product

Calculate the cross product: \\[(1, 1, 0) \times (0, 1, 1) = |\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 0 \ 0 & 1 & 1 \end{array}| = \mathbf{i}(1*1 - 0*1) - \mathbf{j}(1*1 - 0*0) + \mathbf{k}(1*1 - 1*0) = (1, -1, 1)\] \The direction vector for the line is \((1, -1, 1)\).

Step 3: Write Parametric Equations

With the point \((2, 1, 0)\) and direction vector \((1, -1, 1)\), the parametric equations are: \\[ x = 2 + t, \quad y = 1 - t, \quad z = 0 + t \] where \(t\) is the parameter.

Step 4: Formulate Symmetric Equations

Using the parametric equations \( x = 2 + t \), \( y = 1 - t \), and \( z = 0 + t \), the symmetric form is: \\[ \frac{x-2}{1} = \frac{y-1}{-1} = \frac{z-0}{1} \] \These represent the symmetric equations of the line.

Key Concepts

Cross Product

The cross product is a mathematical operation on two vectors in three-dimensional space, often used in physics and engineering. It results in a third vector that is perpendicular to the plane formed by the original two vectors. This is crucial in determining directions in space.

The cross product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is calculated as follows:

• \( \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} + (a_3b_1 - a_1b_3)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k} \)

In our exercise, we wanted to find a line that is perpendicular to both given vectors, \( \mathbf{i} + \mathbf{j} \) and \( \mathbf{j} + \mathbf{k} \). To do this, we took their cross product:

• \((1, 1, 0) \times (0, 1, 1) = (1, -1, 1)\)

This result gives us a direction vector perpendicular to both initial vectors, assuring that our line satisfies the perpendicularity condition.

Direction Vector

A direction vector determines the direction of a line in space. When creating parametric equations for a line, identifying the correct direction vector is a key step.

In our exercise, after finding the cross product, we obtained the direction vector \( \mathbf{d} = (1, -1, 1) \). This vector is crucial for defining the path of the line through space.

The direction vector, combined with a point on the line, helps us to write parametric equations. These equations allow us to express the coordinates of any point on the line using a single parameter, \( t \):

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

For this problem, with the initial point \((2, 1, 0)\) and the direction vector \((1, -1, 1)\), the parametric equations become:

• \( x = 2 + t \)
• \( y = 1 - t \)
• \( z = 0 + t \)

Symmetric Equations

Symmetric equations provide a neat and concise way to express lines in three-dimensional geometry. They are derived from parametric equations and show the relationships between the coordinates of any points on a line.

From the parametric equations we derived earlier, the symmetric form arises by solving each equation for the parameter \( t \) and setting them equal:

• Given, \( x = 2 + t \), \( y = 1 - t \), and \( z = t \).
• Solving for \( t \), we have \( t = x - 2 \), \( t = 1 - y \), and \( t = z \).

We equate these to form the symmetric equations:

• \( \frac{x-2}{1} = \frac{y-1}{-1} = \frac{z-0}{1} \)

These equations show that points \( x \), \( y \), and \( z \) on the line maintain a specific relationship with each other, simplifying visualization and calculations.