Problem 9 - APEX Calculus

Question

Write the vector, parametric and symmetric equations of the lines described. Passes through \(P=(0,1,2)\) and orthogonal to both \(\vec{d}_{1}=\langle 2,-1,7\rangle\) and \(\vec{d}_{2}=\langle 7,1,3\rangle\)

Step-by-Step Solution

Answer

The line's equations are: vector \(\vec{r}(t) = \langle 0,1,2 \rangle + t\langle -10,43,9 \rangle\), parametric \(x = -10t, y = 1+43t, z = 2+9t\), symmetric \(\frac{x}{-10} = \frac{y-1}{43} = \frac{z-2}{9}\).

Step 1: Find the direction vector

The line must be orthogonal to both given direction vectors, \(\vec{d}_{1}=\langle 2,-1,7 \rangle\) and \(\vec{d}_{2}=\langle 7,1,3 \rangle\). Thus, we find the cross product of these two vectors, \(\vec{d} = \vec{d}_{1} \times \vec{d}_{2}\), to find the direction vector of the line. Calculate the cross product:\[\vec{d} = \langle 2,-1,7 \rangle \times \langle 7,1,3 \rangle\]\[= \langle (-1)(3) - (7)(1), (7)(7) - (2)(3), (2)(1) - (-1)(7) \rangle\]\[= \langle -3 - 7, 49 - 6, 2 + 7 \rangle\]\[= \langle -10, 43, 9 \rangle\]

Step 2: Write the vector equation of the line

The vector equation of the line can be expressed as:\[\vec{r}(t) = \vec{r}_{0} + t\vec{d}\]where \(\vec{r}_{0}\) is the position vector of the point \(P=(0,1,2)\) and \(\vec{d}\) is the direction vector \(\langle -10, 43, 9 \rangle\). Thus, the vector equation is:\[\vec{r}(t) = \langle 0,1,2 \rangle + t\langle -10, 43, 9 \rangle\]

Step 3: Write the parametric equations of the line

From the vector equation \(\vec{r}(t) = \langle 0,1,2 \rangle + t\langle -10, 43, 9 \rangle\), we can express the parametric equations by matching the components:\[x = 0 - 10t\]\[y = 1 + 43t\]\[z = 2 + 9t\]

Step 4: Write the symmetric equation of the line

The symmetric equation is derived from the parametric equations by solving for \(t\) in each equation, assuming the direction vector components are not zero:\[x = -10t \rightarrow t = \frac{x}{-10}\]\[y = 1 + 43t \rightarrow t = \frac{y - 1}{43}\]\[z = 2 + 9t \rightarrow t = \frac{z - 2}{9}\]Thus, the symmetric equation is:\[\frac{x}{-10} = \frac{y-1}{43} = \frac{z-2}{9}\]

Key Concepts

Cross Product

The cross product is a fundamental operation in vector mathematics that is primarily used in three dimensions. It results in a vector that is perpendicular to two given vectors, making it very useful for finding orthogonal directions. When you take the cross product of two vectors, say \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), you calculate it using the formula:

• \( \vec{a} \times \vec{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \)

In the context of the provided exercise, the cross product is used to find a direction vector for a line that is orthogonal to two given vectors \( \vec{d}_1 \) and \( \vec{d}_2 \). Since the line must be perpendicular to both, the cross product \( \vec{d}_1 \times \vec{d}_2 \) provides a direction vector precisely aligned with this requirement.
For the exercise, calculating the cross product of \( \langle 2, -1, 7 \rangle \) and \( \langle 7, 1, 3 \rangle \) results in the direction vector \( \langle -10, 43, 9 \rangle \). This vector is the backbone around which the parametric and symmetric equations are constructed.

Parametric Equations

Parametric equations specify the components of a line using one or more parameters—in this case, the parameter \( t \). These equations make it easy to describe a line in three-dimensional space. They are derived from a vector equation of a line, which is expressed as:

• \( \vec{r}(t) = \vec{r}_0 + t \vec{d} \)

Here, \( \vec{r}_0 \) is the position vector of a known point on the line, and \( \vec{d} \) is the direction vector. For the exercise at hand, the position vector is \( \langle 0, 1, 2 \rangle \), and the direction vector computed from the cross product is \( \langle -10, 43, 9 \rangle \).
This gives us the parametric equations:

• \( x = 0 - 10t \)
• \( y = 1 + 43t \)
• \( z = 2 + 9t \)

Each equation corresponds to one of the three spatial dimensions, describing the position of any point along the line as \( t \) varies. This makes it straightforward to compute points on the line by simply plugging different \( t \) values into these equations.

Symmetric Equations

Symmetric equations represent a line by eliminating the parameter \( t \) from the parametric equations, resulting in an equation that directly relates the coordinates \( x, y, \) and \( z \). This type of equation is sometimes more convenient for determining the relative positions of points along the line.
To form symmetric equations, you first isolate \( t \) in each parametric equation:

• From \( x = 0 - 10t \), solve to get \( t = \frac{x}{-10} \)
• From \( y = 1 + 43t \), solve to get \( t = \frac{y - 1}{43} \)
• From \( z = 2 + 9t \), solve to get \( t = \frac{z - 2}{9} \)

The equations are then set equal to each other, resulting in:

• \( \frac{x}{-10} = \frac{y-1}{43} = \frac{z-2}{9} \)

This symmetric form nicely encapsulates the line's geometry by showing how the coordinates are interrelated, without directly involving the parameter \( t \). It can be helpful for intersection and coplanarity problems.