Question

Write the vector, parametric and symmetric equations of the lines described. Passes through \(P=(5,1,9)\) and orthogonal to both \(\vec{d}_{1}=\langle 1,0,1\rangle\) and \(\vec{d}_{2}=\langle 2,0,3\rangle\)

Short answer

The line's vector equation: \(\vec{r}(t) = \langle 5, 1-t, 9 \rangle\). Parametric: \(x=5\), \(y=1-t\), \(z=9\). Symmetric: \(x=5\), \(y+1=t\), \(z=9\).

Step-by-step solution

Find the Direction Vector of the Line

The line we need to find is orthogonal to both direction vectors \( \vec{d}_1 = \langle 1, 0, 1 \rangle \) and \( \vec{d}_2 = \langle 2, 0, 3 \rangle \). To find the direction vector \( \vec{n} \) of the line, we use the cross product: \[\vec{n} = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 1 \ 2 & 0 & 3 \end{vmatrix} = \langle (0\times3 - 0\times1), -(1\times3 - 1\times2), (1\times0 - 0\times2) \rangle = \langle 0, -1, 0 \rangle.\]

Write the Vector Equation of the Line

A vector equation for a line passing through point \( P = (5, 1, 9) \) with direction vector \( \vec{n} = \langle 0, -1, 0 \rangle \) is:\[ \vec{r}(t) = \vec{p} + t \cdot \vec{n} = \langle 5, 1, 9 \rangle + t \cdot \langle 0, -1, 0 \rangle = \langle 5, 1-t, 9 \rangle.\]

Develop the Parametric Equations

The parametric equations can be extracted from the vector equation \( \vec{r}(t) = \langle 5, 1-t, 9 \rangle \):- \( x = 5 \)- \( y = 1 - t \)- \( z = 9 \).

Construct the Symmetric Equations

Since the direction vector \( \vec{n} = \langle 0, -1, 0 \rangle \) causes the \( x \) and \( z \) coordinates to remain constant, the symmetric form involves only the \( y \) component:\[ y - 1 = -t \]However, because the \( x \) and \( z \) components are constant, there is no overall symmetric form involving \( x \) or \( z \). Simply the constant values:\[ x = 5, \quad y + t = 1, \quad z = 9. \]

Key concepts

Cross Product

The cross product is a vector operation that takes two vectors and returns a third vector that is orthogonal to both of the original vectors. This operation is crucial for finding a direction vector that is perpendicular to two given vectors. Imagine you have two vectors, \( \vec{d}_1 = \langle 1, 0, 1 \rangle \) and \( \vec{d}_2 = \langle 2, 0, 3 \rangle \). By taking their cross product, we can find a vector that is perpendicular to both.
The formula for the cross product is similar to finding the determinant of a matrix:

• Calculate the determinant of the following 3x3 matrix, where \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along the x, y, and z axes:

\[\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 1 \ 2 & 0 & 3 \end{vmatrix} = \langle 0, -1, 0 \rangle\]

The resulting vector \( \langle 0, -1, 0 \rangle \) is the direction vector for our line.

Using cross products helps us find paths in space that are completely independent of the directions of our original vectors.

Direction Vector

A direction vector is essential in defining the orientation of a line in space. It tells us in which direction the line points. When given two vectors and asked to find a line orthogonal to both, we must compute a new direction vector using the cross product.
In our problem, the direction vector \( \vec{n} = \langle 0, -1, 0 \rangle \) was found by calculating the cross product of \( \vec{d}_1 \) and \( \vec{d}_2 \).

• Direction vectors can point in any direction depending on their components, \( a_x, a_y, \) and \( a_z \).
• A key feature is that direction vectors remain consistent regardless of where you place the line in space.
• This consistency means you can multiply the vector by any scalar, and the line direction will remain unchanged.

Direction vectors are a cornerstone when defining both vector and parametric equations of a line.

Parametric Equations

Parametric equations describe a line in terms of a parameter, often represented by \( t \). These equations illustrate how each coordinate of a point on the line changes with respect to \( t \). They're derived directly from the vector equation of a line.
Take, for example, the vector equation:\[ \vec{r}(t) = \langle 5, 1-t, 9 \rangle \]From this, we can extract the parametric equations:

• \( x = 5 \)
• \( y = 1 - t \)
• \( z = 9 \)

• The parametric representation is useful because it clearly shows the variable components of the line over \( t \).
• This approach is particularly helpful in understanding how the line behaves as the parameter changes.
• Parametric equations allow for geometry problems to be explored in a step-by-step process, making it easier for students to visualize.

Symmetric Equations

Symmetric equations present a more compact way of showing the equations of a line when the direction vector has non-zero components. However, in our case, the direction vector \( \langle 0, -1, 0 \rangle \) complicates things, as there are zero components.
Normally, symmetric form involves expressing each variable in a formula that equates them through constants and the parameter \( t \). But when components are constant, like in our problem, they remain untouched:

• \( x = 5 \)
• \( y + t = 1 \)
• \( z = 9 \)

Notably, symmetric equations can simplify calculations in some geometric problems by reducing a multidimensional problem into simpler, separate equations.
Despite the constraints, they effectively indicate points where the trajectory doesn't change due to constant values of \( x \) and \( z \).