Question

Write the vector, parametric and symmetric equations of the lines described. Passes through \(P=(1,-2,3)\) and \(Q=(5,5,5)\).

Short answer

The vector equation: \( \mathbf{r} = (1, -2, 3) + t(4, 7, 2) \); parametric: \( x = 1 + 4t, y = -2 + 7t, z = 3 + 2t \); symmetric: \( \frac{x-1}{4} = \frac{y+2}{7} = \frac{z-3}{2} \).

Step-by-step solution

Find the Direction Vector

A line in three-dimensional space can be characterized by a direction vector. The direction vector can be found by subtracting the coordinates of point \(P\) from point \(Q\). Let the direction vector be \(\mathbf{d} = (d_1, d_2, d_3)\). Therefore, \(\mathbf{d} = (5-1, 5-(-2), 5-3) = (4, 7, 2)\).

Write the Vector Equation

The vector equation of a line passing through a point \( \mathbf{a} = (x_0, y_0, z_0) \) with direction vector \( \mathbf{d} = (d_1, d_2, d_3) \) is given by \( \mathbf{r} = \mathbf{a} + t\mathbf{d} \), where \(t\) is a parameter. Using point \(P = (1, -2, 3)\), the vector equation is \( \mathbf{r} = (1, -2, 3) + t(4, 7, 2) \).

Write the Parametric Equations

The parametric equations can be derived from the vector equation. From \(\mathbf{r} = (1, -2, 3) + t(4, 7, 2)\), the parametric equations are: \(x = 1 + 4t\), \(y = -2 + 7t\), and \(z = 3 + 2t\).

Write the Symmetric Equations

The symmetric equations can be derived by eliminating the parameter \(t\) from the parametric equations. We can solve each parametric equation for \(t\): \(t = \frac{x-1}{4}\), \(t = \frac{y+2}{7}\), \(t = \frac{z-3}{2}\). Therefore, the symmetric equations are \(\frac{x-1}{4} = \frac{y+2}{7} = \frac{z-3}{2}\).

Key concepts

Vector Equation

In three-dimensional space, a line can be depicted uniquely by a vector equation. This equation is fundamental as it combines both a point on the line and its direction. To write a vector equation for a line, you need a known point through which the line passes and a direction vector that indicates the line's direction.
The direction vector is calculated by subtracting the coordinates of one point from another point on the line. In our example, we determine the direction vector by using points \( P = (1, -2, 3) \) and \( Q = (5, 5, 5) \).
The direction vector \( \mathbf{d} \) becomes:

• \( d_1 = 5 - 1 = 4 \)
• \( d_2 = 5 - (-2) = 7 \)
• \( d_3 = 5 - 3 = 2 \)

Thus, the vector equation becomes \( \mathbf{r} = (1, -2, 3) + t(4, 7, 2) \), where \( t \) is a scalar parameter that scales the direction vector.

Parametric Equations

Parametric equations transform the vector equation of a line into a set of individual equations that define each coordinate independently. These equations are incredibly useful for analyzing how each component of the position vector \( \mathbf{r} \) changes as the parameter \( t \) changes.
From the vector equation \( \mathbf{r} = (1, -2, 3) + t(4, 7, 2) \), we divide it into three separate equations, one for each spatial dimension:

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

Each of these equations shows how a point’s coordinates change linearly with respect to \( t \). The parameter \( t \) allows for flexibility, showing the path the line takes as \( t \) varies from negative infinity to positive infinity.

Symmetric Equations

Symmetric equations offer a concise form for the equation of a line by eliminating the parameter \( t \) from the parametric equations. This form directly relates the coordinates \( x \), \( y \), and \( z \) that a line passes through.
To derive symmetric equations, solve the parametric equations for \( t \):

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

Equating these expressions for \( t \), we obtain the symmetric equations: \[ \frac{x-1}{4} = \frac{y+2}{7} = \frac{z-3}{2} \]
Symmetric equations are efficient for visualizing the relationship between coordinates along a line without explicitly using \( t \). They are particularly useful for determining whether a point lies on a line or for solving systems of equations involving multiple lines in space.