Find a vector equation and parametric equations for the line. (Use the parameter t.)
the line through the point 
(8, 0, −3)
 and parallel to the line 
x = 4 − 4t, y = −1 + 3t, z = 6 + 9t

To find the vector equation and parametric equations for the line that passes through the point \((8, 0, -3)\) and is parallel to the line given by the parametric equations

\[
x = 4 - 4t, \quad y = -1 + 3t, \quad z = 6 + 9t,
\]

we can follow these steps:

### Step 1: Extract the direction vector
The parametric equations of the given line are written as:
\[
x = 4 - 4t, \quad y = -1 + 3t, \quad z = 6 + 9t.
\]

The direction vector \(\mathbf{d}\) for this line can be read from the coefficients of \(t\), i.e., the vector formed by the values multiplying \(t\) in each equation. Thus, the direction vector \(\mathbf{d}\) is:
\[
\mathbf{d} = \langle -4, 3, 9 \rangle.
\]

Since the new line is parallel to the given line, the direction vector of the new line will also be \(\langle -4, 3, 9 \rangle\).

### Step 2: Write the vector equation of the line
The vector equation of a line can be written as:
\[
\mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{d},
\]
where \(\mathbf{r}_0\) is the position vector of a point on the line, and \(\mathbf{d}\) is the direction vector.

Here, the point on the line is \((8, 0, -3)\), so \(\mathbf{r}_0 = \langle 8, 0, -3 \rangle\). Using \(\mathbf{d} = \langle -4, 3, 9 \rangle\), the vector equation of the line becomes:
\[
\mathbf{r}(t) = \langle 8, 0, -3 \rangle + t \langle -4, 3, 9 \rangle.
\]
Simplifying:
\[
\mathbf{r}(t) = \langle 8 - 4t, 3t, -3 + 9t \rangle.
\]

### Step 3: Write the parametric equations
From the vector equation, we can extract the parametric equations:
\[
x = 8 - 4t,
\]
\[
y = 3t,
\]
\[
z = -3 + 9t.
\]

### Final Answer
- **Vector equation**: 
  \[
  \mathbf{r}(t) = \langle 8 - 4t, 3t, -3 + 9t \rangle
  \]

- **Parametric equations**:
  \[
  x = 8 - 4t, \quad y = 3t, \quad z = -3 + 9t.
  \]

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you find the direction vector for any given parametric line?
2. What are the differences between vector and parametric equations of lines?
3. How do you find the parametric equations of a line given two points?
4. Can a vector equation be expressed using matrices?
5. How would the vector equation change if the direction vector were reversed?

**Tip**: When writing parametric equations for lines, the coefficients of the parameter (like \(t\)) represent the direction ratios of the line.

Find a vector equation and parametric equations for the line through the point (8, 0, -3) and parallel to the line x = 4 − 4t, y = −1 + 3t, z = 6 + 9t.

Find the vector and parametric equations for a line passing through the point (8, 0, -3) and parallel to another given line. This solution involves vector geometry and parametric equations, essential for students in Grades 10-12.

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