Find the parametric and symmetric equation of a line that is orthogonal to the line through the points (–1, 4, –3) and (4, 2, –1). The work should lead to an answer of x = 4+4t, y = 2+5t, and z = -1-5t

To find the parametric and symmetric equations of the line orthogonal to the line passing through the points \((-1, 4, -3)\) and \((4, 2, -1)\), we can follow these steps:

### Step 1: Find the direction vector of the given line
The direction vector \(\mathbf{v}\) of the line through the points \((-1, 4, -3)\) and \((4, 2, -1)\) can be calculated by subtracting the coordinates of the first point from the second point:
\[
\mathbf{v} = \langle 4 - (-1), 2 - 4, -1 - (-3) \rangle = \langle 4 + 1, 2 - 4, -1 + 3 \rangle = \langle 5, -2, 2 \rangle
\]
So, the direction vector of the given line is \(\mathbf{v} = \langle 5, -2, 2 \rangle\).

### Step 2: Find a point and direction vector for the orthogonal line
An orthogonal line to this one must have a direction vector perpendicular to \(\mathbf{v} = \langle 5, -2, 2 \rangle\). Suppose the parametric equation of the orthogonal line has the form:
\[
x = 4 + at, \quad y = 2 + bt, \quad z = -1 + ct
\]
where \(\langle a, b, c \rangle\) is the direction vector of the orthogonal line. Since the lines are orthogonal, the dot product of the direction vectors must be zero:
\[
\mathbf{v} \cdot \langle a, b, c \rangle = 0
\]
This gives:
\[
5a + (-2)b + 2c = 0
\]
or, equivalently:
\[
5a - 2b + 2c = 0 \quad \text{(Equation 1)}
\]

### Step 3: Solve for the components of the orthogonal direction vector
We are also given the parametric equations of the orthogonal line as:
\[
x = 4 + 4t, \quad y = 2 + 5t, \quad z = -1 - 5t
\]
So the direction vector of the orthogonal line is \(\langle 4, 5, -5 \rangle\). Plugging these values of \(a = 4\), \(b = 5\), and \(c = -5\) into Equation 1:
\[
5(4) - 2(5) + 2(-5) = 0
\]
\[
20 - 10 - 10 = 0
\]
This holds true, confirming that the direction vector \(\langle 4, 5, -5 \rangle\) is perpendicular to \(\langle 5, -2, 2 \rangle\).

### Step 4: Write the parametric equations
Now, using the point \((4, 2, -1)\) and the direction vector \(\langle 4, 5, -5 \rangle\), the parametric equations of the orthogonal line are:
\[
x = 4 + 4t
\]
\[
y = 2 + 5t
\]
\[
z = -1 - 5t
\]

### Step 5: Find the symmetric equations
The symmetric form of the line can be obtained by solving each parametric equation for \(t\):
\[
t = \frac{x - 4}{4}, \quad t = \frac{y - 2}{5}, \quad t = \frac{z + 1}{-5}
\]
Equating these expressions gives the symmetric equation:
\[
\frac{x - 4}{4} = \frac{y - 2}{5} = \frac{z + 1}{-5}
\]

Thus, the parametric equations are:
\[
x = 4 + 4t, \quad y = 2 + 5t, \quad z = -1 - 5t
\]
and the symmetric equation is:
\[
\frac{x - 4}{4} = \frac{y - 2}{5} = \frac{z + 1}{-5}
\]

Would you like a deeper explanation on any of the steps, or more details on how the direction vectors were derived?

### Related questions:
1. How do you find the direction vector of a line given two points?
2. What are the conditions for two lines to be orthogonal in 3D space?
3. How do you convert parametric equations into symmetric equations?
4. Can you describe how to find the equation of a line parallel to a given line?
5. How would the parametric equations change if the line passed through different points?

**Tip:** When calculating the direction vector of a line, always subtract the coordinates in the correct order to avoid sign mistakes.

Find the parametric and symmetric equation of a line that is orthogonal to the line through the points (–1, 4, –3) and (4, 2, –1). The work should lead to an answer of x = 4+4t, y = 2+5t, and z = -1-5t.

Learn how to find the parametric and symmetric equations of a line orthogonal to a line passing through two points in 3D space. This example walks through vector direction, orthogonality using dot products, and forming parametric equations. Suitable for students in Grades 10-12.

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