Determine the equation of the line that goes through the origin and forms a right angle with r(t)= <1+2t, 3-2t, 1-t> at their point of intersection

To find the equation of the line that goes through the origin and forms a right angle with the given vector equation \( \mathbf{r}(t) = \langle 1 + 2t, 3 - 2t, 1 - t \rangle \), we need to follow these steps:

### Step 1: Find the point of intersection
Since the line goes through the origin \((0, 0, 0)\), let's first determine when the vector \( \mathbf{r}(t) \) intersects the origin.

We need to solve for \( t \) such that:
\[
1 + 2t = 0, \quad 3 - 2t = 0, \quad 1 - t = 0
\]
Solving these:

- From \( 1 + 2t = 0 \), we get \( t = -\frac{1}{2} \).
- From \( 3 - 2t = 0 \), we get \( t = \frac{3}{2} \).
- From \( 1 - t = 0 \), we get \( t = 1 \).

These values of \( t \) are not consistent. However, the system is over-determined, and this suggests that the vector \( \mathbf{r}(t) \) does not pass through the origin. Instead, we are tasked with finding the line through the origin that is orthogonal to \( \mathbf{r}(t) \) at their point of intersection, which likely refers to the position at a specific \( t_0 \).

### Step 2: Find the direction vector of the line perpendicular to \( \mathbf{r}(t) \)
To find the line perpendicular to \( \mathbf{r}(t) \), we need the direction vector of \( \mathbf{r}'(t) \), which gives the direction of the curve at any point \( t \). We will take the derivative of \( \mathbf{r}(t) \):

\[
\mathbf{r}'(t) = \frac{d}{dt} \langle 1 + 2t, 3 - 2t, 1 - t \rangle = \langle 2, -2, -1 \rangle
\]

Thus, the tangent vector at any point \( t \) is \( \mathbf{r}'(t) = \langle 2, -2, -1 \rangle \).

### Step 3: Equation of the line through the origin
The line we are looking for is orthogonal to this vector. To find the equation of this line, we use the fact that a vector perpendicular to \( \mathbf{r}'(t) = \langle 2, -2, -1 \rangle \) is needed.

One such vector perpendicular to \( \langle 2, -2, -1 \rangle \) can be found using cross products or inspection. But for simplicity, we note that since the line passes through the origin, any scalar multiple of the direction vector will suffice.

Hence, the direction vector for the line is \( \mathbf{v} = \langle 2, -2, -1 \rangle \), and the equation of the line can be written parametrically as:
\[
\mathbf{L}(s) = s \langle 2, -2, -1 \rangle = \langle 2s, -2s, -s \rangle
\]

### Final Answer:
The parametric equation of the line through the origin that forms a right angle with \( \mathbf{r}(t) \) is:
\[
\mathbf{L}(s) = \langle 2s, -2s, -s \rangle
\]

Would you like further details or have any questions about this solution?

Here are some related questions:
1. How can we verify the perpendicularity of two vectors?
2. How do parametric equations of lines differ from their Cartesian forms?
3. What is the geometric interpretation of the derivative of a vector function?
4. How is the cross product used to find perpendicular vectors in 3D?
5. Can a line intersect a curve at more than one point in space?

**Tip:** To check if two vectors are perpendicular, use the dot product: if their dot product is zero, the vectors are orthogonal.

Discover how to determine the equation of a line passing through the origin and perpendicular to a vector function r(t) = <1+2t, 3-2t, 1-t> at their point of intersection. This problem involves vector derivatives, parametric equations, and orthogonality, suitable for advanced high school and college-level students.

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