find a position vector that is parralel to the line x = 4 + 7t y= 5 - 3t z = 2t

The given parametric equations of the line are:

\[
x = 4 + 7t, \quad y = 5 - 3t, \quad z = 2t
\]

A vector parallel to the line can be obtained by looking at the coefficients of \( t \) in these parametric equations, because the direction of the line is determined by these coefficients.

For \(x\), \(y\), and \(z\), the coefficients of \( t \) are \( 7 \), \( -3 \), and \( 2 \) respectively.

Therefore, the direction vector (which is also parallel to the line) is:

\[
\mathbf{v} = \langle 7, -3, 2 \rangle
\]

This is the position vector parallel to the given line.

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Would you like further details or have any questions?

Here are 5 related questions:
1. How do you find the parametric equations of a line given two points?
2. What is the dot product of two vectors and how is it used to find if they are parallel?
3. Can you find a unit vector parallel to the vector \( \langle 7, -3, 2 \rangle \)?
4. How would you determine if two vectors are perpendicular?
5. What is the scalar projection of one vector onto another?

**Tip:** When finding a vector parallel to a line, always focus on the coefficients of the parameter from the parametric form of the line.

Find a position vector that is parallel to the line x = 4 + 7t, y = 5 - 3t, z = 2t.

Discover how to find the position vector parallel to the parametric line x = 4 + 7t, y = 5 - 3t, z = 2t. The solution explains vector algebra, focusing on extracting coefficients from parametric equations and applying them to get the direction vector. Suitable for students in Grades 10-12.

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