Find the unit vector parallel to the line  ݔെܼൌܻൌܼ

The line equation given is \( x - z = y = z \). Let's rewrite this equation in a clearer form for easier interpretation:

\[
x - z = y = z
\]

We can break it down into two parts:
1. \( x - z = y \)
2. \( y = z \)

From equation \( y = z \), we substitute \( z \) for \( y \) in the first equation:

\[
x - z = z
\]

Simplifying this, we get:

\[
x = 2z
\]

Now we have \( x = 2z \) and \( y = z \), which gives us parametric equations:

\[
x = 2t, \quad y = t, \quad z = t
\]

where \( t \) is a parameter.

The direction vector of this line is given by the coefficients of \( t \), which are \( (2, 1, 1) \). To find the **unit vector**, we normalize this direction vector.

The magnitude of the vector \( \vec{v} = (2, 1, 1) \) is:

\[
|\vec{v}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}
\]

The unit vector is:

\[
\hat{v} = \frac{1}{|\vec{v}|} \vec{v} = \frac{1}{\sqrt{6}} (2, 1, 1)
\]

Thus, the unit vector parallel to the line is:

\[
\hat{v} = \left( \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)
\]

Let me know if you'd like more details or have any questions! Here are some related questions to expand your understanding:

1. What is the geometric interpretation of a unit vector?
2. How would you find the unit vector for a line in 2D space?
3. What is the significance of normalizing a vector?
4. Can you express a unit vector in terms of angles?
5. How would the unit vector change if the direction vector were different?

**Tip:** To normalize a vector, divide each component of the vector by its magnitude.

Find the unit vector parallel to the line x - z = y = z

This solution explores how to find the unit vector parallel to the line x - z = y = z. Learn the process of deriving parametric equations, calculating vector magnitudes, and normalizing vectors. Suitable for students in Grades 10-12 studying vectors and 3D geometry.

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