Find a nonzero vector parallel to the line of intersection of the two planes 2x−3y−2z = −1
and y − 5x − 2z = 5.

To find a nonzero vector parallel to the line of intersection of the two planes given by the equations

\[ 2x - 3y - 2z = -1 \]

and

\[ y - 5x - 2z = 5, \]

we need to determine the direction vector that lies along this line of intersection. 

The direction vector parallel to the line of intersection of two planes is given by the cross product of the normal vectors of these planes. Let's first identify the normal vectors of the given planes.

1. For the plane \(2x - 3y - 2z = -1\), the normal vector is \(\mathbf{n_1} = \langle 2, -3, -2 \rangle\).
2. For the plane \(y - 5x - 2z = 5\), the normal vector is \(\mathbf{n_2} = \langle -5, 1, -2 \rangle\).

The cross product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\) will give us a vector that is perpendicular to both normal vectors and thus parallel to the line of intersection of the two planes.

We can calculate the cross product \(\mathbf{n_1} \times \mathbf{n_2}\) as follows:

\[
\mathbf{n_1} \times \mathbf{n_2} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -3 & -2 \\
-5 & 1 & -2
\end{vmatrix}
\]

This determinant can be expanded as:

\[
\mathbf{n_1} \times \mathbf{n_2} = \mathbf{i} \left( (-3)(-2) - (-2)(1) \right) - \mathbf{j} \left( (2)(-2) - (-2)(-5) \right) + \mathbf{k} \left( (2)(1) - (-3)(-5) \right)
\]

Calculating each term:

\[
\mathbf{i} \left( 6 - (-2) \right) = \mathbf{i} \cdot 8
\]

\[
- \mathbf{j} \left( -4 - 10 \right) = - \mathbf{j} \cdot (-14) = \mathbf{j} \cdot 14
\]

\[
\mathbf{k} \left( 2 - 15 \right) = \mathbf{k} \cdot (-13)
\]

Therefore, the cross product \(\mathbf{n_1} \times \mathbf{n_2}\) is:

\[
\mathbf{n_1} \times \mathbf{n_2} = \langle 8, 14, -13 \rangle
\]

So, a nonzero vector parallel to the line of intersection of the two planes is \(\langle 8, 14, -13 \rangle\).

Would you like further details or have any questions?

Here are 8 related questions that expand on this topic:

1. How do you find the angle between two planes?
2. What is the geometric interpretation of the cross product?
3. How can you find the intersection point of two planes?
4. What does it mean if the cross product of two vectors is zero?
5. How can you find the distance between a point and a plane?
6. How do you parametrize the line of intersection of two planes?
7. How do you determine if two planes are parallel or coincident?
8. How can you find the equation of a plane given three points on the plane?

**Tip:** When dealing with cross products, always ensure you use the right-hand rule to determine the direction of the resulting vector.

Learn how to find a nonzero vector parallel to the line of intersection of two planes given by specific equations. This problem involves calculating the cross product of the normal vectors of the planes to determine the direction vector parallel to the intersection line. Suitable for students in Grades 10-12 studying vectors and 3D geometry.

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