91Ó°ÊÓ

To understand the equation of a plane containing two intersecting lines, it's essential to grasp the concept of intersecting lines first. Intersecting lines are lines that meet at a single point. In a 3D space, such lines lie within the same plane.
For example, given the lines: \ \ \( \frac{x}{2}=\frac{y-2}{3}=\frac{z-1}{1} \) and \ \( \frac{x}{1}=\frac{y-2}{-1}=\frac{z-1}{1} \), we need to find the point where they intersect.
This involves setting the corresponding coordinates equal and solving for the parameters, which eventually gives us a common point, an essential step in finding the plane containing these lines. In our example, the intersection point is (0, 2, 1).

The parametric form of a line describes the line using a parameter, making it versatile for calculations. It separates each coordinate (x, y, z) into equations involving a scalar parameter.
For the lines given: \( \frac{x}{2}=\frac{y-2}{3}=\frac{z-1}{1} \) and \( \frac{x}{1}=\frac{y-2}{-1}=\frac{z-1}{1} \), we parameterize them as follows:
For the first line with parameter t:

• \( x = 2t \)
• \( y = 3t + 2 \)
• \( z = t + 1 \)

For the second line with parameter s:

• \( x = s \)
• \( y = -s + 2 \)
• \( z = s + 1 \)

This form helps in finding points on the line and simplifies solving intersection problems.

The normal vector is crucial in defining a plane. It's a vector perpendicular to every line on the plane. To find it, we use the direction vectors of the intersecting lines.
The direction vectors from our parameterized lines are:

• For the first line: \( \ \vec{d_1} = \ \langle 2, 3, 1 \ \rangle \)
• For the second line: \( \ \vec{d_2} = \ \langle 1, -1, 1 \ \rangle \)

The cross product of these direction vectors gives us the normal vector. This result, \( \ \vec{n} = \ \langle 4, -1, -5 \ \rangle \), is then used to derive the plane equation, ensuring it encapsulates the given lines.

The cross product is a vector operation used to find a vector perpendicular to two given vectors in 3D space. It's essential for finding the normal vector to a plane.
For vectors \( \ \vec{d_1} = \ \langle 2, 3, 1 \ \rangle \) and \( \ \vec{d_2} = \ \langle 1, -1, 1 \ \rangle \), their cross product is calculated as follows:
\( \ \vec{n} = \ \vec{d_1} \ \times \ \vec{d_2} = \ \begin{vmatrix} \ i & j & k \ \ 2 & 3 & 1 \ 1 & -1 & 1 \ \ \end{vmatrix} \ = \ i(3 \ \cdot \ 1 - 1 \ \cdot \ (-1)) - j(2 \ \cdot \ 1 - 1 \ \cdot \ 1) + k(2 \ \cdot \ (-1) - 3 \ \cdot \ 1) \ = \ \ \langle 4, -1, -5 \ \rangle \).
This vector \( \ \vec{n} \ \rangle \)is the normal vector needed to form the plane's equation, illustrating its perpendicular nature to the plane containing the intersecting lines.