To find the equation of the plane passing through the point $P(3, -4.5)$ and containing the line $L: \frac{x - 2}{2} = \frac{y + 1}{-3} = \frac{z + 3}{5}$, we can proceed with the following steps:

Step 1: Parametrize the line

The given symmetric equation of the line is: $\frac{x - 2}{2} = \frac{y + 1}{-3} = \frac{z + 3}{5}.$

Let's denote the parameter by $t$. The parametric equations of the line can be written as: $x = 2 + 2t, \quad y = -1 - 3t, \quad z = -3 + 5t.$

So the line $L$ has direction vector: $\mathbf{v_L} = \langle 2, -3, 5 \rangle.$

Step 2: Identify the normal vector of the plane

The plane must contain the point $P(3, -4.5)$ and the line $L$. We need two vectors that lie on the plane to find the normal vector.

• One vector lies on the plane and is parallel to the line. This is the direction vector of the line $\mathbf{v_L} = \langle 2, -3, 5 \rangle$.

• The second vector is from the point $P(3, -4.5)$ to a point on the line. Let's take the point on the line corresponding to $t = 0$, which gives the point $(2, -1, -3)$.

The vector from $P(3, -4.5)$ to the point $(2, -1, -3)$ is: $\mathbf{v_P} = \langle 2 - 3, -1 + 4.5, -3 + 4.5 \rangle = \langle -1, 3.5, 1.5 \rangle.$

Step 3: Find the normal vector to the plane

The normal vector $\mathbf{n}$ to the plane is perpendicular to both $\mathbf{v_L}$ and $\mathbf{v_P}$. We can find it by taking the cross product of these two vectors: $\mathbf{n} = \mathbf{v_L} \times \mathbf{v_P} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 5 \\ -1 & 3.5 & 1.5 \end{vmatrix}.$

Let's compute this determinant: $\mathbf{n} = \mathbf{i} \begin{vmatrix} -3 & 5 \\ 3.5 & 1.5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 5 \\ -1 & 1.5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -3 \\ -1 & 3.5 \end{vmatrix}.$

First, calculate the 2x2 determinants: $\begin{vmatrix} -3 & 5 \\ 3.5 & 1.5 \end{vmatrix} = (-3)(1.5) - (5)(3.5) = -4.5 - 17.5 = -22,$ $\begin{vmatrix} 2 & 5 \\ -1 & 1.5 \end{vmatrix} = (2)(1.5) - (5)(-1) = 3 + 5 = 8,$ $\begin{vmatrix} 2 & -3 \\ -1 & 3.5 \end{vmatrix} = (2)(3.5) - (-3)(-1) = 7 - 3 = 4.$

Thus, the normal vector is: $\mathbf{n} = \langle -22, -8, 4 \rangle.$

Step 4: Find the equation of the plane

Now, the equation of the plane can be written as: $\mathbf{n} \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0,$ where $\mathbf{n} = \langle -22, -8, 4 \rangle$ and $P(3, -4.5)$ is a point on the plane.

Substitute the values: $-22(x - 3) - 8(y + 4.5) + 4(z + 3) = 0.$

Expanding this: $-22x + 66 - 8y - 36 + 4z + 12 = 0,$ $-22x - 8y + 4z + 42 = 0.$

Thus, the equation of the plane is: $22x + 8y - 4z = 42.$

Final Answer:

The equation of the plane is: $22x + 8y - 4z = 42.$

Would you like more details on any of the steps, or have further questions?

Related Questions:

1. How do you find the parametric equations of a line in space?
2. What is the geometric interpretation of the cross product of two vectors?
3. How can you verify if a point lies on a given plane?
4. How do you derive the equation of a plane from three points?
5. What is the significance of the direction vector in the equation of a line?

Tip:

When working with the equation of a plane, always start by finding a normal vector. If you don't have the normal vector, use the cross product of two vectors on the plane to determine it.