Find a plane containing the point (-7,4,-5) and the line of intersection of the planes 4x-y-4z=3 and 4x-5y+8z=-5

To find the equation of a plane containing the point \((-7, 4, -5)\) and the line of intersection of the planes \(4x - y - 4z = 3\) and \(4x - 5y + 8z = -5\), follow these steps:

### Step 1: Find the direction vector of the line of intersection.
The direction vector of the line of intersection is the cross product of the normal vectors of the two planes.

1. **Normal vector of the first plane \(4x - y - 4z = 3\):**
   \[
   \mathbf{n}_1 = \langle 4, -1, -4 \rangle
   \]

2. **Normal vector of the second plane \(4x - 5y + 8z = -5\):**
   \[
   \mathbf{n}_2 = \langle 4, -5, 8 \rangle
   \]

3. **Cross product to find the direction vector \(\mathbf{d}\):**
   \[
   \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2
   \]
   Compute the cross product:
   \[
   \mathbf{d} = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   4 & -1 & -4 \\
   4 & -5 & 8 \\
   \end{vmatrix}
   \]
   Expanding the determinant:
   \[
   \mathbf{d} = \mathbf{i}((-1)(8) - (-4)(-5)) - \mathbf{j}((4)(8) - (-4)(4)) + \mathbf{k}((4)(-5) - (-1)(4))
   \]
   Simplifying:
   \[
   \mathbf{d} = \mathbf{i}(-8 - 20) - \mathbf{j}(32 + 16) + \mathbf{k}(-20 + 4)
   \]
   \[
   \mathbf{d} = \mathbf{i}(-28) - \mathbf{j}(48) + \mathbf{k}(-16)
   \]
   \[
   \mathbf{d} = \langle -28, -48, -16 \rangle
   \]

### Step 2: Use the direction vector and the given point to write the equation of the plane.
The equation of the plane can be written in the form:
\[
a(x + 7) + b(y - 4) + c(z + 5) = 0
\]
where \(\langle a, b, c \rangle\) is a normal vector to the plane. We already know that this plane must be perpendicular to the direction vector \(\mathbf{d} = \langle -28, -48, -16 \rangle\).

Thus, the normal vector \(\langle a, b, c \rangle\) will be parallel to \(\mathbf{d}\).

We can take:
\[
\mathbf{n} = \langle -28, -48, -16 \rangle
\]

### Step 3: Substitute the point and the normal vector into the plane equation.
The plane equation becomes:
\[
-28(x + 7) - 48(y - 4) - 16(z + 5) = 0
\]
Expanding this:
\[
-28x - 196 - 48y + 192 - 16z - 80 = 0
\]
Combine like terms:
\[
-28x - 48y - 16z = 84
\]

Dividing by \(-4\) to simplify:
\[
7x + 12y + 4z = -21
\]

So, the equation of the plane is:
\[
7x + 12y + 4z = -21
\]

This is the required plane that contains the point \((-7, 4, -5)\) and the line of intersection of the given planes.

Would you like any more details or have any questions?

---

1. How do you check if a point lies on a given plane?
2. What are the conditions for two planes to be perpendicular?
3. How can you find the line of intersection of two planes using their equations?
4. What does the magnitude of a cross product tell you about the vectors?
5. How can you represent the line of intersection of two planes parametrically?

**Tip:** When simplifying equations, always check if you can factor out common terms to make the equation simpler and easier to interpret.

Learn how to find the equation of a plane that contains the point (-7, 4, -5) and the line of intersection of planes 4x - y - 4z = 3 and 4x - 5y + 8z = -5. This detailed solution involves calculating the direction vector of the intersection line, using cross product, and writing the plane equation. Suitable for advanced high school students and beyond.

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