Find the equation of the plane through the points A(3,4,2) and B(7,0,6) and perpendicular to the plane `2x-5y=15`.
Hint: The given plane is `2x-5y+0z=15`.

Answer

To find the equation of the plane through the points A(3, 4, 2) and B(7, 0, 6) and perpendicular to the plane given by the equation \(2x - 5y = 15\), we will follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane \(2x - 5y + 0z = 15\) can be expressed in the form \(Ax + By + Cz = D\), where \(A = 2\), \(B = -5\), and \(C = 0\). The normal vector \( \mathbf{n_1} \) of this plane is given by the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n_1} = (2, -5, 0) \]

### Step 2: Find the direction vector of line AB
The direction vector \( \mathbf{d} \) from point A to point B can be calculated as:
\[ \mathbf{d} = B - A = (7 - 3, 0 - 4, 6 - 2) = (4, -4, 4) \]

### Step 3: Find the normal vector of the required plane
Since the required plane is perpendicular to the given plane and also passes through points A and B, its normal vector \( \mathbf{n_2} \) can be found using the cross product of \( \mathbf{n_1} \) and \( \mathbf{d} \):
\[ \mathbf{n_2} = \mathbf{n_1} \times \mathbf{d} \]
Calculating the cross product:
\[ \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -5 & 0 \\ 4 & -4 & 4 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{n_2} = \mathbf{i}((-5)(4) - (0)(-4)) - \mathbf{j}((2)(4) - (0)(4)) + \mathbf{k}((2)(-4) - (-5)(4)) \]
\[ = \mathbf{i}(-20) - \mathbf{j}(8) + \mathbf{k}(-8 + 20) \]
\[ = (-20, -8, 12) \]

### Step 4: Write the equation of the plane
The equation of a plane can be written as:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
Using point A(3, 4, 2) and the normal vector \((-20, -8, 12)\):
\[ -20(x - 3) - 8(y - 4) + 12(z - 2) = 0 \]
Expanding this:
\[ -20x + 60 - 8y + 32 + 12z - 24 = 0 \]
Combining like terms:
\[ -20x - 8y + 12z + 68 = 0 \]
Dividing the entire equation by -4 for simplification:
\[ 5x + 2y - 3z - 17 = 0 \]

### Final Equation of the Plane
The equation of the plane is:
\[ 5x + 2y - 3z = 17 \]