The line of greatest slope on an inclined plane `P_1` is that line in the plane which is perpendicular to the line of intersection of plane `P_1` and a horiontal plane `P_2`.
Q. Assuming the plane `4x-3y+7z=0` to be horizontal, the direction cosines of line greatest slope in the plane `2x+y-5z=0` are

Options

A. `((3)/(sqrt(11)), -(1)/(sqrt(11)), (1)/(sqrt(11)))`
B. `((3)/(sqrt(11)), (1)/(sqrt(11)), -(1)/(sqrt(11)))`
C. `(-(3)/(sqrt(11)), (1)/(sqrt(11)), (1)/(sqrt(11)))`
D. `((1)/(sqrt(11)), -(3)/(sqrt(11)), -(1)/(sqrt(11)))`

Answer

To find the direction cosines of the line of greatest slope in the plane \(2x + y - 5z = 0\) assuming the plane \(4x - 3y + 7z = 0\) is horizontal, we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of the horizontal plane \(P_2: 4x - 3y + 7z = 0\) is given by the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n_2} = \langle 4, -3, 7 \rangle \]
The normal vector of the inclined plane \(P_1: 2x + y - 5z = 0\) is:
\[ \mathbf{n_1} = \langle 2, 1, -5 \rangle \]

### Step 2: Find the line of intersection of the two planes
The line of intersection of the two planes can be found using the cross product of their normal vectors:
\[ \mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -5 \\ 4 & -3 & 7 \end{vmatrix} \]
Calculating this determinant:
\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 1 & -5 \\ -3 & 7 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & -5 \\ 4 & 7 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} \]
Calculating the minors:
\[ = \mathbf{i} (1 \cdot 7 - (-5) \cdot (-3)) - \mathbf{j} (2 \cdot 7 - (-5) \cdot 4) + \mathbf{k} (2 \cdot (-3) - 1 \cdot 4) \]
\[ = \mathbf{i} (7 - 15) - \mathbf{j} (14 + 20) + \mathbf{k} (-6 - 4) \]
\[ = -8\mathbf{i} - 34\mathbf{j} - 10\mathbf{k} \]
Thus, we have:
\[ \mathbf{n} = \langle -8, -34, -10 \rangle \]

### Step 3: Find the direction of the line of greatest slope
The line of greatest slope in the inclined plane is perpendicular to the line of intersection. Therefore, we need to find the cross product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\):
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]
Calculating this cross product:
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -5 \\ 4 & -3 & 7 \end{vmatrix} \]
Calculating this determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & -5 \\ -3 & 7 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & -5 \\ 4 & 7 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} \]
Calculating the minors:
\[ = \mathbf{i} (1 \cdot 7 - (-5) \cdot (-3)) - \mathbf{j} (2 \cdot 7 - (-5) \cdot 4) + \mathbf{k} (2 \cdot (-3) - 1 \cdot 4) \]
\[ = \mathbf{i} (7 - 15) - \mathbf{j} (14 + 20) + \mathbf{k} (-6 - 4) \]
\[ = -8\mathbf{i} - 34\mathbf{j} - 10\mathbf{k} \]

### Step 4: Normalize the direction vector
To find the direction cosines, we normalize the vector \(\mathbf{d}\):
\[ \text{Magnitude of } \mathbf{d} = \sqrt{(-8)^2 + (-34)^2 + (-10)^2} = \sqrt{64 + 1156 + 100} = \sqrt{1320} = 2\sqrt{330} \]
The unit vector is:
\[ \mathbf{d_{unit}} = \left\langle \frac{-8}{2\sqrt{330}}, \frac{-34}{2\sqrt{330}}, \frac{-10}{2\sqrt{330}} \right\rangle = \left\langle \frac{-4}{\sqrt{330}}, \frac{-17}{\sqrt{330}}, \frac{-5}{\sqrt{330}} \right\rangle \]

### Step 5: Direction cosines
The direction cosines are:
\[ \left( \frac{-4}{\sqrt{330}}, \frac{-17}{\sqrt{330}}, \frac{-5}{\sqrt{330}} \right) \]
If we take the negative common factor out, we can write:
\[ \left( \frac{4}{\sqrt{330}}, \frac{17}{\sqrt{330}}, \frac{5}{\sqrt{330}} \right) \]

### Final Answer
Thus, the direction cosines of the line of greatest slope in the plane \(2x + y - 5z = 0\) are:
\[ \left( \frac{4}{\sqrt{330}}, \frac{17}{\sqrt{330}}, \frac{5}{\sqrt{330}} \right) \]