An agle between the plane, `x + y+ z =5` andthe line of intersection of the planes,`3x + 4y + z -1=0 and 5x +8y + 2z+ 14=0,` is :

Options

A. `sin ^(-1) (3//sqrt17)`
B. `cos ^(-1) (3//sqrt7)`
C. `sin ^(-1) (sqrt(3//17))`
D. ` cos ^(-1) (sqrt3 //17)`

Answer

To find the angle between the plane \(x + y + z = 5\) and the line of intersection of the planes \(3x + 4y + z - 1 = 0\) and \(5x + 8y + 2z + 14 = 0\), we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of the first plane \(x + y + z = 5\) can be derived from its coefficients:
- Normal vector \(n_1 = (1, 1, 1)\).

For the second plane \(3x + 4y + z - 1 = 0\):
- Normal vector \(n_2 = (3, 4, 1)\).

For the third plane \(5x + 8y + 2z + 14 = 0\):
- Normal vector \(n_3 = (5, 8, 2)\).

### Step 2: Find the direction vector of the line of intersection
The direction vector of the line of intersection of two planes can be found using the cross product of their normal vectors \(n_2\) and \(n_3\).

\[ \text{Direction vector} = n_2 \times n_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 1 \\ 5 & 8 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(4 \cdot 2 - 1 \cdot 8) - \hat{j}(3 \cdot 2 - 1 \cdot 5) + \hat{k}(3 \cdot 8 - 4 \cdot 5) \]
\[ = \hat{i}(8 - 8) - \hat{j}(6 - 5) + \hat{k}(24 - 20) \]
\[ = 0\hat{i} - 1\hat{j} + 4\hat{k} \]
Thus, the direction vector of the line of intersection is:
\[ \mathbf{d} = (0, -1, 4) \]

### Step 3: Calculate the angle between the plane and the line
The angle \(\theta\) between the plane and the line can be calculated using the formula:
\[ \sin(\theta) = \frac{|\mathbf{n} \cdot \mathbf{d}|}{|\mathbf{n}| |\mathbf{d}|} \]
where \(\mathbf{n}\) is the normal vector of the plane and \(\mathbf{d}\) is the direction vector of the line.

Calculating the dot product:
\[ \mathbf{n} \cdot \mathbf{d} = (1, 1, 1) \cdot (0, -1, 4) = 0 \cdot 1 + 1 \cdot (-1) + 1 \cdot 4 = -1 + 4 = 3 \]

Calculating the magnitudes:
\[ |\mathbf{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
\[ |\mathbf{d}| = \sqrt{0^2 + (-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \]

Substituting into the sine formula:
\[ \sin(\theta) = \frac{|3|}{\sqrt{3} \cdot \sqrt{17}} = \frac{3}{\sqrt{51}} \]

### Step 4: Find the angle
To find the angle \(\theta\):
\[ \theta = \sin^{-1}\left(\frac{3}{\sqrt{51}}\right) \]

This gives us the angle between the plane and the line of intersection.