Prove that the line of intersection of `x + 2y + 3z=0 and 3x + zy+ z=0` is equally inclined to the X and Z axes and that it makes an angle ` theta` with the Y-axis where `sec 2 theta=3.`

Answer

To prove that the line of intersection of the planes \(x + 2y + 3z = 0\) and \(3x + 2y + z = 0\) is equally inclined to the X and Z axes, and that it makes an angle \(\theta\) with the Y-axis where \(\sec 2\theta = 3\), we will follow these steps:

### Step 1: Identify the Normal Vectors of the Planes
The normal vector of a plane given by the equation \(ax + by + cz = 0\) is \((a, b, c)\).

For the first plane \(x + 2y + 3z = 0\):
- The normal vector is \(\mathbf{n_1} = (1, 2, 3)\).

For the second plane \(3x + 2y + z = 0\):
- The normal vector is \(\mathbf{n_2} = (3, 2, 1)\).

### Step 2: Find the Direction Ratios of the Line of Intersection
The direction ratios of the line of intersection of two planes can be found using the cross product of their normal vectors.

\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{d} = \mathbf{i} \begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} = (2)(1) - (3)(2) = 2 - 6 = -4\)
2. \(\begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} = (1)(1) - (3)(3) = 1 - 9 = -8\)
3. \(\begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} = (1)(2) - (2)(3) = 2 - 6 = -4\)

Thus, we have:

\[ \mathbf{d} = -4\mathbf{i} + 8\mathbf{j} - 4\mathbf{k} = (-4, 8, -4) \]

### Step 3: Find the Direction Cosines
The direction cosines \(l, m, n\) can be found from the direction ratios \((-4, 8, -4)\):

\[ l = \frac{-4}{\sqrt{(-4)^2 + 8^2 + (-4)^2}} = \frac{-4}{\sqrt{16 + 64 + 16}} = \frac{-4}{\sqrt{96}} = \frac{-4}{4\sqrt{6}} = \frac{-1}{\sqrt{6}} \]

\[ m = \frac{8}{\sqrt{96}} = \frac{8}{4\sqrt{6}} = \frac{2}{\sqrt{6}} \]

\[ n = \frac{-4}{\sqrt{96}} = \frac{-4}{4\sqrt{6}} = \frac{-1}{\sqrt{6}} \]

### Step 4: Prove the Line is Equally Inclined to X and Z Axes
The angles made with the axes are given by:

\[ \cos \alpha = l, \quad \cos \beta = m, \quad \cos \gamma = n \]

Since \(\cos \alpha = \cos \gamma\), the line is equally inclined to the X and Z axes.

### Step 5: Find the Angle with the Y-axis
We have:

\[ \cos \beta = \frac{2}{\sqrt{6}} \]

To find \(\sec 2\theta\), we first calculate \(\cos 2\theta\):

Using the double angle formula:

\[ \cos 2\theta = 2 \cos^2 \theta - 1 \]

We need to find \(\cos \theta\):

\[ \cos \theta = \frac{2}{\sqrt{6}} \implies \cos^2 \theta = \frac{4}{6} = \frac{2}{3} \]

Thus,

\[ \cos 2\theta = 2 \cdot \frac{2}{3} - 1 = \frac{4}{3} - 1 = \frac{1}{3} \]

Finally, we find \(\sec 2\theta\):

\[ \sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1}{\frac{1}{3}} = 3 \]

### Conclusion
We have proved that the line of intersection of the given planes is equally inclined to the X and Z axes, and that it makes an angle \(\theta\) with the Y-axis where \(\sec 2\theta = 3\).

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