If the angle between the plane `x-3y+2z=1` and the line `(x-1)/(2)=(y-1)/(-1)=(z-1)/(-3)` is `theta`, then `sec 2 theta` is equal to

Options

A. `(107)/(11)`
B. `(49)/(48)`
C. `(100)/(9)`
D. `(87)/(79)`

Answer

To find the value of \( \sec 2\theta \) where \( \theta \) is the angle between the plane \( x - 3y + 2z = 1 \) and the line given by the symmetric equations \( \frac{x-1}{2} = \frac{y-1}{-1} = \frac{z-1}{-3} \), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as:
\[ x - 3y + 2z = 1 \]
The normal vector \( \mathbf{n} \) of the plane can be derived from the coefficients of \( x, y, z \):
\[ \mathbf{n} = \langle 1, -3, 2 \rangle \]

### Step 2: Identify the direction vector of the line
The line is given in symmetric form:
\[ \frac{x-1}{2} = \frac{y-1}{-1} = \frac{z-1}{-3} \]
From this, we can extract the direction vector \( \mathbf{a} \):
\[ \mathbf{a} = \langle 2, -1, -3 \rangle \]

### Step 3: Calculate the cosine of the angle \( \theta \)
The cosine of the angle \( \theta \) between the normal vector \( \mathbf{n} \) and the direction vector \( \mathbf{a} \) can be calculated using the dot product formula:
\[ \cos \theta = \frac{\mathbf{n} \cdot \mathbf{a}}{|\mathbf{n}| |\mathbf{a}|} \]
First, we compute the dot product \( \mathbf{n} \cdot \mathbf{a} \):
\[ \mathbf{n} \cdot \mathbf{a} = 1 \cdot 2 + (-3) \cdot (-1) + 2 \cdot (-3) = 2 + 3 - 6 = -1 \]

Next, we calculate the magnitudes of \( \mathbf{n} \) and \( \mathbf{a} \):
\[ |\mathbf{n}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]
\[ |\mathbf{a}| = \sqrt{2^2 + (-1)^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

Now substituting back into the cosine formula:
\[ \cos \theta = \frac{-1}{\sqrt{14} \cdot \sqrt{14}} = \frac{-1}{14} \]

### Step 4: Calculate \( \sin \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \left(\frac{-1}{14}\right)^2 = 1 - \frac{1}{196} = \frac{196 - 1}{196} = \frac{195}{196} \]
Thus,
\[ \sin \theta = \sqrt{\frac{195}{196}} = \frac{\sqrt{195}}{14} \]

### Step 5: Calculate \( \sec 2\theta \)
Using the double angle formula:
\[ \sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1}{\cos^2 \theta - \sin^2 \theta} \]
We already have \( \cos \theta = \frac{-1}{14} \) and \( \sin \theta = \frac{\sqrt{195}}{14} \):
\[ \cos^2 \theta = \left(\frac{-1}{14}\right)^2 = \frac{1}{196} \]
\[ \sin^2 \theta = \left(\frac{\sqrt{195}}{14}\right)^2 = \frac{195}{196} \]
Now substituting into the double angle formula:
\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \frac{1}{196} - \frac{195}{196} = \frac{1 - 195}{196} = \frac{-194}{196} = -\frac{97}{98} \]
Thus,
\[ \sec 2\theta = \frac{1}{-\frac{97}{98}} = -\frac{98}{97} \]

### Final Answer
The value of \( \sec 2\theta \) is \( -\frac{98}{97} \).