The angle between the line `(x-3)/(2)=(y-1)/(1)=(z+4)/(-2)` and the plane, `x+y+z+5=0` is

Options

A. `sin^(-1)((2)/(sqrt(3)))`
B. `sin^(-1)((1)/(sqrt(3)))`
C. `(pi)/(4)`
D. `sin^(-1)((1)/(3sqrt(3)))`

Explanation

To find the angle between the line given by the equation \((x-3)/(2)=(y-1)/(1)=(z+4)/(-2)\) and the plane given by the equation \(x+y+z+5=0\), we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line is given in symmetric form as:
\[ \frac{x-3}{2} = \frac{y-1}{1} = \frac{z+4}{-2} \]
From this, we can extract the direction ratios of the line, which are:
\[ a_1 = 2, \quad b_1 = 1, \quad c_1 = -2 \]

### Step 2: Identify the normal direction ratios of the plane
The equation of the plane is:
\[ x + y + z + 5 = 0 \]
The coefficients of \(x\), \(y\), and \(z\) give us the normal direction ratios of the plane:
\[ a_2 = 1, \quad b_2 = 1, \quad c_2 = 1 \]

### Step 3: Use the formula for the angle between the line and the plane
The angle \(\theta\) between the line and the plane can be found using the formula:
\[ \sin \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

### Step 4: Calculate the numerator
Substituting the values:
\[ a_1 a_2 = 2 \cdot 1 = 2 \]
\[ b_1 b_2 = 1 \cdot 1 = 1 \]
\[ c_1 c_2 = -2 \cdot 1 = -2 \]
Thus, the numerator becomes:
\[ |2 + 1 - 2| = |1| = 1 \]

### Step 5: Calculate the denominator
Now, we calculate the magnitudes:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]
Thus, the denominator becomes:
\[ 3 \cdot \sqrt{3} \]

### Step 6: Combine the results
Now we can substitute the values into the sine formula:
\[ \sin \theta = \frac{1}{3\sqrt{3}} \]

### Step 7: Find the angle \(\theta\)
To find \(\theta\), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{1}{3\sqrt{3}}\right) \]

### Final Answer
The angle between the line and the plane is:
\[ \theta = \sin^{-1}\left(\frac{1}{3\sqrt{3}}\right) \]