Find the angle between the line `(x-1)/3=(y+1)/2=(z-1)/4` and plane `2x + y - 3z + 4 = 0.`

Answer

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

### Step 1: Identify the direction ratios of the line
The equation of the line can be expressed as:
\[ \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-1}{4} \]
From this, we can identify the direction ratios of the line, which are given by the coefficients in the denominators. Thus, the direction ratios \(b\) of the line are:
\[ b = (3, 2, 4) \]

### Step 2: Identify the normal vector of the plane
The equation of the plane is given by:
\[ 2x + y - 3z + 4 = 0 \]
The coefficients of \(x\), \(y\), and \(z\) in the plane equation give us the normal vector \(n\) of the plane. Therefore, the normal vector is:
\[ n = (2, 1, -3) \]

### Step 3: Use the formula to find the sine of the angle
The sine of the angle \(\theta\) between the line and the plane can be found using the formula:
\[ \sin \theta = \frac{|b \cdot n|}{|b| |n|} \]
where \(b \cdot n\) is the dot product of the direction vector and the normal vector, and \(|b|\) and \(|n|\) are the magnitudes of the vectors \(b\) and \(n\).

### Step 4: Calculate the dot product \(b \cdot n\)
Calculating the dot product:
\[ b \cdot n = (3)(2) + (2)(1) + (4)(-3) = 6 + 2 - 12 = -4 \]

### Step 5: Calculate the magnitudes \(|b|\) and \(|n|\)
Now, we calculate the magnitudes:
\[ |b| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} \]
\[ |n| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

### Step 6: Substitute values into the sine formula
Substituting the values into the sine formula:
\[ \sin \theta = \frac{|-4|}{\sqrt{29} \cdot \sqrt{14}} = \frac{4}{\sqrt{29} \cdot \sqrt{14}} \]

### Step 7: Calculate the angle \(\theta\)
To find the angle \(\theta\), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{4}{\sqrt{29} \cdot \sqrt{14}}\right) \]

### Final Answer
The angle between the line and the plane is:
\[ \theta = \sin^{-1}\left(\frac{4}{\sqrt{29} \cdot \sqrt{14}}\right) \]