Angle between the line `vecr=(2hati-hatj+hatk)+lamda(-hati+hatj+hatk)` and the plane `vecr.(3hati+2hatj-hatk)=4` is

Options

A. `cos^(-1)(2/(sqrt(42)))`
B. `cos^(-1)((-2)/(sqrt(42)))`
C. `sin^(-1)(2/(sqrt(42)))`
D. `sin^(-1)((-2)/(sqrt(42)))`

Explanation

To find the angle between the given line and the plane, we will follow these steps:

1. **Identify the direction vector of the line**:
The line is given in the vector form:
\[ \vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(-\hat{i} + \hat{j} + \hat{k}) \]
The direction vector \( \vec{b} \) of the line can be extracted from the equation:
\[ \vec{b} = -\hat{i} + \hat{j} + \hat{k} \]

2. **Identify the normal vector of the plane**:
The equation of the plane is given by:
\[ \vec{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 4 \]
The normal vector \( \vec{n} \) of the plane is:
\[ \vec{n} = 3\hat{i} + 2\hat{j} - \hat{k} \]

3. **Calculate the dot product \( \vec{b} \cdot \vec{n} \)**:
We compute the dot product:
\[ \vec{b} \cdot \vec{n} = (-1)(3) + (1)(2) + (1)(-1) = -3 + 2 - 1 = -2 \]

4. **Calculate the magnitudes of \( \vec{b} \) and \( \vec{n} \)**:
The magnitude of \( \vec{b} \) is:
\[ |\vec{b}| = \sqrt{(-1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]
The magnitude of \( \vec{n} \) is:
\[ |\vec{n}| = \sqrt{(3)^2 + (2)^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]

5. **Use the formula for the sine of the angle between the line and the plane**:
The formula for the sine of the angle \( \theta \) between the line and the plane is:
\[ \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \]
Substituting the values we calculated:
\[ \sin \theta = \frac{|-2|}{\sqrt{3} \cdot \sqrt{14}} = \frac{2}{\sqrt{42}} \]

6. **Find the angle \( \theta \)**:
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{2}{\sqrt{42}}\right) \]

Thus, the angle between the line and the plane is:
\[ \theta = \sin^{-1}\left(\frac{2}{\sqrt{42}}\right) \]