If `theta` is the between the line `r = ( i + 2j- k) + lambda (i - j + 2k) , lambda in R` and the plane r. (2i - j + k ) = 4. then a value of cos `theta` is :

Options

A. `(sqrt(11))/(6)`
B. `(sqrt(35))/(6)`
C. `(sqrt(13))/(6)`
D. `(sqrt(7))/(3)`

Answer

To find the value of \( \cos \theta \) where \( \theta \) is the angle between the line and the plane, we can follow these steps:

### Step 1: Identify the direction ratios of the line and the normal vector of the plane.

The line is given by:
\[ \mathbf{r} = (1\mathbf{i} + 2\mathbf{j} - 1\mathbf{k}) + \lambda(1\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}) \]
From this, we can identify the direction vector \( \mathbf{b} \) of the line:
\[ \mathbf{b} = (1\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}) = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \]

The equation of the plane is given by:
\[ \mathbf{r} \cdot (2\mathbf{i} - \mathbf{j} + \mathbf{k}) = 4 \]
From this, we can identify the normal vector \( \mathbf{n} \) of the plane:
\[ \mathbf{n} = (2\mathbf{i} - \mathbf{j} + \mathbf{k}) \]

### Step 2: Calculate the dot product \( \mathbf{b} \cdot \mathbf{n} \).

Now, we calculate the dot product of \( \mathbf{b} \) and \( \mathbf{n} \):
\[ \mathbf{b} \cdot \mathbf{n} = (1)(2) + (-1)(-1) + (2)(1) = 2 + 1 + 2 = 5 \]

### Step 3: Calculate the magnitudes of \( \mathbf{b} \) and \( \mathbf{n} \).

Next, we find the magnitudes of \( \mathbf{b} \) and \( \mathbf{n} \):
\[ |\mathbf{b}| = \sqrt{(1)^2 + (-1)^2 + (2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
\[ |\mathbf{n}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 4: Use the sine formula to find \( \sin \theta \).

The sine of the angle \( \theta \) between the line and the plane can be calculated using the formula:
\[ \sin \theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}| |\mathbf{n}|} \]
Substituting the values we found:
\[ \sin \theta = \frac{5}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6} \]

### Step 5: Find \( \cos \theta \).

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36} \]
Thus,
\[ \cos \theta = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6} \]

### Final Answer:
The value of \( \cos \theta \) is:
\[ \cos \theta = \frac{\sqrt{11}}{6} \]