If the angle between the vectors `bar a` and `bar b` having direction ratios 1, 2, 1 and 1, 3k, 1 is `(pi)/(4)`, then k =

Options

A. `(2pm 3sqrt(2))/(3)`
B. `(-2 pm 3sqrt(2))/(3)`
C. `(4pm 3sqrt(2))/(3)`
D. `(- 4 pm 3sqrt(2))/(3)`

Answer

To solve the problem, we need to find the value of \( k \) such that the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{4} \). The direction ratios of the vectors are given as follows:

- For vector \( \mathbf{a} \): \( (1, 2, 1) \)
- For vector \( \mathbf{b} \): \( (1, 3k, 1) \)

### Step 1: Use the formula for the cosine of the angle between two vectors

The cosine of the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be calculated using the formula:

\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]

Where:
- \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of the vectors.
- \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors.

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

The dot product is calculated as follows:

\[ \mathbf{a} \cdot \mathbf{b} = (1)(1) + (2)(3k) + (1)(1) = 1 + 6k + 1 = 2 + 6k \]

### Step 3: Calculate the magnitudes \( |\mathbf{a}| \) and \( |\mathbf{b}| \)

The magnitude of vector \( \mathbf{a} \) is:

\[ |\mathbf{a}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

The magnitude of vector \( \mathbf{b} \) is:

\[ |\mathbf{b}| = \sqrt{1^2 + (3k)^2 + 1^2} = \sqrt{1 + 9k^2 + 1} = \sqrt{2 + 9k^2} \]

### Step 4: Substitute into the cosine formula

Since the angle \( \theta = \frac{\pi}{4} \), we know that:

\[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]

Substituting the values into the cosine formula gives:

\[ \frac{2 + 6k}{\sqrt{6} \cdot \sqrt{2 + 9k^2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Cross-multiply to eliminate the fraction

Cross-multiplying yields:

\[ (2 + 6k) \cdot \sqrt{2} = \sqrt{6} \cdot \sqrt{2 + 9k^2} \]

### Step 6: Square both sides to eliminate the square roots

Squaring both sides results in:

\[ (2 + 6k)^2 \cdot 2 = 6(2 + 9k^2) \]

Expanding both sides gives:

\[ 2(4 + 24k + 36k^2) = 12 + 54k^2 \]

This simplifies to:

\[ 8 + 48k + 72k^2 = 12 + 54k^2 \]

### Step 7: Rearranging the equation

Rearranging the equation leads to:

\[ 72k^2 - 54k^2 + 48k + 8 - 12 = 0 \]

This simplifies to:

\[ 18k^2 + 48k - 4 = 0 \]

### Step 8: Simplifying the quadratic equation

Dividing the entire equation by 2 yields:

\[ 9k^2 + 24k - 2 = 0 \]

### Step 9: Using the quadratic formula to solve for \( k \)

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 9 \), \( b = 24 \), and \( c = -2 \).

Calculating the discriminant:

\[ b^2 - 4ac = 24^2 - 4 \cdot 9 \cdot (-2) = 576 + 72 = 648 \]

Thus, we have:

\[ k = \frac{-24 \pm \sqrt{648}}{2 \cdot 9} \]

Calculating \( \sqrt{648} = 18\sqrt{2} \):

\[ k = \frac{-24 \pm 18\sqrt{2}}{18} = -\frac{4}{3} \pm \sqrt{2} \]

### Final Result

The values of \( k \) are:

\[ k = -4 \pm \frac{3\sqrt{2}}{3} \]