Let the volume of a parallelopiped whose coterminus edges are given by `u = I + j + lambdak, v = I + j + 3k` and w = 2i + j + k be 1 cu. unit. If `theta` be the angle between the edges u and w, then cos `theta` can be:

Options

A. `7/(6sqrt(6))`
B. `5/7`
C. `7/(6sqrt(3))`
D. `5/(3sqrt(3))`

Answer

To solve the problem, we need to find the cosine of the angle \( \theta \) between the vectors \( \mathbf{u} \) and \( \mathbf{w} \) given the volume of the parallelepiped formed by the vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \).

### Step 1: Identify the vectors
The vectors are given as:
- \( \mathbf{u} = \mathbf{i} + \mathbf{j} + \lambda \mathbf{k} \)
- \( \mathbf{v} = \mathbf{i} + \mathbf{j} + 3 \mathbf{k} \)
- \( \mathbf{w} = 2 \mathbf{i} + \mathbf{j} + \mathbf{k} \)

### Step 2: Calculate the volume of the parallelepiped
The volume \( V \) of the parallelepiped formed by the vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) is given by the scalar triple product:
\[ V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \]
We know that the volume is 1 cubic unit, so:
\[ |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| = 1 \]

### Step 3: Calculate \( \mathbf{v} \times \mathbf{w} \)
To find \( \mathbf{v} \times \mathbf{w} \), we can use the determinant method:
\[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ = \mathbf{i} (1 \cdot 1 - 3 \cdot 1) - \mathbf{j} (1 \cdot 1 - 3 \cdot 2) + \mathbf{k} (1 \cdot 1 - 1 \cdot 2) \]
\[ = \mathbf{i} (1 - 3) - \mathbf{j} (1 - 6) + \mathbf{k} (1 - 2) \]
\[ = -2\mathbf{i} + 5\mathbf{j} - \mathbf{k} \]

### Step 4: Calculate \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \)
Now we calculate \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \):
\[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (1\mathbf{i} + 1\mathbf{j} + \lambda\mathbf{k}) \cdot (-2\mathbf{i} + 5\mathbf{j} - 1\mathbf{k}) \]
Calculating the dot product:
\[ = 1 \cdot (-2) + 1 \cdot 5 + \lambda \cdot (-1) \]
\[ = -2 + 5 - \lambda = 3 - \lambda \]
Setting the absolute value equal to 1:
\[ |3 - \lambda| = 1 \]
This gives us two equations:
1. \( 3 - \lambda = 1 \) → \( \lambda = 2 \)
2. \( 3 - \lambda = -1 \) → \( \lambda = 4 \)

### Step 5: Calculate \( \cos \theta \)
Now we find \( \cos \theta \) using the formula:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{w}}{|\mathbf{u}| |\mathbf{w}|} \]

First, calculate \( \mathbf{u} \cdot \mathbf{w} \):
\[ \mathbf{u} \cdot \mathbf{w} = (1\mathbf{i} + 1\mathbf{j} + \lambda\mathbf{k}) \cdot (2\mathbf{i} + 1\mathbf{j} + 1\mathbf{k}) \]
Calculating the dot product:
\[ = 1 \cdot 2 + 1 \cdot 1 + \lambda \cdot 1 = 2 + 1 + \lambda = 3 + \lambda \]

Next, calculate \( |\mathbf{u}| \) and \( |\mathbf{w}| \):
\[ |\mathbf{u}| = \sqrt{1^2 + 1^2 + \lambda^2} = \sqrt{2 + \lambda^2} \]
\[ |\mathbf{w}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

Now substituting back into the cosine formula:
\[ \cos \theta = \frac{3 + \lambda}{\sqrt{2 + \lambda^2} \cdot \sqrt{6}} \]

### Step 6: Substitute values of \( \lambda \)
1. For \( \lambda = 2 \):
\[ \cos \theta = \frac{3 + 2}{\sqrt{2 + 2^2} \cdot \sqrt{6}} = \frac{5}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6} \]

2. For \( \lambda = 4 \):
\[ \cos \theta = \frac{3 + 4}{\sqrt{2 + 4^2} \cdot \sqrt{6}} = \frac{7}{\sqrt{18} \cdot \sqrt{6}} = \frac{7}{6\sqrt{3}} = \frac{7}{6\sqrt{3}} \]

### Final Result
Thus, the possible values for \( \cos \theta \) are:
- \( \frac{5}{6} \)
- \( \frac{7}{6\sqrt{3}} \)