Let the volume of parallelopiped whose coteriminous edges are given by `u=i+j+lambdak, v=i+j+3k" and "w=2i+j+k` be 1 `("unit")^(3)`. If `theta` is angle between the edges u and w, then `cos theta` can be

Options

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

Answer

To solve the problem, we need to find the value of \( \cos \theta \) where \( \theta \) is the angle between the vectors \( \mathbf{u} \) and \( \mathbf{w} \). The volume of the parallelepiped formed by the vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) is given as 1 cubic unit.

### Step 1: Define the vectors
The vectors are defined 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 scalar triple product
The volume \( V \) of the parallelepiped is given by the scalar triple product:
\[ V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \]
We need to compute \( \mathbf{v} \times \mathbf{w} \).

### Step 3: Compute \( \mathbf{v} \times \mathbf{w} \)
Using the determinant form for the cross product:
\[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix} \]
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 each of these 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 compute the dot product:
\[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (1\mathbf{i} + 1\mathbf{j} + \lambda\mathbf{k}) \cdot (-2\mathbf{i} + 5\mathbf{j} - \mathbf{k}) \]
Calculating the dot product:
\[ = 1(-2) + 1(5) + \lambda(-1) = -2 + 5 - \lambda = 3 - \lambda \]

### Step 5: Set the volume equal to 1
Since the volume is given as 1:
\[ |3 - \lambda| = 1 \]
This gives us two equations:
1. \( 3 - \lambda = 1 \) → \( \lambda = 2 \)
2. \( 3 - \lambda = -1 \) → \( \lambda = 4 \)

### Step 6: Find \( \cos \theta \)
Now we need to find \( \cos \theta \) between \( \mathbf{u} \) and \( \mathbf{w} \):
\[ \mathbf{u} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \quad (\text{for } \lambda = 2) \]
\[ \mathbf{w} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} \]
Calculate the dot product:
\[ \mathbf{u} \cdot \mathbf{w} = (1)(2) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 \]

### Step 7: Calculate magnitudes
Calculate the magnitudes:
\[ |\mathbf{u}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
\[ |\mathbf{w}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 8: Use the dot product to find \( \cos \theta \)
Using the formula:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{w}}{|\mathbf{u}| |\mathbf{w}|} = \frac{5}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6} \]

### Final Answer
Thus, the value of \( \cos \theta \) is \( \frac{5}{6} \).