If volume of parallelopiped whose there coterminous edges are `vec u = hat i + hat j + lambda hat k, vec v = 2 hati + hat j + hatk, vecw = hati+hatj+3hatk`, is 1 cubic unit then cosine of angle between `vec u` and `vecv` is

Options

A. `7/(3sqrt10)`
B. `7/(6sqrt3)`
C. `5/(3sqrt3)`
D. `5/7`

Explanation

To find the cosine of the angle between the vectors \(\vec{u}\) and \(\vec{v}\), we first need to determine the value of \(\lambda\) given that the volume of the parallelepiped formed by the vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) is 1 cubic unit.

### Step-by-step Solution:

1. **Write the vectors**:
\[ \vec{u} = \hat{i} + \hat{j} + \lambda \hat{k} \]
\[ \vec{v} = 2 \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{w} = \hat{i} + \hat{j} + 3 \hat{k} \]

2. **Volume of the parallelepiped**:
The volume \(V\) is given by the absolute value of the scalar triple product:
\[ V = |\vec{u} \cdot (\vec{v} \times \vec{w})| \]
We know \(V = 1\).

3. **Calculate \(\vec{v} \times \vec{w}\)**:
To find \(\vec{v} \times \vec{w}\), we use the determinant:
\[ \vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 1 & 3 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot 3 - 1 \cdot 1) - \hat{j} (2 \cdot 3 - 1 \cdot 1) + \hat{k} (2 \cdot 1 - 1 \cdot 1) \]
\[ = \hat{i} (3 - 1) - \hat{j} (6 - 1) + \hat{k} (2 - 1) \]
\[ = 2\hat{i} - 5\hat{j} + 1\hat{k} \]

4. **Calculate \(\vec{u} \cdot (\vec{v} \times \vec{w})\)**:
Now we compute:
\[ \vec{u} \cdot (2\hat{i} - 5\hat{j} + \hat{k}) = (1)(2) + (1)(-5) + (\lambda)(1) \]
\[ = 2 - 5 + \lambda = \lambda - 3 \]

5. **Set the volume equation**:
Since the volume is given as 1,
\[ |\lambda - 3| = 1 \]
This gives us two equations:
\[ \lambda - 3 = 1 \quad \text{or} \quad \lambda - 3 = -1 \]
Solving these:
\[ \lambda = 4 \quad \text{or} \quad \lambda = 2 \]

6. **Find \(\vec{u}\) for both values of \(\lambda\)**:
- If \(\lambda = 4\):
\[ \vec{u} = \hat{i} + \hat{j} + 4\hat{k} \]
- If \(\lambda = 2\):
\[ \vec{u} = \hat{i} + \hat{j} + 2\hat{k} \]

7. **Calculate the cosine of the angle between \(\vec{u}\) and \(\vec{v}\)**:
Using the formula:
\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \]

**For \(\lambda = 4\)**:
\[ \vec{u} = \hat{i} + \hat{j} + 4\hat{k} \]
\[ \vec{v} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{u} \cdot \vec{v} = (1)(2) + (1)(1) + (4)(1) = 2 + 1 + 4 = 7 \]
\[ |\vec{u}| = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} \]
\[ |\vec{v}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]
\[ \cos \theta = \frac{7}{\sqrt{18} \cdot \sqrt{6}} = \frac{7}{\sqrt{108}} = \frac{7}{6\sqrt{3}} \]

**For \(\lambda = 2\)**:
\[ \vec{u} = \hat{i} + \hat{j} + 2\hat{k} \]
\[ \vec{u} \cdot \vec{v} = (1)(2) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 \]
\[ |\vec{u}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
\[ \cos \theta = \frac{5}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6} \]

### Final Results:
- For \(\lambda = 4\), \(\cos \theta = \frac{7}{6\sqrt{3}}\)
- For \(\lambda = 2\), \(\cos \theta = \frac{5}{6}\)