Let `vecu=veci-vecj-2^k, vec v =2vec i +vecj-vec k ,vecv.vecw=2` and `vec v times vec w=vec u+lambda vec v`. Then `vecu .vecw` is equal to

Options

A. 1
B. 2
C. `3/2`
D. `-2/3`

Answer

To solve the problem, we need to find the value of \(\vec{u} \cdot \vec{w}\) given the vectors \(\vec{u}\), \(\vec{v}\), and the conditions provided. Let's break it down step by step.

### Step 1: Define the vectors
We have:
\[ \vec{u} = \hat{i} - \hat{j} - 2\hat{k} \]
\[ \vec{v} = 2\hat{i} + \hat{j} - \hat{k} \]

### Step 2: Calculate \(\vec{v} \cdot \vec{w}\)
We know from the problem that:
\[ \vec{v} \cdot \vec{w} = 2 \]

### Step 3: Use the cross product condition
The problem states:
\[ \vec{v} \times \vec{w} = \vec{u} + \lambda \vec{v} \]
Taking the dot product of both sides with \(\vec{v}\):
\[ (\vec{v} \times \vec{w}) \cdot \vec{v} = (\vec{u} + \lambda \vec{v}) \cdot \vec{v} \]
The left-hand side is zero since the dot product of a vector with a vector perpendicular to it (cross product) is zero:
\[ 0 = \vec{u} \cdot \vec{v} + \lambda (\vec{v} \cdot \vec{v}) \]

### Step 4: Calculate \(\vec{u} \cdot \vec{v}\)
Now, we calculate \(\vec{u} \cdot \vec{v}\):
\[ \vec{u} \cdot \vec{v} = (\hat{i} - \hat{j} - 2\hat{k}) \cdot (2\hat{i} + \hat{j} - \hat{k}) \]
Calculating this:
\[ = 1 \cdot 2 + (-1) \cdot 1 + (-2) \cdot (-1) = 2 - 1 + 2 = 3 \]

### Step 5: Calculate \(\vec{v} \cdot \vec{v}\)
Now, we calculate \(\vec{v} \cdot \vec{v}\):
\[ \vec{v} \cdot \vec{v} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (2\hat{i} + \hat{j} - \hat{k}) = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6 \]

### Step 6: Solve for \(\lambda\)
Substituting back into the equation:
\[ 0 = 3 + \lambda \cdot 6 \]
This gives:
\[ \lambda \cdot 6 = -3 \implies \lambda = -\frac{1}{2} \]

### Step 7: Use the cross product condition again
Now, we take the dot product of the cross product condition with \(\vec{w}\):
\[ (\vec{v} \times \vec{w}) \cdot \vec{w} = (\vec{u} + \lambda \vec{v}) \cdot \vec{w} \]
Again, the left-hand side is zero:
\[ 0 = \vec{u} \cdot \vec{w} + \lambda (\vec{v} \cdot \vec{w}) \]
Substituting \(\lambda = -\frac{1}{2}\) and \(\vec{v} \cdot \vec{w} = 2\):
\[ 0 = \vec{u} \cdot \vec{w} - \frac{1}{2} \cdot 2 \]
This simplifies to:
\[ 0 = \vec{u} \cdot \vec{w} - 1 \implies \vec{u} \cdot \vec{w} = 1 \]

### Final Answer
Thus, the value of \(\vec{u} \cdot \vec{w}\) is:
\[ \boxed{1} \]