If `u=i+j-k, v=2i+j+k" and "w=i+j+2k` then the magnitude of projection of `u times v` on w is given by

Options

A. `sqrt(1/2)`
B. `sqrt(1/3)`
C. `sqrt(3/4)`
D. `sqrt(3/2)`

Answer

To find the magnitude of the projection of \( \mathbf{u} \times \mathbf{v} \) on \( \mathbf{w} \), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \mathbf{u} = \mathbf{i} + \mathbf{j} - \mathbf{k} \]
\[ \mathbf{v} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} \]
\[ \mathbf{w} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \]

### Step 2: Calculate the cross product \( \mathbf{u} \times \mathbf{v} \)
Using the determinant method for the cross product:
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 2 & 1 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 1 + 1 = 2 \)
2. \( \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-1)(2) = 1 + 2 = 3 \)
3. \( \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (1)(2) = 1 - 2 = -1 \)

Thus,
\[ \mathbf{u} \times \mathbf{v} = 2\mathbf{i} - 3\mathbf{j} - \mathbf{k} \]

### Step 3: Find the dot product \( (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} \)
Now we compute the dot product:
\[ \mathbf{w} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \]
\[ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (2\mathbf{i} - 3\mathbf{j} - \mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \]
Calculating the dot product:
\[ = 2 \cdot 1 + (-3) \cdot 1 + (-1) \cdot 2 = 2 - 3 - 2 = -3 \]

### Step 4: Calculate the magnitude of \( \mathbf{w} \)
The magnitude of \( \mathbf{w} \) is:
\[ |\mathbf{w}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 5: Calculate the magnitude of the projection
The magnitude of the projection of \( \mathbf{u} \times \mathbf{v} \) on \( \mathbf{w} \) is given by:
\[ \text{Magnitude of projection} = \frac{|(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}|}{|\mathbf{w}|} \]
Substituting the values we found:
\[ = \frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \]

### Final Answer
Thus, the magnitude of the projection of \( \mathbf{u} \times \mathbf{v} \) on \( \mathbf{w} \) is:
\[ \frac{\sqrt{6}}{2} \]