Let `vec u=hat i+hat j,vec v=hat i-hat j and vec w=hat i+2 hat j+3hat k`.If `hat n` is a unit vector such that `vec u*hat n=0 and vec v*hat n=0 ` then `|vec w*hat n|` is equal to

Options

A. `1`
B. `2`
C. `3`
D. `0`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Define the vectors
Given:
- \(\vec{u} = \hat{i} + \hat{j}\)
- \(\vec{v} = \hat{i} - \hat{j}\)
- \(\vec{w} = \hat{i} + 2\hat{j} + 3\hat{k}\)

### Step 2: Understand the conditions for \(\hat{n}\)
We know that:
- \(\vec{u} \cdot \hat{n} = 0\)
- \(\vec{v} \cdot \hat{n} = 0\)

This means that \(\hat{n}\) is perpendicular to both \(\vec{u}\) and \(\vec{v}\).

### Step 3: Find the cross product \(\vec{u} \times \vec{v}\)
To find a vector that is perpendicular to both \(\vec{u}\) and \(\vec{v}\), we can calculate the cross product:

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} \]

Calculating the determinant, we have:

\[ \vec{u} \times \vec{v} = \hat{i}(1 \cdot 0 - 0 \cdot (-1)) - \hat{j}(1 \cdot 0 - 0 \cdot 1) + \hat{k}(1 \cdot (-1) - 1 \cdot 1) \]

This simplifies to:

\[ \vec{u} \times \vec{v} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-2) = -2\hat{k} \]

### Step 4: Normalize the cross product to find the unit vector \(\hat{n}\)
The magnitude of \(\vec{u} \times \vec{v}\) is:

\[ |\vec{u} \times \vec{v}| = |-2| = 2 \]

Thus, the unit vector \(\hat{n}\) is:

\[ \hat{n} = \frac{\vec{u} \times \vec{v}}{|\vec{u} \times \vec{v}|} = \frac{-2\hat{k}}{2} = -\hat{k} \]

### Step 5: Calculate \(|\vec{w} \cdot \hat{n}|\)
Now we need to compute \(|\vec{w} \cdot \hat{n}|\):

\[ \vec{w} \cdot \hat{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-\hat{k}) = 0 + 0 - 3 = -3 \]

Thus,

\[ |\vec{w} \cdot \hat{n}| = |-3| = 3 \]

### Final Answer
Therefore, the value of \(|\vec{w} \cdot \hat{n}|\) is \(3\).

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