Let `vec(U)=hati,hatj,vecV=hati-hatjand vec(W)=3hati+5hatj+3hatk. ` If `hat(n) =0` then `|vecW.hatn|` is equal to

Answer

To solve the problem step by step, we need to find the value of \(|\vec{W} \cdot \hat{n}|\), given that \(\hat{n} = 0\) means that \(\hat{n}\) is a unit vector perpendicular to both \(\vec{U}\) and \(\vec{V}\).

### Step 1: Identify the vectors
We have:
- \(\vec{U} = \hat{i} + \hat{j}\)
- \(\vec{V} = \hat{i} - \hat{j}\)
- \(\vec{W} = 3\hat{i} + 5\hat{j} + 3\hat{k}\)

### Step 2: Find the cross product \(\vec{U} \times \vec{V}\)
To find a vector \(\hat{n}\) that is perpendicular to both \(\vec{U}\) and \(\vec{V}\), we compute the cross product \(\vec{U} \times \vec{V}\).

Using the determinant form:
\[ \vec{U} \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ -1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \]
\[ = \hat{i}(0) - \hat{j}(0) + \hat{k}(-1 - 1) = -2\hat{k} \]

Thus, \(\vec{U} \times \vec{V} = -2\hat{k}\).

### Step 3: Normalize the cross product to find \(\hat{n}\)
To find the unit vector \(\hat{n}\), we need to normalize \(-2\hat{k}\):
\[ \hat{n} = \frac{-2\hat{k}}{|-2|} = -\hat{k} \]

### Step 4: Calculate the dot product \(\vec{W} \cdot \hat{n}\)
Now, we compute the dot product \(\vec{W} \cdot \hat{n}\):
\[ \vec{W} \cdot \hat{n} = (3\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (-\hat{k}) = 3(0) + 5(0) + 3(-1) = -3 \]

### Step 5: Find the modulus of the dot product
Finally, we find the modulus:
\[ |\vec{W} \cdot \hat{n}| = |-3| = 3 \]

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

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