If `vecP=i+j+2k and vecQ=3i-2j+k`, the unit vector perpendicular to both `vecP and vecQ` is

Options

A. `(hati+hatj-hatk)/(sqrt(3))`
B. `5(hati+hatj-hatk)`
C. `(1)/(sqrt(3))(hati-hatj+hatk)`
D. `(1)/(25)(hati+hatj-hatk)`

Answer

To find the unit vector that is perpendicular to both vectors \(\vec{P}\) and \(\vec{Q}\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{P} = \hat{i} + \hat{j} + 2\hat{k} \]
\[ \vec{Q} = 3\hat{i} - 2\hat{j} + \hat{k} \]

### Step 2: Calculate the cross product \(\vec{P} \times \vec{Q}\)
To find a vector that is perpendicular to both \(\vec{P}\) and \(\vec{Q}\), we calculate the cross product \(\vec{P} \times \vec{Q}\) using the determinant method.

\[ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 3 & -2 & 1 \end{vmatrix} \]

### Step 3: Expand the determinant
Calculating the determinant:
\[ \vec{P} \times \vec{Q} = \hat{i} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = (1)(1) - (2)(-2) = 1 + 4 = 5\)
2. \(\begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (2)(3) = 1 - 6 = -5\)
3. \(\begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} = (1)(-2) - (1)(3) = -2 - 3 = -5\)

Putting it all together:
\[ \vec{P} \times \vec{Q} = 5\hat{i} + 5\hat{j} - 5\hat{k} \]

### Step 4: Simplify the cross product
Thus, we have:
\[ \vec{P} \times \vec{Q} = 5\hat{i} + 5\hat{j} - 5\hat{k} \]

### Step 5: Calculate the magnitude of the cross product
Now, we find the magnitude of \(\vec{P} \times \vec{Q}\):
\[ |\vec{P} \times \vec{Q}| = \sqrt{(5)^2 + (5)^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 6: Find the unit vector
The unit vector \(\hat{n}\) perpendicular to both \(\vec{P}\) and \(\vec{Q}\) is given by:
\[ \hat{n} = \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \frac{5\hat{i} + 5\hat{j} - 5\hat{k}}{5\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k} \]

### Final Answer
Thus, the unit vector perpendicular to both \(\vec{P}\) and \(\vec{Q}\) is:
\[ \hat{n} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} - \frac{1}{\sqrt{3}} \hat{k} \]

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