Find the unit vector perpendicular to two vectors `vecP=hati+hatj+2hatk` and `vecQ=2hati-4hatj+5hatk`.

Answer

To find the unit vector perpendicular to the two vectors \(\vec{P} = \hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{Q} = 2\hat{i} - 4\hat{j} + 5\hat{k}\), we will follow these steps:

### Step 1: Calculate the Cross Product of Vectors \(\vec{P}\) and \(\vec{Q}\)

The cross product \(\vec{P} \times \vec{Q}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{P}\) and \(\vec{Q}\):

\[ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & -4 & 5 \end{vmatrix} \]

Calculating this determinant:

\[ \vec{P} \times \vec{Q} = \hat{i} \begin{vmatrix} 1 & 2 \\ -4 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} \]

Calculating the individual determinants:

1. \(\begin{vmatrix} 1 & 2 \\ -4 & 5 \end{vmatrix} = (1)(5) - (2)(-4) = 5 + 8 = 13\)

2. \(\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = (1)(5) - (2)(2) = 5 - 4 = 1\)

3. \(\begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} = (1)(-4) - (1)(2) = -4 - 2 = -6\)

Now substituting back into the equation for the cross product:

\[ \vec{P} \times \vec{Q} = 13\hat{i} - 1\hat{j} - 6\hat{k} \]

### Step 2: Find the Magnitude of the Cross Product

The magnitude of the vector \(\vec{R} = 13\hat{i} - 1\hat{j} - 6\hat{k}\) is given by:

\[ |\vec{R}| = \sqrt{(13)^2 + (-1)^2 + (-6)^2} \]
\[ = \sqrt{169 + 1 + 36} = \sqrt{206} \]

### Step 3: Calculate the Unit Vector

The unit vector \(\hat{n}\) in the direction of \(\vec{R}\) is given by:

\[ \hat{n} = \frac{\vec{R}}{|\vec{R}|} = \frac{13\hat{i} - 1\hat{j} - 6\hat{k}}{\sqrt{206}} \]

### Final Answer

Thus, the unit vector perpendicular to both \(\vec{P}\) and \(\vec{Q}\) is:

\[ \hat{n} = \frac{13}{\sqrt{206}}\hat{i} - \frac{1}{\sqrt{206}}\hat{j} - \frac{6}{\sqrt{206}}\hat{k} \]

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