Find a unit vector perpendicular to plane ABC when position vectors of A,B,C are `3 hati - hatj + 2hatk, hati - hatj- 3hatk and 4 hati - 3 hatj + hatk` respectively.

Answer

To find a unit vector perpendicular to the plane formed by the points A, B, and C with given position vectors, we can follow these steps:

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \( \vec{A} = 3\hat{i} - \hat{j} + 2\hat{k} \)
- \( \vec{B} = \hat{i} - \hat{j} - 3\hat{k} \)
- \( \vec{C} = 4\hat{i} - 3\hat{j} + \hat{k} \)

### Step 2: Find the vectors \( \vec{BA} \) and \( \vec{BC} \)
To find the vector \( \vec{BA} \):
\[ \vec{BA} = \vec{A} - \vec{B} = (3\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} - \hat{j} - 3\hat{k}) \]
Calculating this gives:
\[ \vec{BA} = (3 - 1)\hat{i} + (-1 + 1)\hat{j} + (2 + 3)\hat{k} = 2\hat{i} + 0\hat{j} + 5\hat{k} = 2\hat{i} + 5\hat{k} \]

To find the vector \( \vec{BC} \):
\[ \vec{BC} = \vec{C} - \vec{B} = (4\hat{i} - 3\hat{j} + \hat{k}) - (\hat{i} - \hat{j} - 3\hat{k}) \]
Calculating this gives:
\[ \vec{BC} = (4 - 1)\hat{i} + (-3 + 1)\hat{j} + (1 + 3)\hat{k} = 3\hat{i} - 2\hat{j} + 4\hat{k} \]

### Step 3: Find the cross product \( \vec{D} = \vec{BA} \times \vec{BC} \)
The cross product can be calculated using the determinant:
\[ \vec{D} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 5 \\ 3 & -2 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{D} = \hat{i} \begin{vmatrix} 0 & 5 \\ -2 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5 \\ 3 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 0 \\ 3 & -2 \end{vmatrix} \]
Calculating each of the minors:
- For \( \hat{i} \): \( 0 \cdot 4 - (-2) \cdot 5 = 0 + 10 = 10 \)
- For \( \hat{j} \): \( 2 \cdot 4 - 3 \cdot 5 = 8 - 15 = -7 \) (but we take the negative, so it becomes \( +7 \))
- For \( \hat{k} \): \( 2 \cdot (-2) - 3 \cdot 0 = -4 - 0 = -4 \)

Thus, we have:
\[ \vec{D} = 10\hat{i} + 7\hat{j} - 4\hat{k} \]

### Step 4: Find the magnitude of \( \vec{D} \)
The magnitude of \( \vec{D} \) is given by:
\[ |\vec{D}| = \sqrt{(10)^2 + (7)^2 + (-4)^2} = \sqrt{100 + 49 + 16} = \sqrt{165} \]

### Step 5: Find the unit vector
The unit vector \( \hat{n} \) in the direction of \( \vec{D} \) is given by:
\[ \hat{n} = \frac{\vec{D}}{|\vec{D}|} = \frac{10\hat{i} + 7\hat{j} - 4\hat{k}}{\sqrt{165}} \]

### Final Answer
The unit vector perpendicular to the plane ABC is:
\[ \hat{n} = \frac{10\hat{i} + 7\hat{j} - 4\hat{k}}{\sqrt{165}} \]

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