Question

If the points (1, 0, 0), (0, 3, 0) and (0, 0, 2) lie on a plane, then the unit normal vector \(\hat n\) to the plane is

• A. \(\frac{1}{\sqrt{14}}(\hat{i}+3\hat{j}+2\hat{k})\)
• B. \(\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\)
• C. \(\frac{1}{\sqrt{14}}(2\hat{i}+3\hat{j}+\hat{k})\)
• D. \(\frac{1}{7}(3\hat{i}+2\hat{j}+6\hat{k})\)
• E. \(\frac{1}{7}(6\hat{i}+2\hat{j}+3\hat{k})\)

Answer 1

Answer: (E)

We are given three points on the plane: \[ A(1, 0, 0), \quad B(0, 3, 0), \quad C(0, 0, 2). \] To find the normal vector \( \mathbf{n} \) to the plane, we need two vectors on the plane. These can be obtained by subtracting the coordinates of the points: \[ \mathbf{AB} = B - A = (0 - 1, 3 - 0, 0 - 0) = (-1, 3, 0), \] \[ \mathbf{AC} = C - A = (0 - 1, 0 - 0, 2 - 0) = (-1, 0, 2). \] Next, we find the cross product \( \mathbf{n} = \mathbf{AB} \times \mathbf{AC} \) to get a vector perpendicular to both \( \mathbf{AB} \) and \( \mathbf{AC} \). The cross product is computed as follows: \[ \mathbf{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 0 \\ -1 & 0 & 2 \end{vmatrix}. \] Expanding this determinant: \[ \mathbf{n} = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 0 \\ -1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 3 \\ -1 & 0 \end{vmatrix} \] \[ \mathbf{n} = \hat{i} (3 \cdot 2 - 0 \cdot 0) - \hat{j} (-1 \cdot 2 - (-1) \cdot 0) + \hat{k} (-1 \cdot 0 - (-1) \cdot 3) \] \[ \mathbf{n} = \hat{i} (6) - \hat{j} (-2) + \hat{k} (3) \] \[ \mathbf{n} = 6 \hat{i} + 2 \hat{j} + 3 \hat{k}. \] ### Step 2: Normalize the normal vector To find the unit normal vector, we divide \( \mathbf{n} \) by its magnitude. The magnitude of \( \mathbf{n} \) is: \[ |\mathbf{n}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7. \] Thus, the unit normal vector is: \[ \hat{n} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} + 3 \hat{k}). \]

The correct option is (E) : \(\frac{1}{7}(6\hat{i}+2\hat{j}+3\hat{k})\)

Answer 2

Let the points be A(1, 0, 0), B(0, 3, 0), and C(0, 0, 2). We can find two vectors lying in the plane by taking the differences between the points:

\(\bar{AB} = (0 - 1, 3 - 0, 0 - 0) = (-1, 3, 0) = -\hat{i} + 3\hat{j} + 0\hat{k}\)

\(\bar{AC} = (0 - 1, 0 - 0, 2 - 0) = (-1, 0, 2) = -\hat{i} + 0\hat{j} + 2\hat{k}\)

The normal vector to the plane is given by the cross product of these two vectors:

\(\bar{n} = \bar{AB} \times \bar{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 0 \\ -1 & 0 & 2 \end{vmatrix} = \hat{i}(3(2) - 0(0)) - \hat{j}((-1)(2) - 0(-1)) + \hat{k}((-1)(0) - 3(-1)) = 6\hat{i} + 2\hat{j} + 3\hat{k}\)

So, \(\bar{n} = 6\hat{i} + 2\hat{j} + 3\hat{k} = (6, 2, 3)\).

To find the unit normal vector, we divide by the magnitude of \(\bar{n}\):

\(|\bar{n}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7\)

Therefore, the unit normal vector is:

\(\hat{n} = \frac{\bar{n}}{|\bar{n}|} = \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = \frac{1}{7}(6\hat{i} + 2\hat{j} + 3\hat{k})\)