Find all vectors of magnitude `10sqrt(3)` that are perpendicular to the plane of `hat(i)+2hat(j)+hat(k)` and `-hat(i)+hat(j)+4hat(k)`.

Answer

To find all vectors of magnitude \( 10\sqrt{3} \) that are perpendicular to the plane defined by the vectors \( \hat{i} + 2\hat{j} + \hat{k} \) and \( -\hat{i} + \hat{j} + 4\hat{k} \), we can follow these steps:

### Step 1: Identify the vectors
Let:
- \( \mathbf{A} = \hat{i} + 2\hat{j} + \hat{k} \)
- \( \mathbf{B} = -\hat{i} + \hat{j} + 4\hat{k} \)

### Step 2: Find the normal vector to the plane
The normal vector \( \mathbf{N} \) to the plane formed by the vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be found using the cross product \( \mathbf{A} \times \mathbf{B} \).

\[ \mathbf{N} = \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -1 & 1 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ \mathbf{N} = \hat{i} \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ -1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = (2)(4) - (1)(1) = 8 - 1 = 7 \)

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

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

Putting it all together:

\[ \mathbf{N} = 7\hat{i} - 5\hat{j} + 3\hat{k} \]

### Step 4: Find the magnitude of the normal vector
Now, we find the magnitude of the normal vector \( \mathbf{N} \):

\[ |\mathbf{N}| = \sqrt{7^2 + (-5)^2 + 3^2} = \sqrt{49 + 25 + 9} = \sqrt{83} \]

### Step 5: Find the unit normal vector
The unit normal vector \( \hat{n} \) is given by:

\[ \hat{n} = \frac{\mathbf{N}}{|\mathbf{N}|} = \frac{7\hat{i} - 5\hat{j} + 3\hat{k}}{\sqrt{83}} \]

### Step 6: Scale the unit normal vector to the desired magnitude
To find the vectors of magnitude \( 10\sqrt{3} \), we scale the unit normal vector:

\[ \mathbf{C} = 10\sqrt{3} \cdot \hat{n} = 10\sqrt{3} \cdot \frac{7\hat{i} - 5\hat{j} + 3\hat{k}}{\sqrt{83}} = \frac{70\sqrt{3}}{\sqrt{83}}\hat{i} - \frac{50\sqrt{3}}{\sqrt{83}}\hat{j} + \frac{30\sqrt{3}}{\sqrt{83}}\hat{k} \]

### Step 7: Consider the opposite direction
Since the vector can point in either direction, we also have:

\[ -\mathbf{C} = -\left(\frac{70\sqrt{3}}{\sqrt{83}}\hat{i} - \frac{50\sqrt{3}}{\sqrt{83}}\hat{j} + \frac{30\sqrt{3}}{\sqrt{83}}\hat{k}\right) \]

### Final Solution
Thus, the vectors of magnitude \( 10\sqrt{3} \) that are perpendicular to the plane are:

\[ \mathbf{C} = \frac{70\sqrt{3}}{\sqrt{83}}\hat{i} - \frac{50\sqrt{3}}{\sqrt{83}}\hat{j} + \frac{30\sqrt{3}}{\sqrt{83}}\hat{k} \]
and
\[ -\mathbf{C} = -\frac{70\sqrt{3}}{\sqrt{83}}\hat{i} + \frac{50\sqrt{3}}{\sqrt{83}}\hat{j} - \frac{30\sqrt{3}}{\sqrt{83}}\hat{k} \]