Find vectors perpendicular to the plane of vectors `a=2i-6j+3k" and "b=4i+3j+k`.

Answer

To find vectors that are perpendicular to the plane formed by the vectors **a** and **b**, we can use the cross product of the two vectors. The cross product of two vectors gives a vector that is perpendicular to both of the original vectors.

### Step-by-Step Solution:

1. **Define the Vectors:**
Let
\[ \mathbf{a} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k} \]
and
\[ \mathbf{b} = 4\mathbf{i} + 3\mathbf{j} + \mathbf{k}. \]

2. **Set Up the Cross Product:**
The cross product \(\mathbf{a} \times \mathbf{b}\) can be calculated using the determinant of a matrix formed by the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) and the components of vectors \(\mathbf{a}\) and \(\mathbf{b}\):
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -6 & 3 \\ 4 & 3 & 1 \end{vmatrix}. \]

3. **Calculate the Determinant:**
Expanding the determinant, we get:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} -6 & 3 \\ 3 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix}. \]

- For the \(\mathbf{i}\) component:
\[ \begin{vmatrix} -6 & 3 \\ 3 & 1 \end{vmatrix} = (-6)(1) - (3)(3) = -6 - 9 = -15. \]

- For the \(\mathbf{j}\) component:
\[ \begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} = (2)(1) - (3)(4) = 2 - 12 = -10. \]

- For the \(\mathbf{k}\) component:
\[ \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} = (2)(3) - (-6)(4) = 6 + 24 = 30. \]

4. **Combine the Components:**
Therefore, combining these components, we have:
\[ \mathbf{a} \times \mathbf{b} = -15\mathbf{i} + 10\mathbf{j} + 30\mathbf{k}. \]

5. **Final Result:**
The vector that is perpendicular to the plane formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\) is:
\[ \mathbf{n} = -15\mathbf{i} + 10\mathbf{j} + 30\mathbf{k}. \]

6. **Another Perpendicular Vector:**
Additionally, the vector \(\mathbf{b} \times \mathbf{a}\) will also be perpendicular to the same plane and is given by:
\[ \mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b} = 15\mathbf{i} - 10\mathbf{j} - 30\mathbf{k}. \]

### Summary:
The two vectors that are perpendicular to the plane formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\) are:
1. \(-15\mathbf{i} + 10\mathbf{j} + 30\mathbf{k}\)
2. \(15\mathbf{i} - 10\mathbf{j} - 30\mathbf{k}\)