if the vector `x i+yj +zk` makes an acute angle with the plane of the two vectors `2, 3,-1 and 1,-1,2` and acute angle is `cot^(-1) sqrt(2)`, then

Options

A. `xy +yz + zx = 0`
B. `x(y +z) = yz`
C. `y(z + x) = zx`
D. `z(x +y) = xy`

Answer

To solve the problem, we need to find the vector \( \mathbf{v} = xi + yj + zk \) that makes an acute angle with the plane formed by the vectors \( \mathbf{a} = 2i + 3j - k \) and \( \mathbf{b} = i - j + 2k \). The angle between the vector and the plane is given as \( \cot^{-1}(\sqrt{2}) \).

### Step-by-Step Solution:

1. **Find the Normal Vector of 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:
\[ \mathbf{n} = \mathbf{a} \times \mathbf{b} \]
where \( \mathbf{a} = (2, 3, -1) \) and \( \mathbf{b} = (1, -1, 2) \).

2. **Calculate the Cross Product**:
Using the determinant method for the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} \]
Expanding this determinant, we get:
\[ \mathbf{n} = \mathbf{i}(3 \cdot 2 - (-1) \cdot (-1)) - \mathbf{j}(2 \cdot 2 - (-1) \cdot 1) + \mathbf{k}(2 \cdot (-1) - 3 \cdot 1) \]
Simplifying:
\[ \mathbf{n} = \mathbf{i}(6 - 1) - \mathbf{j}(4 + 1) + \mathbf{k}(-2 - 3) = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k} \]
Thus, \( \mathbf{n} = 5i - 5j - 5k \).

3. **Find the Angle Between the Vector and the Normal**:
The angle \( \theta \) between the vector \( \mathbf{v} \) and the normal vector \( \mathbf{n} \) can be found using the formula:
\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{n}}{|\mathbf{v}| |\mathbf{n}|} \]
Since the angle with the plane is \( \cot^{-1}(\sqrt{2}) \), we can find \( \sin(\theta) \) and \( \cos(\theta) \):
\[ \cot(\theta) = \sqrt{2} \implies \tan(\theta) = \frac{1}{\sqrt{2}} \implies \sin(\theta) = \frac{1}{\sqrt{3}}, \quad \cos(\theta) = \frac{\sqrt{2}}{\sqrt{3}} \]

4. **Set Up the Equation**:
We know that:
\[ \sin(\theta) = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}| |\mathbf{n}|} \]
Thus, we can set up the equation:
\[ \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}| |\mathbf{n}|} = \frac{1}{\sqrt{3}} \]

5. **Calculate the Magnitude of the Normal Vector**:
The magnitude of \( \mathbf{n} \):
\[ |\mathbf{n}| = \sqrt{5^2 + (-5)^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]

6. **Substitute and Solve**:
Substitute \( |\mathbf{n}| \) into the equation:
\[ |\mathbf{v} \cdot \mathbf{n}| = \frac{1}{\sqrt{3}} |\mathbf{v}| (5\sqrt{3}) \]
Simplifying gives:
\[ |\mathbf{v} \cdot \mathbf{n}| = 5 |\mathbf{v}| \]

7. **Conclusion**:
The vector \( \mathbf{v} \) must satisfy the above equation, which relates the components \( x, y, z \) of the vector \( \mathbf{v} \) to the normal vector \( \mathbf{n} \).