Question

The equation of the plane passing through a point \( A(2, -1, 3) \) and parallel to the vectors \[ \mathbf{a} = (3, 0, -1) \quad \text{and} \quad \mathbf{b} = (-3, 2, 2) \] is:

• A. \( 2x - 3y + 6z + 25 = 0 \)
• B. \( 3x - 2y + 6z - 25 = 0 \)
• C. \( 2x - 3y + 6z - 25 = 0 \)
• D. \( 3x - 2y + 6z = 0 \)

Hint:

To find the equation of a plane, first find the normal vector by taking the cross product of two given vectors parallel to the plane. Then use the point and the normal vector to construct the plane equation.

Answer 1

The Correct Option is C

Step 1: Use the given point and vectors.
We are given the point \( A(2, -1, 3) \) and the vectors \( \mathbf{a} = (3, 0, -1) \) and \( \mathbf{b} = (-3, 2, 2) \), which define the direction of the plane. The general equation of a plane is: \[ \mathbf{r} \cdot (\mathbf{n}) = d \] where \( \mathbf{r} = (x, y, z) \) is the position vector of any point on the plane and \( \mathbf{n} \) is the normal vector to the plane.

Step 2: Find the normal vector.
The normal vector \( \mathbf{n} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Thus, we find \( \mathbf{n} \) by taking the cross product of \( \mathbf{a} \) and \( \mathbf{b} \): \[ \mathbf{n} = \mathbf{a} \times \mathbf{b} \] \[ \mathbf{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 3 & 0 & -1 -3 & 2 & 2 \end{vmatrix} \] \[ \mathbf{n} = \hat{i}(0 \cdot 2 - (-1) \cdot 2) - \hat{j}(3 \cdot 2 - (-1) \cdot (-3)) + \hat{k}(3 \cdot 2 - 0 \cdot (-3)) \] \[ \mathbf{n} = \hat{i}(0 + 2) - \hat{j}(6 - 3) + \hat{k}(6) \] \[ \mathbf{n} = 2\hat{i} - 3\hat{j} + 6\hat{k} \] Thus, \( \mathbf{n} = (2, -3, 6) \).

Step 3: Equation of the plane.
The equation of the plane is: \[ 2(x - 2) - 3(y + 1) + 6(z - 3) = 0 \] Simplifying the equation: \[ 2x - 4 - 3y - 3 + 6z - 18 = 0 \] \[ 2x - 3y + 6z - 25 = 0 \]

Step 4: Conclusion.
Thus, the equation of the plane is \( 2x - 3y + 6z - 25 = 0 \), corresponding to option (C).