Question

The plane, passing through the points (0, -1, 2) and (-1, 2, 1) and parellel to the line passing through (5,1,-7) and (1,-1,-1), also passes through the point

• A. (1,-2, 1)
• B. (-2, 5, 0)
• C. (2, 0, 1)
• D. (0, 5, -2)

Answer 1

The Correct Option is B

We are given that the plane passes through the points \( (0, -1, 2) \) and \( (-1, 2, 1) \), and is parallel to the line passing through \( (5, 1, -7) \) and \( (1, -1, -1) \).
1. Step 1: Find vectors in the plane:
The vector passing through the points \( (0, -1, 2) \) and \( (-1, 2, 1) \) is: \[ \mathbf{a} = (-1 - 0, 2 - (-1), 1 - 2) = (-1, 3, -1) \] The vector representing the direction of the given line is: \[ \mathbf{b} = (1 - 5, -1 - 1, -1 - (-7)) = (-4, -2, 6) \] 2. Step 2: Find the normal vector to the plane:
The normal vector \( \mathbf{n} \) to the plane is the cross product of the two vectors \( \mathbf{a} \) and \( \mathbf{b} \): \[ \mathbf{n} = \mathbf{a} \times \mathbf{b} \] Using the determinant to calculate the cross product: \[ \mathbf{n} = 
\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -1 & 3 & -1 -4 & -2 & 6 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{n} = \hat{i} \left( \begin{vmatrix} 3 & -1 -2 & 6 \end{vmatrix} \right) - \hat{j} \left( \begin{vmatrix} -1 & -1 -4 & 6 \end{vmatrix} \right) + \hat{k} \left( \begin{vmatrix} -1 & 3 -4 & -2 \end{vmatrix} \right) \] Calculating each determinant: \[ \mathbf{n} = \hat{i} \left( (3)(6) - (-1)(-2) \right) - \hat{j} \left( (-1)(6) - (-1)(-4) \right) + \hat{k} \left( (-1)(-2) - (3)(-4) \right) \] \[ \mathbf{n} = \hat{i} (18 - 2) - \hat{j} (-6 - 4) + \hat{k} (2 + 12) \] \[ \mathbf{n} = \hat{i} (16) - \hat{j} (-10) + \hat{k} (14) \] \[ \mathbf{n} = (16, 10, 14) \] 3. Step 3: Equation of the plane:
The equation of the plane is: \[ 16(x - 0) + 10(y + 1) + 14(z - 2) = 0 \] Simplifying: \[ 16x + 10y + 10 + 14z - 28 = 0 \] \[ 16x + 10y + 14z - 18 = 0 \] 4. Step 4: Check the points given in the options:
Substituting \( (-2, 5, 0) \) into the equation of the plane: \[ 16(-2) + 10(5) + 14(0) - 18 = 0 \] \[ -32 + 50 + 0 - 18 = 0 \] \[ 0 = 0 \quad (\text{This satisfies the plane equation}) \] Thus, the correct point that lies on the plane is \( (-2, 5, 0) \).