Question

A plane passes through \( (2, 1, 2) \) and \( (1, 2, 1) \) and parallel to the line \( 2x = 3y \) and \( z = 1 \), then the plane also passes through the point

• A. \( (-6, 2, 0) \)
• B. \( (6, -2, 0) \)
• C. \( (-2, 0, 1) \)
• D. \( (2, 0, 1) \)

Hint:

For $ax=by$, direction ratios are $(b, a, 0)$.

Answer 1

The Correct Option is C

To determine which point the plane passes through, we need to understand how the points given relate to the geometry of the plane. A plane equation in 3D can be generally expressed as:

\(Ax + By + Cz + D = 0\)

Let's break down the problem step-by-step:

1. The plane passes through two points, \( (2, 1, 2) \) and \( (1, 2, 1) \). These points will help define the plane.
2. The plane is also parallel to the line given by the equations \( 2x = 3y \) and \( z = 1 \). Rewriting these, we have \( y = \frac{2}{3}x \) and a constant \( z = 1 \), indicating a line parallel to the xy-plane moving along \( z=1 \).
3. To find the normal vector \( \mathbf{n} \) of the plane, we need a vector direction lying on the plane: from point \( A(2,1,2) \) to point \( B(1,2,1) \), the direction vector is \( \mathbf{AB} = \langle 1-2, 2-1, 1-2 \rangle = \langle -1, 1, -1 \rangle \).
4. The plane is parallel to the line. Since the vector direction in the xy-plane is proportional to \( \langle 3, 2, 0 \rangle \) (due to the line direction indicated by \( 2x = 3y \)), we can cross this with the vector \(\mathbf{AB}\) to find the normal vector of the plane: \(\mathbf{n} = \langle -1, 1, -1 \rangle \times \langle 3, 2, 0 \rangle = \langle 2\cdot(-1) - 0\cdot1, 0\cdot(-1) - (-1)\cdot3, (-1)\cdot2 - 1\cdot3 \rangle\)
5. This results in the plane normal vector\( \mathbf{n} = \langle -2, 3, -5 \rangle \).
6. We now use these to form the plane equation, using point \( (2,1,2) \) as a point on the plane: \(-2(x - 2) + 3(y - 1) - 5(z - 2) = 0\) Simplifying: \(-2x + 4 + 3y - 3 - 5z + 10 = 0 \rightarrow -2x + 3y - 5z + 11 = 0\)
7. Substituting the given points into this plane equation:
   • For option \( (-6, 2, 0) \): \(-2(-6) + 3(2) - 5(0) + 11 = 12 + 6 + 11 = 29 \neq 0\)
   • For option \( (6, -2, 0) \): \(-2(6) + 3(-2) - 5(0) + 11 = -12 - 6 + 11 = -7 \neq 0\)
   • For option \( (-2, 0, 1) \): \(-2(-2) + 3(0) - 5(1) + 11 = 4 - 5 + 11 = 0\)
   • For option \( (2, 0, 1) \): \(-2(2) + 3(0) - 5(1) + 11 = -4 - 5 + 11 = 2 \neq 0\)

Therefore, the plane also passes through the point \( \mathbf{(-2, 0, 1)} \) as the calculation satisfies the equation of the plane with zero result.