Question

Find the vector equation of the plane passing through points \( A(1, 1, 2) \), \( B(0, 2, 3) \), and \( C(4, 5, 6) \).

Hint:

To find the equation of a plane, first find two vectors in the plane, then take their cross product to find the normal vector. Use the point-normal form of the plane equation.

Answer 1

Step 1: Find two vectors in the plane.
The vectors \( \vec{AB} \) and \( \vec{AC} \) are in the plane. They are given by: \[ \vec{AB} = \langle 0 - 1, 2 - 1, 3 - 2 \rangle = \langle -1, 1, 1 \rangle \] \[ \vec{AC} = \langle 4 - 1, 5 - 1, 6 - 2 \rangle = \langle 3, 4, 4 \rangle \]

Step 2: Find the normal vector to the plane.
The normal vector \( \vec{n} \) is given by the cross product of \( \vec{AB} \) and \( \vec{AC} \): \[ \vec{n} = \vec{AB} \times \vec{AC} \] \[ \vec{n} = \langle -1, 1, 1 \rangle \times \langle 3, 4, 4 \rangle \] Using the formula for the cross product: \[ \vec{n} = \langle (1 \cdot 4 - 1 \cdot 4), (1 \cdot 3 - (-1) \cdot 4), (-1 \cdot 4 - 1 \cdot 3) \rangle \] \[ \vec{n} = \langle 0, 7, -7 \rangle \]

Step 3: Write the vector equation of the plane.
The vector equation of the plane is: \[ \vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n} \] Substituting \( \vec{A} = \langle 1, 1, 2 \rangle \) and \( \vec{n} = \langle 0, 7, -7 \rangle \): \[ \vec{r} \cdot \langle 0, 7, -7 \rangle = \langle 1, 1, 2 \rangle \cdot \langle 0, 7, -7 \rangle \] \[ \vec{r} \cdot \langle 0, 7, -7 \rangle = 7 - 14 = -7 \] Thus, the vector equation of the plane is: \[ \vec{r} \cdot \langle 0, 7, -7 \rangle = -7 \]

Final Answer: The vector equation of the plane is: \[ \vec{r} \cdot \langle 0, 7, -7 \rangle = -7 \]