Planes in 3D Math Example 2

Solution

1. 1
   Find two vectors in the plane: $\vec{AB} = \langle -1, 2, 0 \rangle$ and $\vec{AC} = \langle -1, 0, 3 \rangle$.
2. 2
   The normal vector is $\vec{AB} \times \vec{AC} = \langle (2)(3)-(0)(0), (0)(-1)-(−1)(3), (-1)(0)-(2)(-1) \rangle = \langle 6, 3, 2 \rangle$.
3. 3
   Using point $A(1,0,0)$: $6(x-1) + 3(y-0) + 2(z-0) = 0$, so $6x + 3y + 2z = 6$.
4. 4
   Verify: $B$: $6(0) + 3(2) + 2(0) = 6$ ✓. $C$: $6(0) + 3(0) + 2(3) = 6$ ✓.