15. Find a scalar equation of the plane that contains the origin and the point $(2,-3,2)$ and is perpendicular to the plane $x+2 y-z+3=0$
[5 marks]
Write the scalar equation of the plane using the normal vector and the fact that the plane contains the origin: \(\boxed{x - 4y - 7z = 0}\)
Step 1 :Find the normal vector of the given plane: \(\textbf{n}_{\text{given}} = (1, 2, -1)\)
Step 2 :Find a vector in the plane containing the origin and the point (2, -3, 2): \(\textbf{v} = (2, -3, 2)\)
Step 3 :Find the normal vector of the new plane by taking the cross product of the normal vector of the given plane and the vector in the plane: \(\textbf{n}_{\text{new}} = (1, -4, -7)\)
Step 4 :Write the scalar equation of the plane using the normal vector and the fact that the plane contains the origin: \(\boxed{x - 4y - 7z = 0}\)