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- Homework Statement
- The planes ABCD and EFGH are parallel to the horizontal surface represented by the xy-plane. The triangles APD and BQC are congruent isosceles triangles and the lengths of the pillars AE, BF, CG, and DH are of the same height. The equation of the plane ABQP is given by -2x-y+3z = 2 and the point P has position vector 4i + 2j + 4k.

Explain why plane CDPQ is perpendicular to 2i + j + 3k. Hence show that the equation of the plane CDPQ is 2x + y + 3z = 22

- Relevant Equations
- Equation of plane: r . n = c

I know the normal of plane ABQP is -2i - j + 3k but I don't know how to prove that 2i + j + 3k is the normal vector of plane CDPQ

Thanks