Perpendicularity in a Rhombic Parallelepiped

★ 7/10solid geometry Triangle Geometry vectors

Problem

In parallelepiped $ABCD\text{-}A_1B_1C_1D_1$ the base is a rhombus, and the three edges meeting at vertex $C$ make equal angles: $\angle C_1CB=\angle C_1CD=\angle BCD=60^\circ$ .

(1) Prove that $C_1C\perp BD$ .

(2) For what value of $\dfrac{CD}{CC_1}$ is $A_1C\perp$ plane $C_1BD$ ?

Answer

Solution

Difficulty	7/10
Topics	solid geometry, Triangle Geometry, vectors

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