Given a parallelepiped $ABCDA1B1C1D1$, with vector $a = AB, b = AD, c = AA1$, find vectors $AC1, B1D1, DB1$

Question

Given a parallelepiped $ABCDA_1B_1C_1D_1$, with vector $a = AB, b = AD, c = AA_1$, find vectors $AC_1, B_1D_1, DB_1$.

Answers should be:

$AC_1 = a + b + c$

$B_1D_1 = b - a$

$DB_1 = a + c - b$

The issue is, that I'm failing to figure out how to solve it.

Answer 1

Hint: Opposite sides are parallel and equal, so $\vec a = \overrightarrow{AB} = \overrightarrow{A_1B_1} = \overrightarrow{DC} = \overrightarrow{D_1C_1}$. Similar equalities hold for $\vec b,\vec c$. Then for example $\overrightarrow{AC_1} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CC_1} = \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{AA_1} = \vec a + \vec b + \vec c$.