Vectors & 3D Geometry
Parallelogram with projections condition — finding AB and AC
MJAT_TS6_P1
Grade 12
Question:
In the parallelepiped $ABP'CC'DB'P$, edges $\overrightarrow{AB}=\vec{a}$, $\overrightarrow{AC}=\vec{b}$, $\overrightarrow{AD}=\vec{c}$, $|\vec{a}|=\sqrt{3}$, $\overrightarrow{AB}\times\overrightarrow{AC}=\vec{b}\times\vec{a}$, $\overrightarrow{AB}\times\overrightarrow{AD}=\vec{c}$. Projections of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ on diagonal $\overrightarrow{AP}$ both equal 1. Then which of the following can be correct?
A) $\overrightarrow{AC}=\dfrac{\vec{a}}{3}-\dfrac{1}{9}(\vec{a}\times(2\vec{b}-\vec{c}))$
B) $\overrightarrow{AD}=\dfrac{\vec{a}}{3}+\dfrac{1}{9}(\vec{a}\times(\vec{b}-\vec{c}))$
C) $\overrightarrow{AB}=\vec{a}+\dfrac{1}{3}(\vec{a}\times(2\vec{b}+\vec{c}))$
D) $\overrightarrow{CD}=\dfrac{\vec{a}}{3}+\dfrac{1}{9}(\vec{a}\times(\vec{b}-2\vec{c}))$
Step-by-Step Solution
Key Concept: The diagonal $\overrightarrow{AP}=\vec{a}+\vec{b}+\vec{c}$ (sum of three adjacent edges). The projection of $\overrightarrow{AB}=\vec{a}$ on $\overrightarrow{AP}$: $\frac{\vec{a}\cdot(\vec{a}+\vec{b}+\vec{c})}{|\vec{a}+\vec{b}+\vec{c}|}=1$. Similarly for $\overrightarrow{AC}=\vec{b}$.
Answer: A, B, D.
Correct Answer: ABD