Vectors
Non-coplanar vectors — perpendicularity and minimum norm
MJAT_TS3_P2
Grade 12
Question:
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector perpendicular to $\vec{a}+\vec{b}+\vec{c}$. If $\vec{d} = (\vec{a}\times\vec{b})\sin x + (\vec{b}\times\vec{c})\cos y + 2(\vec{c}\times\vec{a})$, then:
A) $\dfrac{\vec{d}\cdot(\vec{a}+\vec{c})}{[\vec{a}\vec{b}\vec{c}]} = 2$
B) $\dfrac{\vec{d}\cdot(\vec{a}+\vec{c})}{[\vec{a}\vec{b}\vec{c}]} = -2$
C) Minimum value of $x^2+y^2 = \dfrac{\pi^2}{4}$
D) Minimum value of $x^2+y^2 = \dfrac{5\pi^2}{4}$
Step-by-Step Solution
Key Concept: $\vec{d}\perp(\vec{a}+\vec{b}+\vec{c})$: compute $\vec{d}\cdot\vec{a}$, $\vec{d}\cdot\vec{b}$, $\vec{d}\cdot\vec{c}$ using $(\vec{a}\times\vec{b})\cdot\vec{c}=[\vec{a}\vec{b}\vec{c}]$. Then $\vec{d}\cdot(\vec{a}+\vec{b}+\vec{c})=[\vec{a}\vec{b}\vec{c}](\sin x+\cos y+2)=0\Rightarrow\sin x+\cos y=-2\Rightarrow\sin x=-1$ and $\cos y=-1$.
B ✓ (ratio $=-2$), D ✓ (min $=5\pi^2/4$). Answer: B, D.
Correct Answer: BD