Vector Algebra
Vector Perpendicularity and Reflection
Grade 12
Question:
<p>Let \(\vec{a},\vec{b},\vec{c}\) be unit vectors such that
\(\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\sqrt{3}}{2}(\vec{b}+\vec{c})\).
If \(\vec{b}\) is not parallel to \(\vec{c}\), find the angle between \(\vec{a}\)
and \(\vec{b}\).</p>
<li>\(\dfrac{5\pi}{6}\)</li>
<li>\(\dfrac{\pi}{3}\)</li>
<li>\(\dfrac{\pi}{2}\)</li>
<li>\(\dfrac{2\pi}{3}\)</li>
Step-by-Step Solution
Key Concept: Apply BAC–CAB: a \times (b \times c) = (a \cdot c)b - (a \cdot b)c. Since b and c are not parallel, compare coefficients on both sides.
BAC-CAB: $\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}
=\dfrac{\sqrt{3}}{2}(\vec{b}+\vec{c})$.
Comparing coefficients (b and c linearly independent):
$\vec{a}\cdot\vec{c}=\dfrac{\sqrt{3}}{2}$ and $-\vec{a}\cdot\vec{b}=\dfrac{\sqrt{3}}{2}$
$\Rightarrow\vec{a}\cdot\vec{b}=-\dfrac{\sqrt{3}}{2}$.
$\cos\theta_{ab}=-\dfrac{\sqrt{3}}{2}\Rightarrow\theta_{ab}=\dfrac{5\pi}{6}$.
Answer: A .
Correct Answer: A