Question:
<p>Let \(\vec{a},\vec{b},\vec{c}\) be non-coplanar unit vectors such that
\(\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\vec{b}+\vec{c}}{\sqrt{2}}\).
Find the angle between \(\vec{a}\) and \(\vec{b}\).</p>
\(\dfrac{3\pi}{4}\)
\(\dfrac{\pi}{4}\)
\(\dfrac{\pi}{2}\)
\(\dfrac{\pi}{3}\)
Step-by-Step Solution
Key Concept: BAC–CAB: a \times (b \times c) = (a \cdot c)b - (a \cdot b)c = (b+c)/\sqrt{2.} Since b,c independent, compare coefficients: a \cdot c = 1/\sqrt{2} and a \cdot b = -1/\sqrt{2.}
BAC-CAB: $(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}=\dfrac{1}{\sqrt{2}}\vec{b}+\dfrac{1}{\sqrt{2}}\vec{c}$.
Comparing: $\vec{a}\cdot\vec{c}=\dfrac{1}{\sqrt{2}}$ and $-\vec{a}\cdot\vec{b}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow\vec{a}\cdot\vec{b}=-\dfrac{1}{\sqrt{2}}$.
Since all unit vectors: $\cos\theta_{ab}=-\dfrac{1}{\sqrt{2}}\Rightarrow\theta_{ab}=\dfrac{3\pi}{4}$.
Answer: A .
Correct Answer: A