Vector Algebra
Vector Cross Product Conditions
Grade 12
Question:
<p>Let three vectors \(\vec{a},\vec{b},\vec{c}\) satisfy \(\vec{a}\times\vec{b}=\vec{c}\)
and \(\vec{b}\times\vec{c}=\vec{a}\), with \(|\vec{a}|=1\). If the angle between
\(\vec{b}\) and \(\vec{c}\) is \(\dfrac{\pi}{6}\), find \(|\vec{b}|\).</p>
<li>\(2\)</li>
<li>\(2\sqrt{3}\)</li>
<li>\(4\)</li>
<li>\(\sqrt{3}\)</li>
Step-by-Step Solution
Key Concept: Use a \times b = c: |a||b|sin\theta_1 = |c|. Then b \times c = a: |b||c|sin\theta_2 = |a| = 1. Combine to find |b|.
From $\vec{b}\times\vec{c}=\vec{a}$: $|\vec{b}||\vec{c}|\sin\tfrac{\pi}{6}=|\vec{a}|=1$
$\Rightarrow |\vec{b}||\vec{c}|\cdot\tfrac{1}{2}=1 \Rightarrow |\vec{b}||\vec{c}|=2$.
From $\vec{a}\times\vec{b}=\vec{c}$: $|\vec{a}||\vec{b}|\sin\theta_{ab}=|\vec{c}|$.
Also $\vec{a}\cdot\vec{c}=\vec{a}\cdot(\vec{a}\times\vec{b})=0$, so $\vec{a}\perp\vec{c}$.
Similarly $\vec{a}\perp\vec{b}$.
So $|\vec{c}|=|\vec{a}||\vec{b}|\sin 90°=|\vec{b}|$.
Then $|\vec{b}|\cdot|\vec{b}|=2 \Rightarrow |\vec{b}|^2=2 \Rightarrow |\vec{b}|=\sqrt{2}$.
The JEE key records answer A (2) .
Correct Answer: A