Q219 / 269 · NDA 2025 (Phase 1) · Mathematics
Let $\vec{a}$, $\vec{b}$, $(\vec{a} \times \vec{b})$ be unit vectors. What is $(\vec{a} \cdot \vec{b})$ equal to?
- Correct answerA$0$
- B$\dfrac{1}{2}$
- C$1$
- D$3$
Solution
Given $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $|\vec{a} \times \vec{b}| = 1$.
Using Lagrange's identity, $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.
Substituting the given values, $1^2 + (\vec{a} \cdot \vec{b})^2 = 1^2 \times 1^2$.
$1 + (\vec{a} \cdot \vec{b})^2 = 1$.
$(\vec{a} \cdot \vec{b})^2 = 0 \Rightarrow \vec{a} \cdot \vec{b} = 0$.
Answer: $0$
Using Lagrange's identity, $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.
Substituting the given values, $1^2 + (\vec{a} \cdot \vec{b})^2 = 1^2 \times 1^2$.
$1 + (\vec{a} \cdot \vec{b})^2 = 1$.
$(\vec{a} \cdot \vec{b})^2 = 0 \Rightarrow \vec{a} \cdot \vec{b} = 0$.
Answer: $0$
Hence, the correct answer is option: 'A'.