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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Vector Algebra

NDA 2026 · Mathematics · Vector Algebra

Passage: Let a , b , c , a + b , b + c , a + b + c be unit vectors....

Passage: Let a⃗\vec{a}, b⃗\vec{b}, c⃗\vec{c}, a⃗+b⃗\vec{a}+\vec{b}, b⃗+c⃗\vec{b}+\vec{c}, a⃗+b⃗+c⃗\vec{a}+\vec{b}+\vec{c} be unit vectors.

Question: What is the angle between a⃗\vec{a} and c⃗\vec{c} ?

Options

  1. Aπ6\dfrac{\pi}{6}
  2. Bπ4\dfrac{\pi}{4}
  3. Cπ2\dfrac{\pi}{2}
  4. D2π3\dfrac{2\pi}{3}

Correct answer

C. π2\dfrac{\pi}{2}

Step-by-step solution

Given ∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec{a}| = |\vec{b}| = |\vec{c}| = 1

∣a⃗+b⃗∣=1⇒∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗=1⇒1+1+2a⃗⋅b⃗=1⇒a⃗⋅b⃗=−12|\vec{a}+\vec{b}| = 1 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = 1 \Rightarrow 1 + 1 + 2\vec{a}\cdot\vec{b} = 1 \Rightarrow \vec{a}\cdot\vec{b} = -\dfrac{1}{2}

∣b⃗+c⃗∣=1⇒∣b⃗∣2+∣c⃗∣2+2b⃗⋅c⃗=1⇒1+1+2b⃗⋅c⃗=1⇒b⃗⋅c⃗=−12|\vec{b}+\vec{c}| = 1 \Rightarrow |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b}\cdot\vec{c} = 1 \Rightarrow 1 + 1 + 2\vec{b}\cdot\vec{c} = 1 \Rightarrow \vec{b}\cdot\vec{c} = -\dfrac{1}{2}

∣a⃗+b⃗+c⃗∣=1⇒∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=1|\vec{a}+\vec{b}+\vec{c}| = 1 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 1

1+1+1+2(−12−12+c⃗⋅a⃗)=11 + 1 + 1 + 2\left(-\dfrac{1}{2} - \dfrac{1}{2} + \vec{c}\cdot\vec{a}\right) = 1

3+2(−1+c⃗⋅a⃗)=13 + 2(-1 + \vec{c}\cdot\vec{a}) = 1

1+2c⃗⋅a⃗=1⇒c⃗⋅a⃗=01 + 2\vec{c}\cdot\vec{a} = 1 \Rightarrow \vec{c}\cdot\vec{a} = 0

Since the dot product of a⃗\vec{a} and c⃗\vec{c} is 00, the angle between them is π2\dfrac{\pi}{2}.

Answer: π2\dfrac{\pi}{2}
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