Passage: Let a , b , c , a + b , b + c , a + b + c be unit vectors. Question: What is the angle between a and b ?

Correct answer: D. (2π)/(3) Given | a | = 1 , | b | = 1 , and | a + b | = 1 . Squaring the magnitude of a + b , we get: | a + b |^2 = | a |^2 + | b |^2 + 2| a || b | θ Substituting the given values: 1^2 = 1^2 + 1^2 + 2(1)(1) θ 1 = 2 + 2 θ 2 θ = -1 θ = -(1)/(2) θ = (2π)/(3) Answer: (2π)/(3)

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

\vec{a}

\vec{b}

\vec{c}

\vec{a}+\vec{b}

\vec{b}+\vec{c}

\vec{a}+\vec{b}+\vec{c}

be unit vectors.

Question: What is the angle between

\vec{a}

and

\vec{b}

Options

A $\dfrac{\pi}{6}$
B $\dfrac{\pi}{4}$
C $\dfrac{\pi}{2}$
D $\dfrac{2\pi}{3}$

Correct answer

D. $\dfrac{2\pi}{3}$

Step-by-step solution

Given

|\vec{a}| = 1

|\vec{b}| = 1

, and

|\vec{a}+\vec{b}| = 1

.

Squaring the magnitude of

\vec{a}+\vec{b}

, we get:

|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta

Substituting the given values:

1^2 = 1^2 + 1^2 + 2(1)(1)\cos\theta

1 = 2 + 2\cos\theta

2\cos\theta = -1

\cos\theta = -\dfrac{1}{2}

\theta = \dfrac{2\pi}{3}

Answer:

\dfrac{2\pi}{3}

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