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UPSC NDA I 2026 Mathematics Paper

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For the next two (02) items that follow:

Let

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

be unit vectors.

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Question : 65 of 120

Marks: +1, -0

What is the angle between

\vec{a}

\vec{b}

?

$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\frac{2 \pi}{3}$

Solution:

Concept:

Since

\vec a+\vec b

is a unit vector, square its magnitude to find

\vec a\cdot\vec b

.

Explanation:

Given

|\vec a|=|\vec b|=1

and

|\vec a+\vec b|=1

.

Squaring:

|\vec a+\vec b|^2=1

|\vec a|^2+|\vec b|^2+2(\vec a\cdot\vec b)=1

1+1+2(\vec a\cdot\vec b)=1

\vec a\cdot\vec b=-\frac12

If

\theta

is the angle between

\vec a

and

\vec b

, then:

\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}=-\frac12

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

Answer:

\frac{2\pi}{3}

, i.e. option D.

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