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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 : 66 of 120

Marks: +1, -0

What is the angle between

\vec{a}

\vec{c}

?

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

Solution:

Concept:

Use

|\vec{x}|^2 = \vec{x}\cdot\vec{x}

for all given unit vectors to determine dot products.

Explanation:

Given

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

and

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

.

From

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

:

|\vec{a}+\vec{b}|^2 = 1 + 1 + 2\vec{a}\cdot\vec{b} = 1

\Rightarrow \vec{a}\cdot\vec{b} = -\frac{1}{2}

From

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

:

\Rightarrow \vec{b}\cdot\vec{c} = -\frac{1}{2}

From

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

:

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

3 + 2\left(-\frac{1}{2} - \frac{1}{2} + \vec{a}\cdot\vec{c}\right) = 1

1 + 2\vec{a}\cdot\vec{c} = 1

\vec{a}\cdot\vec{c} = 0

Since

|\vec{a}| = |\vec{c}| = 1

,

\cos\theta = \vec{a}\cdot\vec{c} = 0

, so

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

.

Answer:

\frac{\pi}{2}

, i.e. option C.

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