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UPSC NDA II 2025 Mathematics Paper

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Question : 66 of 120

Marks: +1, -0

Let

\vec{p}=\vec{a}-\vec{b},\vec{q}=\vec{a}+\vec{b}

| \vec{a} | = | \vec{b} | =2

\vec{a}\cdot \vec{b}=2

, then what is the value of

| \vec{p}\times \vec{q} |

?

$\sqrt{3}$
$\sqrt{6}$
$2\sqrt{3}$
$4\sqrt{3}$

Solution:

Concept:

Use vector cross product formula:

|\vec{p}\times\vec{q}| = |\vec{p}||\vec{q}|\sin\theta

.

First find magnitudes of

\vec{p}

and

\vec{q}

, and the angle between them.

Explanation:

\vec{p} = \vec{a}-\vec{b}

,

\vec{q} = \vec{a}+\vec{b}

.

Given

|\vec{a}|=|\vec{b}|=2

and

\vec{a}\cdot\vec{b}=2

.

|\vec{p}|^2 = |\vec{a}|^2+|\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 4+4-4=4 \implies |\vec{p}|=2

.

|\vec{q}|^2 = |\vec{a}|^2+|\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = 4+4+4=12 \implies |\vec{q}|=2\sqrt{3}

.

Find dot product:

\vec{p}\cdot\vec{q} = |\vec{a}|^2 - |\vec{b}|^2 = 4-4=0

.

Hence

\vec{p}\perp\vec{q}

, so

\theta=90^\circ

and

\sin\theta=1

.

Now

|\vec{p}\times\vec{q}| = |\vec{p}||\vec{q}|\sin\theta = 2 \cdot 2\sqrt{3} \cdot 1 = 4\sqrt{3}

.

Answer:

Option D.

4\sqrt{3}

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