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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}

be unit vectors. Further,

\vec{a}

is perpendicular to

\vec{b} ; \vec{c}

makes an angle

\frac{\pi}{3}

with both

\vec{a}

and

\vec{b}

; and

\vec{c}=p \vec{a}+q \vec{b}+r(\vec{a} \times \vec{b})

.

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Question : 69 of 120

Marks: +1, -0

What is the value of

(p+q)

?

$\frac{1}{2}$
1
$\frac{3}{2}$
2

Solution:

Concept:

Use dot product with

\vec{a}

and

\vec{b}

, since

\vec{a}, \vec{b}, \vec{a}\times\vec{b}

are mutually perpendicular unit vectors.

Explanation:

Given

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

and

\vec{a}\perp\vec{b}

, so

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

and

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

.

Also

\vec{c}

makes angle

\frac{\pi}{3}

with both

\vec{a}

and

\vec{b}

, so

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

and

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

.

Dot

\vec{c}=p\vec{a}+q\vec{b}+r(\vec{a}\times\vec{b})

with

\vec{a}

:

\vec{c}\cdot\vec{a}=p(\vec{a}\cdot\vec{a})+q(\vec{b}\cdot\vec{a})+r((\vec{a}\times\vec{b})\cdot\vec{a})

Since

(\vec{a}\times\vec{b})\cdot\vec{a}=0

, we get

p=\frac{1}{2}

.

Dot with

\vec{b}

:

\vec{c}\cdot\vec{b}=p(\vec{a}\cdot\vec{b})+q(\vec{b}\cdot\vec{b})+r((\vec{a}\times\vec{b})\cdot\vec{b})

Thus

q=\frac{1}{2}

.

Therefore,

p+q=\frac{1}{2}+\frac{1}{2}=1

.

Answer:

p+q=1

, which is option B.

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