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UPSC NDA II 2015 Mathematics Paper

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Question : 100 of 120

Marks: +1, -0

If

\left | {{\rm{\vec a}}} \right | = 2

\left | {{\rm{\vec b}}} \right | = 3

{\left | {{\rm{\vec a}} \times {\rm{\vec b}}} \right | ^2} + {\left | {{\rm{\vec a}} \cdot {\rm{\vec b}}} \right | ^2}

is equal to

72
64
48
36

Solution:

Concept:

The expression

\left| \vec{a} \times \vec{b} \right|^2 + \left| \vec{a} \cdot \vec{b} \right|^2

simplifies using the magnitudes given as

|\vec{a}| = 2

and

|\vec{b}| = 3

, and the identity

\sin^2\theta + \cos^2\theta = 1

.

Explanation:

We know:

\left| \vec{a} \times \vec{b} \right| = |\vec{a}|\,|\vec{b}| \sin\theta

.

And:

\left| \vec{a} \cdot \vec{b} \right| = |\vec{a}|\,|\vec{b}| \cos\theta

.

Substitute these into the given expression:

\left| \vec{a} \times \vec{b} \right|^2 + \left| \vec{a} \cdot \vec{b} \right|^2 = (|\vec{a}|\,|\vec{b}| \sin\theta)^2 + (|\vec{a}|\,|\vec{b}| \cos\theta)^2

.

This simplifies to:

|\vec{a}|^2 |\vec{b}|^2 (\sin^2\theta + \cos^2\theta)

.

Since

\sin^2\theta + \cos^2\theta = 1

, the expression becomes

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

.

Now plug in the given values:

|\vec{a}| = 2

and

|\vec{b}| = 3

.

Thus,

|\vec{a}|^2 |\vec{b}|^2 = (2)^2 \cdot (3)^2 = 4 \cdot 9 = 36

.

Answer:

36

Hence, option (D) is correct.

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