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UPSC NDA II 2023 Mathematics Solved Paper

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Question : 92 of 120

Marks: +1, -0

If

(\vec{a}\times \vec{b})^{2}+(\vec{a}\cdot \vec{b})^{2}=144

| \vec{b} | =4

, then what is the value of

| \vec{a} |

?

3
4
5
6

Solution:

Concept:

Use the identity

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

.

This is derived from

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

.

Explanation:

Given:

(\vec{a} \times \vec{b})^{2} + (\vec{a} \cdot \vec{b})^{2} = 144

and

| \vec{b} | = 4

.

The squared magnitude of the cross product is

| \vec{a} \times \vec{b} |^{2} = ( | \vec{a} | | \vec{b} | \sin\theta )^{2}

.

The dot product squared is

(\vec{a} \cdot \vec{b})^{2} = ( | \vec{a} | | \vec{b} | \cos\theta )^{2}

.

Adding them:

| \vec{a} |^{2} | \vec{b} |^{2} ( \sin^{2}\theta + \cos^{2}\theta ) = | \vec{a} |^{2} | \vec{b} |^{2} = 144

.

Since

| \vec{b} | = 4

, we have

| \vec{b} |^{2} = 16

.

Thus

| \vec{a} |^{2} \cdot 16 = 144

.

Solving:

| \vec{a} |^{2} = \frac{144}{16} = 9

.

Therefore

| \vec{a} | = \sqrt{9} = 3

.

Answer:

| \vec{a} | = 3

Thus option A is correct.

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