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UPSC NDA I and II 2020 Math Solved Paper

Question : 99 of 120

Marks: +1, -0

\left| \vec{a} \times \vec{b} \right|^{2} + \left| \vec{a} \cdot \vec{b} \right|^{2} = 144

\left| \vec{a} \right| = 4

\left| \vec{b} \right|

equal to?

Solution:

Concept:

The sum of squares of the magnitudes of the cross product and dot product equals the square of the product of the magnitudes of the vectors.

Explanation:

Given:

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

and

|\vec{a}| = 4

We know:

|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| \cos\theta

and

|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta

Square both expressions and add them:

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

Since

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

, this simplifies to

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

Substitute the given values:

144 = (4)^{2} |\vec{b}|^{2} = 16 |\vec{b}|^{2}

Thus,

|\vec{b}|^{2} = 9

Taking the positive magnitude,

|\vec{b}| = 3

Answer:

3

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