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UPSC NDA 2 2016 Mathematics Solved Paper

Question Numbers: 67-68

Direction: Consider the following for the next two (02) items that follow:

Let

{\rm{\vec a}},{\rm{\;\vec b}}

and

{\rm{\vec c}}

be three vectors such that

{\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0

, and

\left | {{\rm{\vec a}}} \right | = 10,{\rm{\;}}\left | {{\rm{\vec b}}} \right | = 6

and

\left | {{\rm{\vec c}}} \right | = 14

Question : 68 of 120

Marks: +1, -0

{\rm{\vec a}}

{\rm{\vec b}}

Solution:

Concept:

Using dot product properties and the condition

\vec{a}+\vec{b}+\vec{c}=0

to find the angle between two vectors.

Explanation:

Given:

\vec{a}+\vec{b}+\vec{c}=0

|\vec{a}|=10

|\vec{b}|=6

|\vec{c}|=14

, and

\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=-166

First, take dot product of

\vec{c}

with the zero vector equation:

\vec{c}\cdot(\vec{a}+\vec{b}+\vec{c})=0

\vec{c}\cdot\vec{a}+\vec{c}\cdot\vec{b}+|\vec{c}|^{2}=0

\vec{c}\cdot\vec{a}+\vec{b}\cdot\vec{c}+196=0

Thus

\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=-196

Now use the known sum of all dot products:

\vec{a}\cdot\vec{b}+(\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=-166

\vec{a}\cdot\vec{b}-196=-166

\vec{a}\cdot\vec{b}=30

Let

\theta

be angle between

\vec{a}

and

\vec{b}

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

30=10\times6\times\cos\theta=60\cos\theta

\cos\theta=\frac{30}{60}=\frac{1}{2}

\theta=60^{\circ}

Answer:

60° (Option C).

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