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UPSC NDA I 2014 Mathematics Paper

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For the next two (02) items that follow:

Let

\left|\overset{\rightarrow}{a}\right|=7,\left|\overset{\rightarrow}{b}\right|=11,\left|\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right|=10\sqrt{3}

[NDA I 2014]

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Question : 74 of 120

Marks: +1, -0

What is the angle between

\left(\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right)

\left(\overset{\rightarrow}{a}-\overset{\rightarrow}{b}\right)

$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
None of the above

Solution:

\left|\left(\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right)\right|^2=\left|\overset{\rightarrow}{a}\right|^2+\left|\overset{\rightarrow}{b}\right|^2+2\left|\overset{\rightarrow}{a}\right|\cdot\left|\overset{\rightarrow}{b}\right|\cos\theta

\left(\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right)\cdot\left(\overset{\rightarrow}{a}-\overset{\rightarrow}{b}\right)=\left|\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right|\left|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}\right|\cos\alpha

\Rightarrow\left|\overset{\rightarrow}{a}\right|^2+\left|\overset{\rightarrow}{b}\right|^2=\left|\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right|\left|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}\right|\cos\alpha

\Rightarrow\cos\alpha=\frac{\left|\overset{\rightarrow}{a}\right|^2+\left|\overset{\rightarrow}{b}\right|^2}{\left|\overset{\rightarrow}{a}+\overset{\rightarrow}{b}\right|\left|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}\right|}=\frac{49+121}{10\sqrt{3}\times2\sqrt{10}}=\frac{17}{2\sqrt{30}}

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