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NDA Math Vector Algebra

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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 : 101 of 101

Marks: +1, -0

What is the angle between

(\vec{a}+\vec{b})

(\vec{a}-\vec{b})

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

Solution:

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

(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=|\vec{a}+\vec{b}||\vec{a}-\vec{b}|\cos\alpha

\Rightarrow |\vec{a}|^2+|\vec{b}|^2=|\vec{a}+\vec{b}||\vec{a}-\vec{b}|\cos\alpha

\Rightarrow \cos\alpha = \frac{|\vec{a}|^2+|\vec{b}|^2}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|} = \frac{49+121}{10\sqrt{3}\times 2\sqrt{10}} = \frac{17}{2\sqrt{30}}

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