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UPSC NDA Math Model Paper 1

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Question : 38 of 120

Marks: +1, -0

If

\vec{a}+\vec{b}+\vec{c}=\vec{0},|\vec{a}|=3,|\vec{b}|=5

|\vec{c}|=7,

find the angle between

\vec{a}

\vec{b}

.

$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{4}$

Solution:

consider, the angle between

\vec{a}

and

\vec{b}

is

\theta

Given,

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

\Rightarrow\vec{a}+\vec{b}=-\vec{c}

\Rightarrow\left|\vec{a}+\vec{b}\right|=\left|-\vec{c}\right|

Squaring on both side, we get

\Rightarrow\left|\vec{a}+\vec{b}\right|^{2}=\left|-\vec{c}\right|^{2}

\Rightarrow\left|\vec{a}\right|^{2}+2\vec{a}\cdot\vec{b}+\left|\vec{b}\right|^{2}=\left|-\vec{c}\right|^{2}

\Rightarrow\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}+2ab\cos\theta=\left|-\vec{c}\right|^{2}

\Rightarrow(3)|^{2}+(5)^{2}+2(3)(5)\cos\theta=(7)^{2}

\Rightarrow30\cos\theta=15

\Rightarrow\cos\theta=\frac{1}{2}

\Rightarrow\theta=\frac{\pi}{3}

Hence, If

\vec{a}+\vec{b}+\vec{c}=\vec{0},|\vec{a}|=3,|\vec{b}|=5

and

|\vec{c}|=7,

then the angle between

\vec{a}

and

\vec{b}

is

\frac{\pi}{3}

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