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If $\vec{a},\vec{b},\vec{c}$ are vectors show that $\vec{a}+\vec{b}+\vec{c}=0$ and $\left | \vec{a} \right |=7,\left | \vec{b} \right |=5,\left | \vec{c} \right |=3,$ then angle between vector $\vec{b}\; \; and \; \; \vec{c}$ is

Option 1)
$60^{\circ}\;$

Option 2)
$\; 30^{\circ}\;$

Option 3)
$\; 45^{\circ}\;$

Option 4)
$\; 90^{\circ}$

Answers (1)

As we learnt in

Scalar Product of two vectors (dot product) -

$\vec{a}\vec{b}=\left | a \right |\left | b \right |Cos\theta$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\: and\:\vec{b}$

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

We need to find angle between $\vec{b}$ and $\vec{c}$

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

Squaring both sides

$\left | \vec{b} \right |^{2}+\left | \vec{c}\right |^{2}+2\vec{b}.\vec{c}\:=\:\left | \vec{a} \right |^{2}$

$2\left | \vec{b} \right |\left | \vec{c} \right |cos\theta \:=\:\left | \vec{a} \right |^{2}-\left | \vec{b} \right |^{2}-\left | \vec{c} \right |^{2}$

$2\times5\times3cos\theta =49-25-9$

$cos\theta = \frac{1}{2} => \theta = \frac{\pi }{3}=60^{\circ}$

Option 1)

$60^{\circ}\;$

This option is correct.

Option 2)

$\; 30^{\circ}\;$

This option is incorrect.

Option 3)

$\; 45^{\circ}\;$

This option is incorrect.

Option 4)

$\; 90^{\circ}$

This option is incorrect.

Posted by

Sabhrant Ambastha

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