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.

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