Three vectors \vec{a} ,\vec{b} ,and \: \vec{c} are such that  \left | \vec{a} \right |=1 ,\left | \vec{b} \right |=2 , \: \left | \vec{c} \right |=4 and \vec{a}+\vec{b}+\vec{c}=\vec{0}

 Then the value of    4\; \vec{a}\cdot \vec{b}+3\; \vec{b}\cdot \; \vec{c}+3\; \vec{c}\cdot \; \vec{a} is equal to :

  • Option 1)

    27

  • Option 2)

    -68

  • Option 3)

    -26

  • Option 4)

    - 34

 

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}

 

 \left | \vec{a} \right |=1,\left | b \right |=2,\left | \vec{c} \right |=4

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

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

Squaring 

\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta_{1} =\left | \vec{c} \right |^{2}

1+4+2\times \vec{a}.\vec{b}=16

\vec{a}.\vec{b}=\frac{11}{2}

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

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

\vec{b}.\vec{c}\:=\frac{1}{2}\:(1-4-16)\:=\:-\frac{19}{2}

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

\vec{a}.\vec{c}=\frac{1}{2}(4-16-1)=-\frac{13}{2}\\ 4\vec{a}.\vec{b}+3\vec{b}.\vec{c}+3\vec{c}.\vec{a}

=4\times \frac{11}{2}+3\times -\frac{19}{2}+3\times -\frac{13}{2}=22-3\times 16 =-26


Option 1)

27

This option is incorrect 

Option 2)

-68

This option is incorrect 

Option 3)

-26

This option is correct 

Option 4)

- 34

This option is incorrect 

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