Let \vec{a}\: and \: \vec{b} be two unit vectors such that \left | \vec{a} +\vec{b}\right |=\sqrt{3}

if\vec{c}= \vec{a}+2\vec{b}+3\left ( \vec{a} \times \vec{b}\right )    then   2\left | \vec{c} \right |    is equal to:

  • Option 1)

    \sqrt{55}

  • Option 2)

    \sqrt{51}

  • Option 3)

    \sqrt{43}

  • Option 4)

    \sqrt{37}

 

Answers (1)
A Aadil Khan

As we learnt in 

Vector Product of two vectors(cross product) -

If \vec{a} and \vec{b} are two vectors and \Theta is the angle between them , then \vec{a}\times \vec{b}=\left |\vec{a} \left | \right |\vec{b} \right |Sin\Theta \hat{n}

- wherein

\hat{n} is unit vector perpendicular to both \vec{a} \: and \: \vec{b}

 

and 

 

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}+\vec{b} \right |=\sqrt{3}

Squaring both sides

\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |cos \theta =3

cos \theta=\frac{1}{2}    \Rightarrow \theta=\frac{\pi }{3}

\vec{c}=\vec{a}+2\vec{b}+3\left | \vec{a} \right |\left | \vec{b} \right | sin \frac{\pi }{3}\hat{n}

\vec{c}=\vec{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{n}

\vec{c}\cdot \vec{c}=\left ( \vec{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{n} \right )\cdot \left (\vec{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{n} \right )

\left | \vec{c} \right |^{2}=\left | \vec{a} \right |^{2}+4\left | \vec{b} \right |^{2}+\frac{27}{4}+4\left ( \vec{a}\cdot \vec{b} \right )

\left | \vec{c} \right |^{2}=1+4+\frac{27}{4}+4\times 1\times 1\times \frac{1}{2}

\left | \vec{c} \right |^{2}=7+\frac{27}{4}=\frac{55}{4}

\left | \vec{c} \right |^{2}=\frac{\sqrt{55}}{2}

2\left | \vec{c} \right |=\sqrt{55} 

 

 


Option 1)

\sqrt{55}

This option is correct.

Option 2)

\sqrt{51}

This option is incorrect.

Option 3)

\sqrt{43}

This option is incorrect.

Option 4)

\sqrt{37}

This option is incorrect.

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