MCQ
Let $\vec a$ and $\vec b$ be two unit vectors such that $\left| {\vec a\, + \,\vec b} \right| = \sqrt 3 .$ If $\vec c = \vec a\, + \,2\vec b + 3\,(\vec a \times \vec b),$ then $2\left| {\vec c} \right|$ is equal to
  • $\sqrt {55}$
  • B
    $\sqrt {37}$
  • C
    $\sqrt {51}$
  • D
    $\sqrt {43}$

Answer

Correct option: A.
$\sqrt {55}$
a
$\left| {\vec a + \vec b} \right| = \sqrt 3 $

angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is $60^{\circ} .$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ is $\perp^{\mathrm{r}}$ to plane containing $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$

$\vec{c}=\vec{a}+2 \vec{b}+3(\vec{a} \times \vec{b})$

$\overrightarrow {\rm{c}}  = |{\rm{\vec a}}{|^2} + 4|\overrightarrow {\rm{b}} {|^2} + 2.2|{\rm{\vec a}}{|^2}\cos {\rm{ }}60^\circ {\rm{ }}{{\rm{n}}_1} + 3|{\rm{\vec a}}|||\overrightarrow {\rm{b}} |\sin {\rm{ }}60^\circ {\rm{ n}}2$

$+||3 \overrightarrow{\mathrm{b}}|| \overrightarrow{\mathrm{b}} | \sin 60^{\circ} \cdot \overrightarrow{\mathrm{n}} 2$

$\overrightarrow{\mathrm{n}}_{1} \perp^{\mathrm{r}} \overrightarrow{\mathrm{n}}_{2}$

$|\vec c{|^2} = (1 + 4 + 2) + 9 \times \frac{3}{4}$

$|\overrightarrow{\mathrm{c}}|^{2}=7+27 / 4=55 / 4$

$2|\overrightarrow{\mathrm{c}}|=\sqrt{55}$

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