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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Let $\hat{a}, \hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{ a }$ and $\overrightarrow{ c }$ is $\frac{\pi}{12}$, and $\hat{ b }=\overrightarrow{ c }+2(\overrightarrow{ c } \times \hat{ a })$, then $|6 \overrightarrow{ c }|^{2}$ is equal to
  • A
    $6(3-\sqrt{3})$
  • B
    $3+\sqrt{3}$
  • ✓
    $6(3+\sqrt{3})$
  • D
    $6(\sqrt{3}+1)$
✓

Answer

Correct option: C.
$6(3+\sqrt{3})$
c
$|\hat{b}|^{2}=|\vec{c}+2(\vec{c} \times \hat{a})|^{2}$

$|\hat{ b }|^{2}=| c |^{2}+4|\overrightarrow{ c } \times \hat{ a }|^{2}+4 \overrightarrow{ c } \cdot(\overrightarrow{ c } \times \hat{ a })$

$1=|c|^{2}+4|c|^{2} \sin ^{2} \frac{\pi}{12}+0$

$1=|c|^{2}+4|c|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2}$

$|c|^{2}=\frac{1}{3-\sqrt{3}}=\frac{3+\sqrt{3}}{6}$

So $6^{2}|c|^{2}=6(3+\sqrt{3})$

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