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STD 12 - 10. vector algebra — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. vector algebra4 Marks
MCQ
If $\vec a,\vec b$ and  $\vec c$ are unit vectors such that $\vec a + 2\vec b + 2\vec c = \vec 0$, then $\left| {\vec a \times \vec c} \right|$ is equal to
  • A
    $\frac{1}{{4}}$
  • ✓
    $\frac{{\sqrt {15} }}{4}$
  • C
    $\frac{15}{{16}}$
  • D
    $\frac{{\sqrt {15} }}{16}$
✓

Answer

Correct option: B.
$\frac{{\sqrt {15} }}{4}$
b
(b) $\quad \because \vec{a}+2 \vec{b}+2 \vec{c}=\overrightarrow{0}$ [Given]

$\Rightarrow \vec{a}+2 \overrightarrow{\mathrm{c}}=-2 \vec{b}$

$ \Rightarrow (\vec a + 2\vec c) \cdot (\vec a + 2\overrightarrow {\rm{c}} ) = ( - 2\vec b)( - 2\overrightarrow {\rm{b}} )$

$\Rightarrow \vec{a} \cdot \vec{a}+4 \vec{c} \cdot \vec{c}+4 \vec{a} \cdot \vec{c}=4 \vec{b} \cdot \vec{b}$

$\Rightarrow 1+4+4 \vec{a} \cdot \overrightarrow{\mathrm{c}}=4$

$\Rightarrow \vec{a} \cdot \vec{c}=\frac{-1}{4}$

$\because|\vec{a} \cdot \overrightarrow{\mathrm{c}}|^{2}+|\vec{a} \times \vec{c}|^{2}=1(\vec{a} \text { is unit vector })$

$\frac{1}{16}+|\bar{a} \times \bar{c}|^{2} =1 $

$|\vec{a} \times \vec{c}|^{2} =\frac{15}{16} $

$|\vec{a} \times \vec{c}| =\frac{\sqrt{15}}{4}$

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