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Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVectors2 Marks
MCQ
If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and $\theta$ is angle between $\vec{a}$ and $\vec{c}$ and $\vec{a}+2 \vec{b}+2 \vec{c}=\overrightarrow{0}$, then $|\vec{a} \times \vec{c}|=$
  • A
    $\frac{\sqrt{15}}{2}$
  • ✓
    $\frac{\sqrt{15}}{4}$
  • C
    $\sqrt{15}$
  • D
    $\frac{\sqrt{15}}{3}$
✓

Answer

Correct option: B.
$\frac{\sqrt{15}}{4}$
We have, $\vec{a}+2 \vec{b}+2 \vec{c}=0$
$\Rightarrow \vec{a}+2 \vec{c}=-2 \vec{b} $
$\Rightarrow|\vec{a}+2 \vec{c}|^2=|2 \vec{b}|^2$
$\Rightarrow|\vec{a}|^2+4|\vec{c}|^2+4 \vec{a} \cdot \vec{c}=4|\vec{b}|^2$
$\Rightarrow 5+4 \vec{a} \cdot \vec{c}=4 \quad[\because \vec{a}, \vec{b}$ and $\vec{c} $ are unit vectors $]$
$\Rightarrow \vec{a} \cdot \vec{c}=\frac{-1}{4}$
Now, $|\vec{a} \times \vec{c}|^2+(\vec{a} \cdot \vec{c})^2=|\vec{a}|^2|\vec{c}|^2$
$\Rightarrow|\vec{a} \times \vec{c}|^2+\frac{1}{16}=1 $
$\Rightarrow|\vec{a} \times \vec{c}|^2=1-\frac{1}{16}$
$=\frac{15}{16}$
$\Rightarrow|\vec{a} \times \vec{c}|=\frac{\sqrt{15}}{4}$

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