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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 three vectors such that $|\vec{a}+\vec{b}+\vec{c}|=1$, $\vec{c}=\lambda(\vec{a} \times \vec{b})$ and $|\vec{a}|=\frac{1}{\sqrt{3}},|\vec{b}|=\frac{1}{\sqrt{2}},|\vec{c}|=\frac{1}{\sqrt{6}}$, then the angle between $\vec{a}$ and $\vec{b}$ is
  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{3}$
  • ✓
    $\frac{\pi}{2}$
✓

Answer

Correct option: D.
$\frac{\pi}{2}$
Given, $|\vec{a}+\vec{b}+\vec{c}|=1$
$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^2=1$
$\Rightarrow|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}=1$
$\Rightarrow \frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot(\lambda(\vec{a} \times \vec{b}))$
$+2 \vec{a} \cdot(\lambda(\vec{a} \times \vec{b}))=1$
$\Rightarrow \frac{2+3+1}{6}+2 \vec{a} \cdot \vec{b}+2 \lambda(\vec{b} \cdot(\vec{a} \times \vec{b}))+2 \lambda(\vec{a} \cdot(\vec{a} \times \vec{b}))=1$
$\Rightarrow 1+2 \vec{a} \cdot \vec{b}=1 $
$\Rightarrow 2 \vec{a} \cdot \vec{b}=0 $
$\Rightarrow \vec{a} \cdot \vec{b}=0$
$\Rightarrow$ Angle between $\vec{a} $ and $\vec{b} $  is $ \frac{\pi}{2}$

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