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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
If $\vec a$ and $\vec b$ are non-zero vectors which are linearly dependent such that $\frac{{\left| {\vec a + \vec b} \right|}}{{\left| {\vec a - \vec b} \right|}}\, = \,2,\,\left| {\vec b} \right|\, > \,\left| {\vec a} \right|$ Then 
  • ✓
    $\vec b = 3\vec a$
  • B
    $\vec b =  - 3\vec a$
  • C
    $\vec b = 2\vec a$
  • D
    $\vec b =  - 2\vec a$
✓

Answer

Correct option: A.
$\vec b = 3\vec a$
a
Let $\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}}$

$ \Rightarrow \frac{{|\vec a + \lambda \vec a|}}{{|\vec a - \lambda \vec a|}} = 2\quad  \Rightarrow \quad \frac{{1 + \lambda }}{{1 - \lambda }} =  \pm 2$

$ \Rightarrow \lambda  = 3,\frac{1}{3}\quad  \Rightarrow \quad \vec b\left\langle {\begin{array}{*{20}{c}}
{3\vec a}\\
{\frac{{\vec a}}{3}}
\end{array}} \right.$

$ \Rightarrow \overrightarrow {\rm{b}}  = 3\vec a\quad $     ($\because $ $|\overrightarrow {\rm{b}} | > |\overrightarrow {\rm{a}} |$)

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