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જો સદિશ $(\hat a +2\hat b )$ એ સદિશ $(5 \hat a -4 \hat b )$ ને લંબ હોય તો , $\hat a $ અને $\hat b $ વચ્ચેનો ખૂણો ........ $^o$
$30$
$45$
$60$
$90$
Solution
$\begin{array}{l}
\left( {{\rm{\hat a}} + \;{\rm{2\hat b}}} \right)\, \bot \left( {5\hat a- 4\hat b} \right)\,\,\,\,\\ \therefore \left( {{\rm{\hat a}}\,\, + \;{\rm{2\hat b}}\,} \right)\,.\,\,\left( {\,5\hat a\,\, – 4\hat b\,} \right)\,\, = 0 \\ \Rightarrow 5\hat a.\,\hat a – 4\hat a.\,\hat b\,\, + \;10\hat b.\,\hat a – 8\hat b.\,\hat b\,\, = 0\\ \Rightarrow \,5+ \;\,6\hat a\,.\hat b\,\, – 8 = 0 \Rightarrow \,6\hat a\,.\,\hat b= 3\Rightarrow \hat a\,.\hat b = \frac{1}{2} \\ \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = 60^\circ \end{array}$
Similar Questions
જો $\left| {\vec A } \right|\, = \,2$ અને $\left| {\vec B } \right|\, = \,4$ હોય, તો કોલમ $-II$માં આપેલા ખૂણાને અનુરૂપ કોલમ $-I$માં આપેલા યોગ્ય સંબંધ સાથે જોડો.
| કોલમ $-I$ | કોલમ $-II$ |
| $(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$ | $(i)$ $\theta = \,{0^o}$ |
| $(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$ | $(ii)$ $\theta = \,{90^o}$ |
| $(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$ | $(iii)$ $\theta = \,{180^o}$ |
| $(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$ | $(iv)$ $\theta = \,{60^o}$ |
