MCQ
જો સદીશ $(\vec{a}+3 \vec{b})$ એ $(7 \vec{a}-5 \vec{b})$ અને $(\vec{a}-4 \vec{b})$ એ $(7 \vec{a}-2 \vec{b})$ લંબ હોય તો સદીશ $\vec{a}$ અને $\vec{b}$ વચ્ચેનો ખૂણો મેળવો. (ડિગ્રીમાં )
- A$40$
- ✓$60$
- C$15$
- D$75$
$(\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b})=0$
$7|\vec{a}|^{2}-15|\vec{b}|^{2}+16 \vec{a} \cdot \vec{b}=0 \ldots(1)$
$(\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b})=0$
$7|\vec{a}|^{2}+8|\vec{b}|^{2}-30 \vec{a} \cdot \vec{b}=0 \ldots(2)$
From $(1)\,\&\,(2)$
$\mid\overrightarrow{\mathrm{a}} \mid=\mid \overrightarrow{\mathrm{b}} \mid$
as $|\vec{a}|=|\vec{b}|$
$\therefore 7|\vec{a}|^{2}-15|\vec{a}|^{2}+16 a \vec{b}=0$
$\Rightarrow \vec{a} \cdot \vec{b}=\frac{|\vec{a}|^{2}}{2}$
$\therefore \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{|\vec{a}|^{2}}{2|\vec{a}||\vec{a}|}$
$\therefore \cos \theta=\frac{1}{2}$
$\Rightarrow \theta=60^{\circ}$
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