MCQ
જો $a + b + c = 0,\,\,\left| {\vec a} \right| = 3,\,\left| {\vec b} \right| = 5$ અને $\left| {\vec c} \right| = 7,$ હોય તો $\vec a$ અને $\vec b$ વચ્ચેનો ખૂણો મેળવો.
- A$\frac {\pi }{3}$
- B$\frac {\pi }{4}$
- C$\frac {\pi }{6}$
- D$\frac {\pi }{2}$
$\Rightarrow \quad(a+b)^{2}=c^{2}$
$\Rightarrow a^{2}+b^{2}+2 a b=c^{2}$
$\Rightarrow 9+25+2.3 .5 \cos \theta=49$
$(\because|\vec{a}|=3,|\vec{b}|=5 \text { and }|\vec{c}|=7)$
$\therefore \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$
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વિધાન $II:$ દરેક $x \in R ,$ માટે ${\sin ^{ - 1}}\,x + {\cos ^{ - 1}}\,x = \frac{\pi }{2}$ અને $0 \le {\left( {{{\sin }^{ - 1}}\,x - \frac{\pi }{4}} \right)^2} \le \frac{{9{\pi ^2}}}{{16}}$ થાય.