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3-1.Vectors
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$\mathop A\limits^ \to = 2\hat i + 4\hat j + 4\hat k$ तथा $\overrightarrow B = 4\hat i + 2\hat j - 4\hat k$ दो सदिश हैं। उनके मध्य कोण ........ $^o$ होगा
A
$0$
B
$45$
C
$90$
D
$60$
Solution
(c) $\cos \theta = \frac{{\mathop A\limits^ \to \,.\,\mathop B\limits^ \to }}{{|\mathop A\limits^ \to \,|\,.\,|\,\mathop B\limits^ \to |}} = \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{|\mathop A\limits^ \to \,|\,.\,|\mathop B\limits^ \to |\,}}$
$ = \frac{{2 \times 4 + 4 \times 2 – 4 \times 4}}{{|\mathop A\limits^ \to |\,.\,|\mathop B\limits^ \to |}} = 0$
$\therefore \theta = {\cos ^{ – 1}}(0^\circ )$ $ \Rightarrow \,\,\,\theta = 90^\circ $
Standard 11
Physics
