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3-1.Vectors
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सदिश $A = \hat i + \hat j$ द्वारा $x-$ अक्ष के साथ बनाया गया कोण ....... $^o$ होगा
A$90$
B$45$
C$22.5$
D$30$
Solution
(b) $\mathop A\limits^ \to = \hat i + \hat j$
$⇒$ $|A| = \sqrt {{1^2} + {1^2}} = \sqrt 2 $
$\cos \alpha = \frac{{{A_x}}}{{|A|}} = \frac{1}{{\sqrt 2 }} = \cos 45^\circ $
$⇒$ $\alpha = 45^\circ $
$⇒$ $|A| = \sqrt {{1^2} + {1^2}} = \sqrt 2 $
$\cos \alpha = \frac{{{A_x}}}{{|A|}} = \frac{1}{{\sqrt 2 }} = \cos 45^\circ $
$⇒$ $\alpha = 45^\circ $
Standard 11
Physics
