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3-1.Vectors
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दिया है सदिश $\mathop A\limits^ \to = 2\hat i + 3\hat j,$$\mathop A\limits^ \to $व y-अक्ष के बीच कोण होगा
A${\tan ^{ - 1}}3/2$
B${\tan ^{ - 1}}2/3$
C${\sin ^{ - 1}}2/3$
D${\cos ^{ - 1}}2/3$
Solution
(b) $\vec{A}=2 \hat{i}+3 \hat{j}$
$A=|\vec{A}|=\sqrt{4+9}=\sqrt{13}$
$\cos \beta=\frac{y}{A}=\frac{3}{\sqrt{13}}$
$\cos \beta=\frac{3}{\sqrt{13}}$
$\sin \beta=\frac{x}{A}=\frac{2}{\sqrt{13}}$
$\operatorname{Tan} \beta =\frac{x}{y}=\frac{2}{3}$
$=\tan ^{-1}\left(\frac{2}{3}\right)$
$A=|\vec{A}|=\sqrt{4+9}=\sqrt{13}$
$\cos \beta=\frac{y}{A}=\frac{3}{\sqrt{13}}$
$\cos \beta=\frac{3}{\sqrt{13}}$
$\sin \beta=\frac{x}{A}=\frac{2}{\sqrt{13}}$
$\operatorname{Tan} \beta =\frac{x}{y}=\frac{2}{3}$
$=\tan ^{-1}\left(\frac{2}{3}\right)$
Standard 11
Physics
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