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JEE Main & Advanced Physics Vectors Lami's Theorem

Lami's Theorem

Category : JEE Main & Advanced

In any \[\Delta \,A\,B\,C\] with sides \[\overrightarrow{a},\,\overrightarrow{b},\,\overrightarrow{c}\] \[\frac{\sin \alpha }{a}=\frac{\sin \beta }{b}=\frac{\sin \gamma }{c}\]

 

i.e.for any triangle the ratio of the sine of the angle containing the side to the length of the side is a constant. For a triangle whose three sides are in the same order we establish the Lami's theorem in the following manner. For the triangle shown

\[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\] [All three sides are taken in order] ?

(i) \[\Rightarrow \]\[\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\] ?

(ii) Pre-multiplying both sides by \[\overrightarrow{a}\]\[\overrightarrow{a}\times (\overrightarrow{a}+\overrightarrow{b})=-\overrightarrow{a}\times \overrightarrow{c}\]

\[\Rightarrow \]\[\overrightarrow{0}+\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{a}\times \overrightarrow{c}\]

\[\Rightarrow \,\,\,\,\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}\] ?(iii) Pre-multiplying both sides of

(ii) by \[\overrightarrow{b}\] \[\overrightarrow{b}\times (\overrightarrow{a}+\overrightarrow{b})=-\,\overrightarrow{b}\times \overrightarrow{c}\]

\[\Rightarrow \,\,\,\,\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{c}\]

\[\Rightarrow \,\,\,\,-\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{c}\]\[\Rightarrow \,\,\,\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}\] ?

(iv) From (iii) and (iv), we get \[\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}\] Taking magnitude, we get \[|\overrightarrow{a}\times \overrightarrow{b}|\,=\,|\overrightarrow{b}\times \overrightarrow{c}|\,=\,|\overrightarrow{c}\times \overrightarrow{a}|\] \[\Rightarrow \,\,\,ab\sin (180-\gamma )=bc\sin (180-\alpha )=ca\sin (180-\beta )\] \[\Rightarrow \,\,\,ab\sin \gamma =bc\sin \alpha =ca\sin \beta \] Dividing through out byabc,we have \[\Rightarrow \,\,\,\,\frac{\sin \alpha }{a}=\frac{\sin \beta }{b}=\frac{\sin \gamma }{c}\]



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