# 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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