Question Number 73222 by Tanmay chaudhury last updated on 08/Nov/19
Answered by Tanmay chaudhury last updated on 08/Nov/19
![[a^→ b^→ c^→ ]=2=(a^→ ×b^→ ).c^→ l(b^→ ×c^→ ).c^→ +m(c^→ ×a^→ ).c^→ +n(a^→ ×b^→ ).c^→ =4i.c^→ l×0+m×0+n[abc]=4i.(ic_x +jc_y +kc_z )=4c_x 2n=4c_x similarly 2l=4a_x 2m=4b_x i(a_x +b_x +c_x )=3i→a_x +b_x +c_x =3 2(l+m+n)=4(a_x +b_x +c_x )=4×3 l+m+n=6 plz check](Q73228.png)
$$\left[\overset{\rightarrow} {{a}}\overset{\rightarrow} {{b}}\overset{\rightarrow} {{c}}\right]=\mathrm{2}=\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}} \\ $$$${l}\left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right).\overset{\rightarrow} {{c}}+{m}\left(\overset{\rightarrow} {{c}}×\overset{\rightarrow} {{a}}\right).\overset{\rightarrow} {{c}}+{n}\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}}=\mathrm{4}{i}.\overset{\rightarrow} {{c}} \\ $$$${l}×\mathrm{0}+{m}×\mathrm{0}+{n}\left[{abc}\right]=\mathrm{4}{i}.\left({ic}_{{x}} +{jc}_{{y}} +{kc}_{{z}} \right)=\mathrm{4}{c}_{{x}} \\ $$$$\mathrm{2}{n}=\mathrm{4}{c}_{{x}} \:{similarly}\:\mathrm{2}{l}=\mathrm{4}{a}_{{x}} \:\:\:\mathrm{2}{m}=\mathrm{4}{b}_{{x}} \\ $$$${i}\left({a}_{{x}} +{b}_{{x}} +{c}_{{x}} \right)=\mathrm{3}{i}\rightarrow{a}_{{x}} +{b}_{{x}} +{c}_{{x}} =\mathrm{3} \\ $$$$\mathrm{2}\left({l}+{m}+{n}\right)=\mathrm{4}\left({a}_{{x}} +{b}_{{x}} +{c}_{{x}} \right)=\mathrm{4}×\mathrm{3} \\ $$$${l}+{m}+{n}=\mathrm{6} \\ $$$${plz}\:{check} \\ $$