Question Number 44020 by rahul 19 last updated on 20/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18
$$\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{a}}\mid{cos}\theta_{\mathrm{1}} \:\:\:\:\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{b}}\mid{cos}\theta_{\mathrm{2}} \:\:\:\overset{\rightarrow} {{c}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{c}}\mid{cos}\theta_{\mathrm{3}} \\ $$$${acos}\theta_{\mathrm{1}} \:\:\:\:\:{bcos}\theta_{\mathrm{2}} \:\:\:\:\:{ccos}\theta_{\mathrm{3}} \:{where}\:\:\mid\overset{\rightarrow} {{a}}\mid={a} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid={b}\:\:\:\:\mid\overset{\rightarrow} {{c}}\mid={c} \\ $$$${sogiven}\:{expression}\:{is} \\ $$$$\mid{acos}\theta_{\mathrm{1}} \left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right)+{bcos}\theta_{\mathrm{2}} \left(\overset{\rightarrow} {{c}}×\overset{\rightarrow} {{a}}\right)+{ccos}\theta_{\mathrm{3}} \left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right)\mid \\ $$$${is}\:{independent}\:{of}\:\overset{\rightarrow} {{d}} \\ $$
Commented by rahul 19 last updated on 20/Sep/18
$${Ok}\:{sir},\:\mathrm{thanks}. \\ $$