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Question-44020




Question Number 44020 by rahul 19 last updated on 20/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Sep/18
a^→ .d^→ =∣a^→ ∣cosθ_1     b^→ .d^→ =∣b^→ ∣cosθ_2    c^→ .d^→ =∣c^→ ∣cosθ_3   acosθ_1      bcosθ_2      ccosθ_3  where  ∣a^→ ∣=a  ∣b^→ ∣=b    ∣c^→ ∣=c  sogiven expression is  ∣acosθ_1 (b^→ ×c^→ )+bcosθ_2 (c^→ ×a^→ )+ccosθ_3 (a^→ ×b^→ )∣  is independent of d^→
$$\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{a}}\mid{cos}\theta_{\mathrm{1}} \:\:\:\:\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{b}}\mid{cos}\theta_{\mathrm{2}} \:\:\:\overset{\rightarrow} {{c}}.\overset{\rightarrow} {{d}}=\mid\overset{\rightarrow} {{c}}\mid{cos}\theta_{\mathrm{3}} \\ $$$${acos}\theta_{\mathrm{1}} \:\:\:\:\:{bcos}\theta_{\mathrm{2}} \:\:\:\:\:{ccos}\theta_{\mathrm{3}} \:{where}\:\:\mid\overset{\rightarrow} {{a}}\mid={a} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid={b}\:\:\:\:\mid\overset{\rightarrow} {{c}}\mid={c} \\ $$$${sogiven}\:{expression}\:{is} \\ $$$$\mid{acos}\theta_{\mathrm{1}} \left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right)+{bcos}\theta_{\mathrm{2}} \left(\overset{\rightarrow} {{c}}×\overset{\rightarrow} {{a}}\right)+{ccos}\theta_{\mathrm{3}} \left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right)\mid \\ $$$${is}\:{independent}\:{of}\:\overset{\rightarrow} {{d}} \\ $$
Commented by rahul 19 last updated on 20/Sep/18
Ok sir, thanks.
$${Ok}\:{sir},\:\mathrm{thanks}. \\ $$

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