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Question Number 147162 by nadovic last updated on 18/Jul/21

Answered by liberty last updated on 18/Jul/21

(1)p^→ =(2/5)a^→ +(1/5)b^→ +(2/5)c^→ (2)p^→ .b^→ =p^→ .a^→ (2/5)a^→ .b^→ +(1/5)∣b^→ ∣^2 +(2/5)b^→ .c^→ = (2/5)∣a^→ ∣^2 +(1/5)a^→ .b^→ +(2/5)a^→ .c^→ (2/5)b^→ .c^→ +(1/5)∣b^→ ∣^2 =(2/5)∣a^→ ∣^2 +(2/5)a^→ .c^→ 2b^→ .c^→ −2a^→ .c^→ =2∣a^→ ∣^2 −∣b^→ ∣^2 (3)a^→ .b^→ = 0 then c^→ .(b^→ −a^→ )= b^→ .c^→ −a^→ .c^→ = ∣a^→ ∣^2 −(1/2)∣b^→ ∣^2

$$\left(\mathrm{1}\right)\overset{\rightarrow} {{p}}=\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{a}}+\frac{\mathrm{1}}{\mathrm{5}}\overset{\rightarrow} {{b}}+\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{c}} \\ $$$$\left(\mathrm{2}\right)\overset{\rightarrow} {{p}}.\overset{\rightarrow} {{b}}=\overset{\rightarrow} {{p}}.\overset{\rightarrow} {{a}} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}+\frac{\mathrm{1}}{\mathrm{5}}\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}\:= \\ $$$$\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{5}}\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{5}}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}+\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}} \\ $$$$\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}\:+\frac{\mathrm{1}}{\mathrm{5}}\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{5}}\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{5}}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}} \\ $$$$\:\:\:\:\:\mathrm{2}\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}−\mathrm{2}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}\:=\mathrm{2}\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} −\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}=\:\mathrm{0} \\ $$$${then}\:\overset{\rightarrow} {{c}}.\left(\overset{\rightarrow} {{b}}−\overset{\rightarrow} {{a}}\right)=\:\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}−\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} \\ $$

Commented by nadovic last updated on 18/Jul/21

Thank you very much Sir

$${Thank}\:{you}\:{very}\:{much}\:{Sir} \\ $$

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