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Question Number 133398 by rexford last updated on 21/Feb/21

Answered by EDWIN88 last updated on 22/Feb/21

a^→ .b^→ +b^→ .c^→ =(a^→ +c^→ ).b^→ =−b^→ .b^→ =−∣b^→ ∣^2 c^→ .a^→ =− ∣c^→ ∣∣a^→ ∣{((∣a^→ ∣^2 +∣c^→ ∣^2 −∣b^→ ∣^2 )/(2∣a^→ ∣∣c^→ ∣))} = ((∣b^→ ∣^2 −∣a^→ ∣^2 −∣c^→ ∣^2 )/2) Then a^→ .b^→ +b^→ .c^→ +c^→ .a^→ = ((−∣a^→ ∣^2 −∣b^→ ∣^2 −∣c^→ ∣^2 )/2) = −(3/2)

a→.b→+b→.c→=(a→+c→).b→=−b→.b→=−∣b→∣2c→.a→=−∣c→∣∣a→∣{∣a→∣2+∣c→∣2−∣b→∣22∣a→∣∣c→∣}=∣b→∣2−∣a→∣2−∣c→∣22Thena→.b→+b→.c→+c→.a→=−∣a→∣2−∣b→∣2−∣c→∣22=−32

Answered by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

a∙b=b∙c=c∙a=−1×1×cos 60°=−(1/2) Σa∙b=−(3/2)

a⋅b=b⋅c=c⋅a=−1×1×cos60°=−12Σa⋅b=−32

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