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Question Number 117543 by bemath last updated on 12/Oct/20

If vector a^→ +b^→ +c^→ =0 ∣a^→ ∣=7, ∣b^→ ∣=3 and ∣c^→ ∣=5 find the angle vector a^→ and c^→ ?

Ifvectora→+b→+c→=0∣a→∣=7,∣b→∣=3and∣c→∣=5findtheanglevectora→andc→?

Answered by bobhans last updated on 12/Oct/20

let α=∡(a^→ ,c^→ ) by Cosine of law ⇒cos (π−α) = ((7^2 +5^2 −3^2 )/(2.7.5)) ⇒−cos α= ((65)/(70)) ; cos α = −((13)/(14)) ⇒α = cos^(−1) (−((13)/(14))) = 158.21°

letα=∡(a→,c→)byCosineoflaw⇒cos(π−α)=72+52−322.7.5⇒−cosα=6570;cosα=−1314⇒α=cos−1(−1314)=158.21°

Answered by AbduraufKodiriy last updated on 12/Oct/20

a^→ +c^→ =−b^→ ⇒ ∣a^→ +c^→ ∣=∣b^→ ∣ ⇒ ⇒ (√(∣a^→ ∣^2 +2∙a^→ ∙c^→ +∣c^→ ∣^2 ))=3 (√(7^2 +2∙∣a^→ ∣∣b^→ ∣cos(𝛟)+5^2 ))=3 74+2∙7∙5cos(𝛟)=9 ⇒ 70cos(𝛟)=−65 ⇒ ⇒ cos(𝛟)=−((13)/(14)) ⇒ 𝛟=π−arccos(((13)/(14))) 𝛟 is angle vector a^→ and c^→

a→+c→=−b→⇒∣a→+c→∣=∣b→∣⇒⇒∣a→∣2+2⋅a→⋅c→+∣c→∣2=372+2⋅∣a→∣∣b→∣cos(φ)+52=374+2⋅7⋅5cos(φ)=9⇒70cos(φ)=−65⇒⇒cos(φ)=−1314⇒φ=π−arccos(1314)φisanglevectora→andc→

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