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Ocheana
@Ocheana
January 2021
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Bonjour je suis en 1ère S et j’ai besoin de vous en maths sur les produits scalaires. Quelqu’un sait calculer AB.AC dans ces deux cas svp ?
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ProfdeMaths1
Verified answer
1er cas :
les droites (AB) et (AC) se coupent en A(1;4)
de plus B(-3;0) et C(3;0)
donc AB²=(-3-1)²+(0-4)²=16+16=32
BC=6 et AC²=(3-1)²+(4-0)²=4+16=20
ainsi
AB.AC
=1/2(AB²+AC²-BC²)
=1/2(32+20-36)
=
8
2eme cas :
AB.AC
=1/2(AD²-AB²-AC²)
=1/2(5²-4²-2²)
=
2,5
0 votes
Thanks 1
Ocheana
Merci beaucoup !!
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Verified answer
1er cas :les droites (AB) et (AC) se coupent en A(1;4)
de plus B(-3;0) et C(3;0)
donc AB²=(-3-1)²+(0-4)²=16+16=32
BC=6 et AC²=(3-1)²+(4-0)²=4+16=20
ainsi AB.AC=1/2(AB²+AC²-BC²)
=1/2(32+20-36)
=8
2eme cas :
AB.AC=1/2(AD²-AB²-AC²)
=1/2(5²-4²-2²)
=2,5