Vecteurs et points du plan : fiche d'exercices corrigés Seconde C

⃗u −⃗u
∥⃗u +⃗v∥≤∥⃗u∥+∥⃗v∥⃗u, ⃗v (⃗u, ⃗v)
det(⃗u, ⃗v)̸= 0
B∈[AC]
⃗u =−−→ AB +−−→ BC +−−→ CD +−−→ DE =−→ AE
EF GH −−→ EF =−−→ HG ⃗v =−−→ EB +−−→ HG+−−→ AH +−−→ GE =−−→ EB +(−−→ AH +−−→ HG+−−→ GE) = −−→ EB +−→ AE =−−→ AB
ABCD O −−→ AB =−−→ DC −−→ AD =−−→ BC −−→ BO =−−→ OD
−−→
AB −−−→ OC =−−→ DC +−−→ CO =−−→ DO −−→ DC −−−→ CB =−−→ AB +−−→ BC =−→ AC −−→ BO −−−→ OD =⃗ 0
−−→
OC −−−→ AD =−−→ OC +−−→ CB =−−→ OB −−→ DA =−−→ CB −−→ AD −−→ AO =−→ OA+−−→ AD =−−→ OD −−→ DO −−−→ AB =−−→ OB +−−→ BA =−→ OA
⃗a = 7⃗u −4⃗v + 3⃗u + 9 ⃗w −10 ⃗w + 5⃗v = 10⃗u +⃗v −⃗w
⃗
b= 8⃗u −4⃗v −12⃗u + 4⃗v + 3⃗u =−⃗u
⃗c = 2⃗u −15⃗v + 3⃗u + 4 ⃗w + 8⃗v = 5⃗u −7⃗v + 4 ⃗w
⃗
d=⃗u + 6⃗v + 2 ⃗w −10⃗u −15⃗v =−9⃗u −9⃗v + 2 ⃗w
ABC AB =BC AB2+BC2=AC22AB2= 12 AB2= 6 
−−→
AB
=√6
∥−⃗u∥=∥⃗u∥= 2 0 ≤ ∥⃗u +⃗v∥≤∥⃗u∥+∥⃗v∥0≤ ∥⃗u +⃗v∥ ≤ 2,5⃗u
∥⃗u∥=1
2⃗u
2⃗u +⃗v = 4⃗a + 2⃗ b−⃗a −2⃗ b= 3⃗a ⃗a =2
3⃗u +1 3⃗v
⃗u + 2⃗v = 2⃗a +⃗ b−2⃗a −4⃗ b=−3⃗ b⃗ b=−1
3⃗u −2 3⃗v
2⃗u + 2⃗x −21⃗x + 12⃗u =⃗ 0⇐⇒ 14⃗u = 19⃗x ⇐⇒ ⃗x =14
19 ⃗u
5(2⃗u −4⃗x) = 3(⃗u −3⃗x)⇐⇒ 10⃗u −20⃗x = 3⃗u −9⃗x ⇐⇒ 7⃗u = 11⃗x ⇐⇒ ⃗x =7
11 ⃗u
BC =AC −AB = 5 −2=3 −→ BI =1
3−−→
BC ⇒BI =1
3×3=1 AI =AB +BI = 3 IA =−3
F B =−6⇒BF = 6 EF =EB +BF =−3 + 6 = 3
M[AB]A N [AC)AN = 3AC
−−→
NB =−−→ NA +−−→ AB =−3−→ AC +−−→ AB −−→ CM =−→ CA +−−→ AM =−−→ AC +1
3−−→
AB −−→ NB = 3−−→ CM
(BN)∥(CM )B /∈(AC)
−−→
CE =−2−→ CA = 2−→ AC −−→ CE =−→ CA +−→ AE −→ AE = 2−→ AC +−→ AC = 3−→ AC −→ AC =1
3−→
AE
−−→
DC =−−→ DA +−→ AC =−1
3−−→
AB +1
3−→
AE =1
3(−→
AE −−−→ AB) = 1
3−−→
BE (BE)∥(CD)
−−→
F E =−→ AE −−→ AF =2
5−−→
AB −2
5−→
AC =2
5(−−→
AB −−→ AC) = 2
5−−→
CB −−→ F E −−→ CB (EF )∥(BC)
2−−→ CD = 9⃗u −6⃗v −3−−→ AB = 9⃗u −6⃗v 2−−→ CD =−3−−→ AB
−−→
AB =−2
3−−→
CD (AB)∥(CD)

−−→
OD =−−→ OC +−−→ CD = 4−→ OA + 4−−→ AB = 4(−→ OA +−−→ AB)=4−−→ OB −−→ OD −−→ OB O, B, D
E∈(AB)AE =3
2AB B F ∈(AD)AF = 3AD D
−−→
EC =−→ EA +−→ AC =−3
2−−→
AB + (−−→ AB +−−→ BC) = −1
2−−→
AB +−−→ AD −−→ BC =−−→ AD
−−→
CF =−−→ CD +−−→ DF =−−−→ AB + (−−→ DA +−→ AF ) = −−−→ AB −−−→ AD + 3−−→ AD =−−−→ AB + 2−−→ AD
−−→
CF = 2−−→ EC E, C, F
3⃗u 9
−6−2⃗v 4 −2−3
2⃗u −9/2
3⃗u +⃗v 1 −1⃗v −⃗u −5 3⃗u −2⃗v 7 −4
∥⃗u∥=√5 + 4 = 3 ∥3⃗u∥= 3 ×3=9 ∥⃗v∥=√4 + 5 = 3 ∥−2⃗v∥= 6
⃗u −⃗v √5−2
2 + √5∥⃗u −⃗v∥2= (9 −4√5) + (9 + 4√5) = 18 ∥⃗u −⃗v∥= 3√2
AB =√1 + 5 = √6AC =√2 + 4 = √6BC =√3 + 3 = √6ABC
⃗u −1
3⃗v 1 −2⃗w 3 a
det(⃗u, ⃗v)=(−1)(−2) −3×1 = −1̸= 0
det(⃗v, ⃗w) = 1 ×a−(−2) ×3 = a+ 6 ⇐⇒ a+ 6 = 0 ⇐⇒ a=−6
det2
1,3 2= 2 ×2−1×3=1̸= 0
det2
1,3 1,5= 2 ×1,5−1×3=0
−−→
AB 2
−2−−→
AM 23
x−3M, A, B ⇐⇒ det = 0 ⇐⇒ 23 ×(−2) −2(x−3) = 0 ⇐⇒ −2x−40 = 0 ⇐⇒
x=−20
I2+3
2;−3+4 2=5 2;1 2G−4+2+3 3;1−3+4 3=1 3;2 3−→
IG 1
3−5 2
2
3−1 2=−13/6
1/6
−→
AG 13/3
−1/3−→
AI 13/2
−1/2det = 13
3×−1 2−−1 3×13 2=−13 6+13 6= 0 A, G, I
ABCD ⇐⇒ −−→ AB =−−→ DC −−→ AB 6
−43−xD= 6 4 −yD=−4D(−3; 8)
A′−3+3
2;2+2 2= (0; 2)
−−→
GA′=1
3−−→
AA′−−→ AA′−2
5−xG=−2
32−yG=5 3G2 3;1 3
A′[BC]G[AA′]A′G ABC
−→
GA 4/3
−10/3−−→
GB −11/3
5/3−−→
GC 7/3
5/3  (4 −11 + 7)/3 (−10 + 5 + 5)/3=0 0
OA =√4 + 9 = √13 OB =√9 + 4 = √13 OC =√9 + 4 = √13 O
ABC
H(2 −3 + 3; −3 + 2 + 2) = (2; 1)
I[AC] [BD]−→ CI =−→ IA −→ IB =−−→ ID −→ DI =−−→ ID
⃗u =−2(−→ CI +−→ IB) + 7(−→ CI +−→ ID) + 3(−→ DI +−→ IB)−4(−→ CI +−→ IA) = −→ CI +−→ IB + 7−→ ID + 3−→ DI −4−→ IA

=−→ IA −−→ ID + 7−→ ID −3−→ ID −4−→ IA =−3−→ IA + 3−→ ID
−−→ IA =−→ IC −−→ IM = 3(−→ IC +−→ ID)ICP D −→ IP =−→ IC +−→ ID M
−−→
IM = 3−→ IP
(A,⃗ı,⃗ȷ)⃗ı =−−→ AB ⃗ȷ =−−→ AD AB = 1 A(0; 0) B(1; 0) D(0; 1) C(1; 1)
I H I (AB) [AB]xI=1
2[IH]
yI=√3
2I1 2;√3 2
V V √3
2(BC) [BC]V1 + √3 2;1 2
−→
DI 1/2
√3
2−1−−→
DV 1 + √3
2
−1/2
det(−→ DI, −−→ DV ) = 1
2×−1 2−√3 2−11 + √3 2=−1 4−3 4−1=−1 4+1 4= 0
−→
DI −−→ DV D I V