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If two vectors are added =0, then is the sum of their abscissa equal to 0?
0
Two Proofs of Vector Algorithm (λa)·b=λ(a·b)=a·(λb) Two Proofs of Vector Algorithm (λA)·b=λ(a·b)=a·(λb) A·b=a·ca⊥(b-c)
Let a=(x, y), b (m, n)
(λA) b (λx,λy)(m, n)(λxm yn)λ(xm+ym)λ(ab)
(Xλm+yλn)(x, y)(λm,λn) a (λb)
Ab=acab-ac=0a (b-c)=0a⊥(b-c)
Given the plane vector a=(2,4), b=(-1,2). If c=a-(a·b) b, then |c |=_____
C=a-(a·b) b
=(2,4)-(-1×2+2×4)(-1,2)
=(2,4)-6(-1,2)
=(2+6,4-12)
=(8,-8)
|C|= square of √8+ square of (-8)
=8√2
Given plane vector a=(2,4) b=(-1,2) if c=a-(a.b) b then/c/= Given the plane vector a=(2,4) b=(-1,2) if c=a-(a.b) b then/c/=
A·b=2×(-1)+4×2=6c=a-(a·b) b=a-6b=(2,4)-6(-1,2)=(2,4)-(-6,12)=(2+6,4-12)=(8,8)|c|=√82+(-8)2=8√2 A:|c| is 8√2. PS: if vector a=(x1, y1), b=(x2, y2) number product coordinate operation: a·b=x1x2+y1y2 number multiplication operation:λa=(λ...
Let A, B, C of the internal angles of the acute triangle ABC be a, b, c, a=2bsinA (I) Find the size of B; (II) If a=3 3, C=5, find b.
(I) by a =2bsinA,
According to the sine theorem, sinA =2sinBsinA,, so sinB =1.
2,
From △ABC as an acute triangle, B=π
6.
(II) According to the cosine theorem, b2=a2+c2-2accosB=27+25-45=7.
So, b=
7.
Given that the reciprocal of three sides abc of triangle ABC is an equal difference sequence, it is proved that angle B is an acute angle. Given that the reciprocal of the three sides abc of the triangle ABC is an equal difference sequence, it is proved that the angle B is an acute angle.
Let three sides a, b, c
Then cos∠B =(a^2+ c^2-b^2)/(2ac)
By
1/A-1/b=1/b-1/c
Get
(A+c) b=2ac
Because a+c≥2√(ac)
So b (ac)
So b^2≤ac <2ac≤a^2+b^2
So cos∠B >0, so ∠B is an acute angle
Let three sides a, b, c
Then cos∠B=(a^2+c^2-b^2)/(2ac)
By
1/A-1/b=1/b-1/c
Get
(A+c) b=2ac
Because a+c≥2√(ac)
So b (ac)
So b^2≤ac <2ac≤a^2+b^2
So cos∠B >0, so ∠B is an acute angle
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