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In the equilateral triangle ABC with 1 side length, let BC vector be a vector, CA vector be b vector, AB vector be c vector, then a.b+b.c+c.a=? Why the included angle is 120
A·b+b·c+c·a=BC·CA+CA·AB+AC·BC
=|BC CA|*cos (π-C)+|CA AB|*cos (π-A)+|AC BC|*cos (π-B)
= Cos (2π/3)+ cos (2π/3)+ cos (2π/3)=-1/2-1/2-1/2=-3/2
Why the included angle is 2π/3 mainly depends on the starting point position of the vector
The side length of the equilateral triangle ABC is 1, vector BC=a, vector CA=b, vector AB=c, then ab+bc+ca is
AB+BC+CA=0 vector
Square on both sides:
(AB+BC+CA)2=0
AB|2 BC|2++|CA|2+2AB ● BC+2BC ● CA+2CA ● AB=0
The side length of the equilateral triangle ABC is 1
2 A B ● BC+2 BC ● CA+2 CA ● AB=-3
BC●CA+CA●AB+AB●BC=-3/2
Ab+bc+ca
=BC●CA+CA●AB+AB●BC
=-3/2
Given that the three sides of △ABC are AB=7, BC=5, CA=6, then AB • The value of BC is ______.
From the cosine theorem, cosB = AB2+BC2−AC2
AB×BC=19
35,
AB •
BC =|
AB||
BC|cos (π−B)=−7×5×19
35=−19
Therefore, the answer is:-19
Let AB=c, BC=a, CA=b, then ab+bc+ca equals? A.0 B.-3/2 C.3 D.-3
Their included angle is 120°, cos120°=-1/2, and the side length is √2.
√2 2×(-1/2)×3=-3, Select D
Given that abc is a unit vector and ab=0, and (a-c)(b-c) is less than or equal to 0, then /a+b-c/max is? Very anxious to get the answer. Thanks a lot.
(A-c)(b-c)=ab-bc-ac+c^2=c^2-ba-ca ≤0 1≤Ba+ca
|A+b-c|^2=a^2+b^2+c^2+2+2ab-2bc-2ac=a^2+b^2+c^2-2bc-2ac≤1+1+1-2=1
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Line surface perpendicularity: can prove that the normal vector of the plane is parallel to the straight line. Face perpendicular: can prove that the normal vector of the two planes is perpendicular. Dihedral angle: for example, the vector a (x, y, z), b (m, n, p) of the straight line that knows perpendicular to the intersection line of the plane (sorry, I don't know the sign of vector), the size of dihedral angle =(vector) a*b...
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