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If the three sides of triangle ABC are AB=7, BC=5, CA=6, then vector AB is () Don't worry about it. If the three sides of triangle ABC are AB=7, BC=5, CA=6, then vector AB is () Don't be so long-winded.
Cos∠ABC=(AB^2+BC^2-AC^2)/2AB*BC
=(49+25-36)/2*7*5
=19/35
The angle between vector AB and vector BC is 180°-∠ABC
Vector AB×Vector BC
=|AB BC cos (180°-∠ABC)
=-| AB BC cos∠ABC
=-| AB BC (|AB 2 BC|^2-||AC|^2)/(2 AB BC|)
=-7*5*19/35
=-19
In the vector △ABC, BC=a, CA=b, AB=c, and a·b=b·c=c·a, what graph is the triangle ABC? This is a vector problem. The upper and lower letters of the face are vectors. Please don't give me the final result. In the vector △ABC, BC=a, CA=b, AB=c, and a·b=b·c=c·a, what is the graph of the test triangle ABC? This is a vector problem. The upper and lower letters of the face are vectors. Please don't give me the final result.
Because in triangle ABC, a+b+c=0
Then a =-(b+c), b =-(a+c)
Because a·b=b·c=c·a, a·b+c·a=a·b+b·c
With a·(b+c)=b·(a+c)
-A^2=-b^2
A^2= b^2
|A|=|b|
Similarly,|a|=|b|=|c|
So it's a regular triangle.
In triangle ABC, if the vector AB^2= AB*AC+BA*BC+CA*CB, what triangle is this? 1 Equilateral triangle 2 acute triangle 3 right triangle 4 obtuse triangle In triangle ABC, if the vector AB^2=AB*AC+BA*BC+CA*CB, what triangle is this? 1 Equilateral triangle 2 acute triangle 3 right triangle 4 obtuse triangle
Option 1
Steps:
AB^2=AB (AC+BC)+AC×BC
AB^2-AB (AC+BC)-AC×BC=0
(AB-AC)(AB+BC)=0 or (AB+AC)(AB-BC)=0
So choose 1
In the triangle ABC, if the vector AB*CA=BA*CB=-1, prove that the triangle ABC is an isosceles triangle, find the length of AB, if the module of the vector AB+AC=root 6, Find the area of the triangle and write as much as you know.
AB•CA=BA•CB=-AB•CB
Therefore, AB•CA+AB•CB=0
AB•(CA+CB)/2=0
Let D be the midpoint of AB, then CD =(CA+CB)/2
Therefore AB•CD=0
AB⊥CD
CD is both the center line and the high line.
So triangle ABC is isosceles triangle, CA=CB
-1=AB•CA=AB•(CD+DA)=AB•CD+AB•DA=0+AB•((-1/2) AB)
=(-1/2) AB•AB =(-1/2)|AB |^2
| AB |= Root 2
AB+AC=AB+AD+DC=3/2AB+DC
Root 6=|AB+AC|=|3/2AB+DC|
Because of AB⊥CD, the Pythagorean theorem
(Root 6)^2=|3/2AB+DC |^2=|3/2AB |^2 DC |^2=9/2+|DC |^2
Solution |DC |= Root 1.5
Area of triangle ABC =|||CD|/2=(root 2)*(root 1.5)/2=(root 3)/2
Given that the area of triangle ABC is S and the vector AB vector AC=S1. Find the value of tan2A 2. If B=π/4[ vector CB-vector CA ]=3, find S Just answer the second question.
1
AB·AC=|AB AC|cosA, and: S=(1/2)|AB AC|sinA, so:|AB AC|=2S/sinA
Therefore: AB·AC=(2S/sinA)*cosA=S, i.e. tanA=sinA/cosA=2
Therefore: tan2A=2tanA/(1-tanA^2)=4/(-3)=-4/3
2
TanA=2> sqrt (3), so:π/3
Given triangle ABC,(vector AB)^2= vector AB * vector AC + vector BA * vector BC + vector CA * vector CB, let a, b, c be divided into three sides of the triangle Given that △ABC is a right triangle, angle C=90°, let a, b, c be the three sides of triangle ABC respectively, if the inequality a^2(b+c)+b^(c+a)+c^2(a+b)>=kabc holds for any a, b, c, find the range of k
Angle C =90°,
A=csinA, b=ccosA, the inequality becomes
K <=(sinA)^2(cosA+1)+(cosA)^2(1+sinA)+sinA+cosA, constant 1
Let sinA+cosA=t,0k <=1+t [(t^2-1)/2+1]=(1/2)(t^3+t+2), denoted by f (t),
F (t) is an increasing function, f (t)> f (1)=2,
K <=2, is required.
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