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Given that triangle ABC and point M satisfy vector MA + vector MB + vector MC =0 vector, if vector AB + vector AC = M * vector AM holds in real number M, then M =?
Why are real numbers and dots represented by the same letter? A point is a coordinate of two numbers, and a real number is just a number. To separate, write the real number in m
AB+AC=AM+MB+AM+MC
MA+MB+MC=0
MB+MC=-MA
AB+AC=2AM-MA=3AM.m=3
Given three vectors in the plane a=(3,2), b=(-1,2), c=(4,1),(1) Find the value of the real number m, n satisfying a=mb+nc (2) If vector d=(x, y) satisfies d//(a+b) and /d/=5, find vector d
(1) A = mb + nc (3,2)= m (-1,2)+ n (4,1)(3,2)=(-m +4n,2m + n)-m +4n =3,2m + n =2 The simultaneous equation m =5/9, n =8/9(2) a + kc =(3,2)+ k (4,1)=(3+4k,2+ k)2b -a =2(-1,2)-(3,2)=(-5, 2)(A + kc)//(2b - a)2(3+4k)-(-5)(2+ k)=0, k =-16/13(3) d -c =(x, y)-(4,1)=(x -4, y -1) a + b =(3,2)+(-1,2)=(2,4)(d - c)//(a + b)4(x -4)-2(y -1)=0 2X-y-7=0~(1) and d-c =1[(x-4)2+(y-1)2]=1 square on both sides,(x-4)2+(y-1)2=1~(2) simultaneous (1),(2) formula, x=4 5/5y=1±2√5/5 solution, d=(4 5/5,1+2√5/5) or (4-√5/5,1-2√5/5)
Given the non-zero vector a, b, and a//b, vector |a|=2, vector |b|=1, find the value of real number t when |a+tb| takes the minimum value
A//b, therefore: a·b=|a b|cos (0)=2
Or: a·b=|a b|cos (π)=-2
|A+tb|^2=|a|^2+t^2||b|^2+2ta·b
When a is in the same direction as b:
|A+tb|^2=4+t^2+4t=(t+2)^2, i.e. when t=-2|a+tb|get minimum value 0
When a and b are reversed:
|A+tb|^2=4+t^2-4t=(t-2)^2, i.e. when t=2|a+tb|get minimum value 0
Given that M and N are the midpoint of any two line segments AB and CD respectively, verify that vector MN=1/2(vector AD + vector BC) Note that any two segments... ……… ……… ……… … Given that M and N are the midpoint of any two line segments AB and CD, verify that vector MN =1/2(vector AD + vector BC) Note that any two segments... ……… ……… ……… …
Vector MN=1/2(vector MC + vector MD)
Vector MC=vector MA+vector AC
Vector MD = Vector MB + Vector BD
Vector MA+vector MB=0
So vector MN=1/2(vector AD + vector BC)
In any quadrilateral ABCD, E is the midpoint of AD and F is the midpoint of BC.
Vector EF = Vector ED + Vector DC + Vector CF
Vector EF = vector CA + vector AB + vector BF
2 Vector EF =(vector ED + vector DC + vector CF)+(vector CA + vector AB + vector BF)
Because E is the midpoint of AD and F is the midpoint of BC
Therefore: vector ED=-vector EA, vector CF=-vector FB
2 Vector EF = vector AB + vector CD
Vector EF =1/2(vector AB + vector CD)
Given any quadrilateral ABCD, E is the midpoint of AD and F is the midpoint of BC, prove that:(both are vectors) EF+FE+AB+DC
E and F are respectively the midpoint of AD and BC, EA+ED=0, FB+FC=0, BF+BF+FE+EA=0, EF=AB+BF+EA1, and EF=ED+DC+CF2 are obtained from 1+2, EF=AB+DC+EA+ED+BF+CF=AB+DC.: AB+DC=...
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