rule
a^2+b^2=4....................( 1)
x^2+y^2= 1....................(2)
|a vector +tb vector | = √ [(a+tx) 2+(b+ty) 2]
=√[a^2+2atx+(tx)^2+b^2+2tby+(ty)^2]
=√[(a^2+b^2)+t^2(x^2+y^2)+2t(ax+by)]
=√[t^2+2t(ax+by)+4]
It's not good here
So, I suggest you draw a picture. It can be said blatantly from the figure that the minimum value of |a vector +tb vector | is taken if and only if (A vector +tb vector) is perpendicular to B vector. ...
To achieve this condition, as long as the following conditions are met.
|a|^2|b|^2=|tb|^2+|b|^2
That is to say,
2^2=t^2* 1- 1
At this time, t=√5 or t=-√5.
Reply: No sword and pen-Jianghu beginner level 3 12- 1 23:39
|a vector +tb vector | = radical sign [(A vector +tb vector) 2]
therefore
|a vector +tb vector | 2
=(a vector +tb vector) 2
=|a|^2+t^2*|b|^2+2t|a||b|cos<； a，b & gt
= 1+4t^2+4cos<； a，b & gt*t
The above formula is a quadratic function about t!
therefore
When t =-8/4 cos
|a vector +tb vector | 2 takes the minimum value, that is |a vector +tb vector | takes the minimum value.
Note: Well, I have done this problem.
Responder: Dew in the sky-Magician Level 5 12-2 08:52.
There is no explanation upstairs why | a+tb | takes the minimum value if and only if (a+tb) is perpendicular to the B vector.
Answer:
When |a+tb| is the smallest, | a+TB | 2 is also the smallest. () 2 represents a square.
|a+tb|^2=|a|^2+|t*b|^2+2*|a|*|tb|*cos(a,t*b)
Because cos (a, t * b) = cos (a, b), | a | = 2, | b | = 1.
therefore
|a+tb|^2=t^2+4*t*cos(a,b)+4=(t+2*cos(a,b))^2+4-(2*cos(a,b))^2
| a+TB | 2 has a minimum value if and only if t+2*cos(a, b)=0, that is, t=-2*cos(a, b).
Therefore, when t=-2*cos(a, b), |a+tb| has a minimum value.
Respondent: mumian 1985- first grade of probation 12-3 14:0 1.
Solution: Let a vector a(a, b) and a vector b(x, y).
rule
a^2+b^2=4....................( 1)
x^2+y^2= 1....................(2)
|a vector +tb vector | = √ [(a+tx) 2+(b+ty) 2]
=√[a^2+2atx+(tx)^2+b^2+2tby+(ty)^2]
=√[(a^2+b^2)+t^2(x^2+y^2)+2t(ax+by)]
=√[t^2+2t(ax+by)+4]
= ...
Unfinished. I wonder if Cauchy inequality is possible?
Interviewee: wzzju- Magician Level 4 12-3 19:57.
When t=- 1, the minimum value of the original formula is equal to 1, which means that the directions of A and B vectors are opposite at this time.
Respondent: atponearth- probation level 1 12-4 17:32.
Agree with sth completely.
Liuyifei2372, mumian 1985 friend's solution! 、
It seems that my friend forgot to substitute the values of a and b.
Respondent: zza lex- assistant level 3 12-4 20: 19.
Vector point integral distribution law
t=-2a b
Responder: Fall into the shadow of darkness-Trainee Magician II 12-6 20: 17
Take point O as the starting point, make two vectors to represent vectors A and B respectively, and let point A and point B be a b respectively, then the vector to be obtained is the vector starting from point A and ending at a point on the straight line BO. So we can know that the minimum value should be perpendicular to point A, and the intersection with OB can get the end point of vector tb. So we can know that (vector A -tb vector) is perpendicular to tb vector, that is, vector B, so the dot product is zero, and t = vector A multiplied by vector B/B, that is, you can substitute the touch of A multiplied by cosine of included angle divided by B into data, or use quadratic function to find the extreme value, and the answer is the same. The teacher just gave it to me before I went home, hehe.
Respondents: livey _ Liwei- Trainee Magician Level 3 12-7 10: 14.
The fourth floor doesn't seem right. Imagine if A and B are perpendicular, t=0, and the modulus of a+tb is the smallest! So the value of t should be related to the positional relationship between a and B.
Draw a picture and take the minimum value of |a+tb| if and only if (a+tb) is perpendicular to the B vector! You can see from the geometric relationship when drawing a graph. You don't need to use formulas to calculate. This should be the easiest way. )
If a = (m, n) and b = (x, y), then
A+tb=(m+tx, n+ty), perpendicular to b, then
(a+tb)*b=0 means MX+TX 2+NY+TY 2 = 0,
(where x 2 and y 2 represent the squares of x and y), so
t(x^2+y^2)=-(mx+ny)=-(a*b)=-|a|*|b|*cos(#)
Let the included angle between vectors A and B be angle #, because | A | = 2 and | B | = 1.
So x 2+y 2 =1,so there is:
t=-2*cos(#)
Is the value of t (related to the angle between a and b vectors, that is, the position of a and b! )
References:
Sophomore students are really original!
Interviewee: liuyifei 2372- assistant level 3 12-7 14:04.
Solution: Let a vector a(a, b) and a vector b(x, y).
rule
a^2+b^2=4....................( 1)
x^2+y^2= 1....................(2)
|a vector +tb vector | = √ [(a+tx) 2+(b+ty) 2]
=√[a^2+2atx+(tx)^2+b^2+2tby+(ty)^2]
=√[(a^2+b^2)+t^2(x^2+y^2)+2t(ax+by)]
=√[t^2+2t(ax+by)+4]
It's not good here
So, I suggest you draw a picture. It can be said blatantly from the figure that the minimum value of |a vector +tb vector | is taken if and only if (A vector +tb vector) is perpendicular to B vector. ...
To achieve this condition, as long as the following conditions are met.
|a|^2|b|^2=|tb|^2+|b|^2
That is to say,
2^2=t^2* 1- 1
At this time, t=√5 or t=-√5. It can be seen that there is no need for formula calculation, which should be the simplest method)
If a = (m, n) and b = (x, y), then
A+tb=(m+tx, n+ty), perpendicular to b, then
(a+tb)*b=0 means MX+TX 2+NY+TY 2 = 0,
(where x 2 and y 2 represent the squares of x and y), so
t(x^2+y^2)=-(mx+ny)=-(a*b)=-|a|*|b|*cos(#)
Let the included angle between vectors A and B be angle #, because | A | = 2 and | B | = 1.
So x 2+y 2 =1,so there is:
t=-2*cos(#)
Is the value of t (related to the angle between a and b vectors, that is, the position of a and b! )
Respondent: Intellectual's Commitment-Beginner in Jianghu Level 2 12-8 12:35
Our teacher said that there was something wrong with the topic and it was not rigorous. The above solution was correct.
Respondent: Your 22- Scholar Level 2 12-8 19:38.
There are two cases of this problem. One is that the A vector and the B vector are on the same straight line, and the other is that the B vector is perpendicular to the A vector +tb vector. In both cases, you can use the minimum value. Try it yourself. I have done this question before, and the original question is two questions. One of them is the case 1: t=- 1. I can't remember the second answer, but I haven't forgotten the method. Try it yourself first.
Respondent: Yushanchao 52 1- Juren IV 12-8 22:0 1.
Setting vectors is a bit stupid.
You can just square the back,
Then we get a function about the included angle of the vector.
Just ask for scope.