Senior 1 math vector questions, ask for answers, thank you

Answer: 1) Knowing a=( 3,-1), so |a|=2, if|c|=2|a|, so |c|=4

and c a, so c = a

i.e. c(x,y) = (3,-1).

So x= 3, y= -

Because |c|=√｛（3）^2+（-2｝=4

i.e. 3 2 + 2 = 16

So = 2

So x= - 2 3 y=2 or x=2 3 y= - 2 so c=(- 2 3 ,2 ) or c=(2 3 ,-2 )2) If 12a+7b is perpendicular to a-b, then (12a+7b) (a-b)=0

i.e. 12a 2-5ab-7b 2=0

And because |a|=2

So 7|b|^2-5|b|-48=0

So (b|-3)(7|b|+16)=0

b|>0

, so |b|=3

It is known that a, b, and c are three vectors in the same plane, a=( 3,-1)1)if|c|=2|a|and c a, find the coordinates of c.

2) If 12a+7b is perpendicular to a-b, and the angle between b and a is 120°, find |b|.

Solution: (1) By.

a=(√3,-1)

Gotta |a|=2, thus |c|=2|a|=4, let c = (x,y) then x 2 + y 2 = 16 (1).

From a = ( 3, -1), and c a, x 3 = y -1 i.e. -x 3 = y (2).

Coupling (1) and (2) solution.

x=±2√3,y=±2

c = (2 3, -2) or c = (-2 3, 2) and b = (x, y).

by a = ( 3, -1), and the angle between b and a is 120°

It is obtained from the formula of the included angle.

cos120°=(√3x-y)/2|b|

Gotta |b|=y-√3x

y=|b|+√3x (1)

12a+7b=12(3,-1)+7(x,y)(7x+12 3,7y-12).

a-b = ( 3,-1)-(x,y) = ( 3-x,-1-y) is perpendicular to a-b by 12a+7b.

Gotta |7x+12√3)(√3-x)+(7y-12)(-1-y)=0 (2)

Substituting (1) into (2) to get a quadratic equation about x.

It has solid roots, and it has δ 0

Sorted out about|b|of quadratic equations.

The solution is available|b|。(omitted).

1) Let c=(x,y)|a|=2,x 2+y 2=8,y x=-1 root number 3, the two forms are synapoid y 2=2, y=+- root number 2, x = -+ root number 6

2）（12a+7b）.(a-b)=0, we get 12|a|, angle ab

Two-way confederation, can be obtained|b|

I suspect you're doing a wrong question, you shouldn't have a 2 in front of your 2m, and if you remove 2, you should choose b

The following vector " " " is omitted, and the quantity product is represented by * The quantity product of the vector a*b=(4i+3j)x(3i-4j).

4i×3i-4ix4j+3ix3j-3jx4j=12i²-7ij-12j²

ij is a single vector.

i²=j²=1

ij is perpendicular to each other (the product of the number of vectors perpendicular to each other is zero) 7ij=0

a*b=0, if you understand well, you can also solve it like this.

Since ij is a unit vector.

So let's do the XY axis separately in IJ.

At this time, the coordinates of the vector ab are (4,3) (3,-4) (the coordinate expression of the quantity product of the vector: (x,y)*(m,n)=xm+ny) then we have a*b=4x3+3x(-4)=0

a*b = 9i square - 16j square = 9 - 16 = 7. i square = j square = 1

1、|a＋b|=√3|a－b|, so |a＋b|²=3|a－b|²，a|²＋2a*b＋|b|²=3(|a|²－2a*b＋|b|Reuse|a|=|b|=1, we get 4a*b=1. Again|3a－2b|²=9|a|²－12a*b＋4|b|=10, so |3a－2b|=√10。

2. If the angle is a right angle, then a*b=1 2 =0, and the solution is = 1 2;If the angle is obtuse, then a*b <0 and a and b are not parallel. < 1 2 is obtained by a*b<0, and ≠2 is obtained by a being and b are not parallel. Therefore, when the angle is obtuse, <1 2 (in this problem, if the angle is obtuse, then a*b<0, and vice versa, it is incorrect); If the angle is acute, then a*b>0 and a and b are not parallel, the solution is > 1 2 and ≠2 (this question also needs to round off the case where the angle is 0).

Vector a=(-1 2, 3 2), vector oa = vector a - vector b, vector ob = vector a + vector b

The triangle AOB is an isosceles right triangle at right angles to O.

0a=ob，oa⊥ob

Let the vector b=(x,y).

Vector oa=(-1 2-x, 3 2-y), vector ob=(-1 2+x, 3 2+y).

Vector oa vector ob =1 4-x 2+3 4-y 2 =0==> x 2+y 2=1 (1).

0a=ob again

1/2+x)^2+(√3/2-y)^2=(x-1/2)^2+(√3/2+y)^2

1/4+x+x^2+3/4-√3y+y^2=x^2-x+1/4+3/4+√3y+y2

x=√3y (2)

1) (2) The simultaneous solution yields x= 3 2, y = 1 2

Vector b (3 2, 1 2).

Vector oa=(-1 2- 3 2, 3 2-1 2), vector oa|=√2

Vector ob=(-1 take 2+ 3 2, 3 2+1 2), vector ob|=√2

s△aob=1/2•2=1

Vector b (3 2, 1 2).

Vector oa=(-1 2- 3 2, 3 2-1 2), vector oa|=√2

Vector ob=(-1 2- Xiaoqing3 2, 3 2-1 Chayu2), vector ob|=√2

s△aob=1/2•2=1

The triangular AOB is an isosceles right-angled triangle Zhihu beam at right angles to O.

0a=ob，oa⊥ob

Let the vector b=(m,n).

oa ob vector b · vector a = 0

m2 + root number 3n 2 = 0

0a=ob again

m²+n²=1

The solution is m = root transport number 3 2, n = 1 2 or.

m=-Gen Qing No. 3 2, n=-1 2

i.e. vector b (root number 3 2, 1 2) or (- root number 3 2, -1 2) s aob = 1·1 2 = 1 2

The landlord, the first floor said that the vertical vector uses the dot product, who told you that it is equal to zero? It's not the same as A*B. But I don't know why Hu Sui wants to be pure.

1.Let the point p(x,y) then the vector ap=(x-2,y-3)ab=(3,1)ac=(5,7).

ab + ac=(3+5, 1+7).

Because ap=ab+into ac

So (x-2, y-3) = (3+5 in, 1+7 in) and x-2 = 3+5 in, y-3 = 1+7 in.

x=5+5 y=4+7.

Because p is in the third quadrant of Chaqingwang, x<0 y<0 and 5+5 into <0 and 4+7 into <0

The solution can be <-1

2 (1)a+kc=(3+4k, 2+k) 2b-a=(-5, 2)

Because (a+kc) 2b-a) then 2 x (4k+3)=-5 x (2+k) gives k=-16 13

2)d-c=(x-4, y-1) a+b=(2,4)

Because (d-c) is poorly resistant to a+b so (x-4) x 4=2 x(y-1) y=2x-7 so d=(x,2x-7 ) d-c=(x-4, 2x-8).

And because 丨d-c丨=1 so (x-4) 2=(2x-8) 2 solution gives x=4 d=(4,1).

3 k×a+b=(k-3,2k+2) a-3b=(10,-4)

Since k a+b is parallel to a-3b, (k-3)x (-4)=(2k+2) x 10 gives k=-1 3

Bring k into k a+b=(-10 3,4 3) and a-3b= -1 3(k a+b) as reverse parallelism.

4 丨a+b丨= 3 丨a+b丨 square a 2 + 2 ab + b 2 = 0 and -2ab = a 2 + b 2

丨3a-2b丨 2=9a 2-12ab+4b 2=15a 2+10b 2

Because 丨a丨=丨b丨=1 so 丨3a-2b丨=5

1) For the convenience of "in", it is represented by m. Let p be (a,b),ab=(3,1) ac=(5,7).

then ap=(a-2,b-3)=ab+mac=(3+5m,1+7m)

a-2=3+5m, get a=5+5m

b-3=1+7m, get b=4+7m

p(a,b) is in the third quadrant, so a<0 and b<0

That is: 5+5m<0, 4+7m<0

m<-1,m<-4/7

The same is small, so m<-1

2)2b-a=(-5,2)

a+kc=(3,2)+(4k,k)=(4k+3,k+2)

a+kc∥2b-a

So -5 = m(4k+3) and 2=m(k+2).

That is: mk=2-2m, 4mk=-3m-5

0=13-5m

m=13/5

So k=-16 is 13

3)d-c=(x-4,y-1),a+b=(2,4)

d-c a+b, so (x-4) = 2m, (y-1) = 4m

(y-1) (x-4)=2 (*.)

Because |b-c|=1

i.e. (x-4) 2+(y-1) 2=1

x-4)^2+4(x-4)^2=1

x-4)�0�5=1/5

x= 5 5+4, and then substitute (*) to get y, and then d(x,y) to get (note: the values of x,y should be one-to-one).

4) Knowing a=(1,2)huwei,b=(-3,2), when k is what value, k a+b is parallel to a-3b, are they in the same direction or opposite when parallel?

Solution: a-3b = (10,-4).

Let ka+b=m(a-3b)=(10m,-4m).

k-3,2k+2)=(10m,-4m)

k-3=10m

2k+2=-4m

The solution gives k = -1 3

m=-1/3

Because m is negative, it is reversed.

5）4.It is known that a and b meet the requirements of |a+b|= 3 丨a+b丨,丨a丨=丨b丨=1,find 丨3a-2b丨.

This question is incorrect. - Scattered key ———

6）.It is known that the angles of the three vectors a, b, c, and the two are all 120°, 丨a丨=1, 丨b丨=2, 丨c丨=3, and find the angle between a and a of the vector (a+b+c).

Solution: (a+b+c) a=a 0 5+ab+ac=1-1 2-1 2=0, so (a+b+c) a angle is 90

a*b=|a|*|b|*cos60=1

a+xb)*(xa-2b)=xaa-2ab+xxab-2xbb=x*|a|*|a|+(xx-2)ab-2x|b|*|b|

Put the number of generations in, sort it out, and express the range of obtuse angles, and you can find the range of the value of x