Mathematics High One Vector Calculation Problem, how to do this math problem High One Vector Calcula

Establish a planar Cartesian coordinate system with c as the original center, there is.

Points a(0,m),b(n,0),c(0,0),d(n2,m2),e(n4,m4).

then the vector ae = (n 4, -3m 4).

Let the coordinates of point f be (x,0), and the vector af=(x,-m) vector af is collinear with ae, according to the collinear theorem of plane vectors: (n 4) * (m) + (3m 4) * x = 0

x=n 3 and the coordinates of the point f are (n 3,0).

then the vector af=(n 3,-m).

So |af|= under the root number (m2+n2 9).

Is point D the midpoint of AB?

If so, it should be answered like this.

As eg perpendicular to cb and dh perpendicular to cb, obviously vector eg = 1 2 vector dh = 1 4 vector ac, so vector fe = 1 4 vector fa, so vector ea = 3 4 vector fa.

And the vector EA = vector AC + vector CE = vector AC + 1 2 vector CD = vector AC + 1 4 (vector CA + vector CB) = vector M-1 4 (vector M + vector N) = 3 4 vector M-1 4 vector N

So the vector af = -4 3 * vector ea=-4 3 * (4 3 vector m-1 4 vector n) = 1 3 vector n - vector m

So the length af = under the root number (1 9n + m).

Note: n = square of n.

Crossing the point F is fk bc, passing ae to h, and passing ab to k f is the midpoint of cd, fk bc

KH is the triangular ABE median.

H is the midpoint of AE, and Kh=1 2BE=1 2CE FGH BGE

ge:gh=be:fh=2:1

and eh=ah

eg:ga=2:(1+3)=1:2

Vector ae = a + 1 2 * b

Vector ag=2 3 vector ge=2 3*a+1 3*b

Vector symbols are inconvenient to type, and alphabetical order represents vector directions.

Easy to know: EF=EA+AB+BF

ef=ed+dc+cf

then 2ef=(ea+ed) +ab+dc + bf+cf) all add up to 0 in parentheses

2ef=ab+dc

ef=1/2 (a -b)

Solution: ea + bf = (-a - b) 2;

b + 2*ea + a + 2bf = 0;

ef = ea + a + bf = a + a - b)/2 = (a - b)/2

Then the vector of EF = (a - b) 2

All right? Hope you can adopt!

Hello! This way:

Proof: Vector OC = Vector OA + Vector AC

Vector OA+ vector BC

Vector OA+ (Vector OB-Vector OC).

i.e. (1+) vector oc = vector oa + vector ob i.e. vector oc = vector oa + vector ob 1+

Because (2a-3b) point times (2a+b) = 61, so 4a squared - 3b square - 4a point times b = 61, and a modulo = 4 b modulo = 3 , so a square = 16, b square = 9, so 4a point times b = 24, so a point multiplies b = point b=a modulo multiplies b modulo times cos, so cos = so ab angle = 60 degrees and (a + b) square = a square + b square + 2a point is b = 16 + 9 + 2 * 6 = 37

So the absolute value of a + b = root number 37

1.(2a-3b) dot times (2a+b) = 61

Because modulo of vector a = 4 modulo of b = 3

Substituting the above equation yields 64-27-4ab=61

The solution yields ab=-6

3.|a+b|^2=(a+b)^2=a^2+b^2+2ab=4^2+3^2+2*(-6)=13

Solution: Easy to know, a = 2, b = should have (a+tb)(ta+b) 0===>3t²+11t+3＜0.===>(-11-√85)/6＜t＜(-11+√85)/6

a+tb）^2＝a^2+t^2b^2+2ta·b=a^2+t^2b^2+2t|a||b|cos

t|b|+|a|cos＜a,b＞）^2+|a|^2（1-cos^2＜a,b＞）

When t -|a|cos noisy a,b |b|time.|a+tb|There is a minimum value |a|√（1-cos^2＜a,b＞）。

At this time, (a+tb) book is trapped·b a·b+[-a|cos＜a,b＞/|b|]|b|^2＝a·b-a·b＝0

i.e. (a+tb) b, the angle is 90 degrees.

If the vertical multiplication is zero, then (k-3)*10-(2k+2)*4=0 then k=19

In parallel, the value of k is the same, i.e. (2k+2) (k-3)=-4 10 then (k-3)*10+(2k+2)*4=0 of k=11 9

ka+b=(k-3,2k+2), a-3b=(10,-4), because the two vectors are perpendicular, the product of the quantity is 0

So 10(k-3)-4(2k+2)=0, the solution is k=19 because the two vectors are parallel, so -4(k-3)-10(2k+2)=0 is solved to k=-1 3

1.Because ka=(k,2k) 3b=(-9,6)ka+b=(k-3,2k+2) a-3b=(2,0) when ka+b is perpendicular to a-3b, (k-3)*2+(2k+2)*0=0 can be solved to k=3

2.When ka+b is parallel to a-3b, (k-3)*0-(2k+2)*2=0 can be obtained

The solution is k=-1

First of all, it's hard to write clearly, but if you can talk about it in person, you'll understand it a little bit. I try though. The first question, through the graph, is enough.

Draw the vector a and vector b of the co-intersection point, and then draw the vector a + vector b on this basis, you should be able to do this, right? Then, a(a+b)=0 means that vector a is perpendicular to vector a+b, and then, solved in a right-angled triangle by the parallelogram rule, vector b is hypotenuse and has a length of |b|, the vector a is the right-angled edge, and the length is |a|。And because the angle between vector A and vector B is 120, and the angle between vector A+B and vector A is 90, the angle between vector A+B and vector B is 30, so, |a|=1/2|b|, so the answer is 1 2.

The second question, by drawing a picture, draw a feasible domain, this, you will definitely be. This feasible domain is a triangle. Then, first calculate y x, so that k=y x, then y=kx, and then, draw a straight line through the origin, swing the straight line clockwise, the first contact angle, this point is (1, 3 2), bring this point into the above equation, then calculate k, and then add a one is the maximum value.

The answer is 5 2 This method is correct, and I haven't calculated it seriously, so I'll do it again when you do it. What can be said clearly in a few sentences, it is very hard to write, but I hope it can help you!

a(a+b)=a moves toward about |a|/|b|=1/4

The second question is to draw a picture to know that you need to do more.

The angle between the vectors a and b in the first question is 120°, and the |a||b|cos120=-1/2|a||b|=ab

A*2+ab=0 is obtained from a(a+b)=0

Then we get a*2=-ab

So |a|*2=1/2|a||b|

So |a|=1/2|b|

then |a|/|b|=1/2