High One Vector Question !! High School Vector Fundamentals Problem!!!! Help with the solution

m*n=(a+kb)(ka+b)

Vector multiplication assigns the ratio ka 2 + kb 2 + 2 ka*b because the vertical a*b = 0

The original formula is equal to KA2 + KB2

Because m2 = m 2 * cos zero degree = m 2 is the same as above.

m = under the root number (ka+b) 2 = a 2*k 2 +b 2 under the root number in the same way n = a 2 +k 2*b 2 under the root number, m = n = 1+k 2 under the root number, m*n=2k and because m*n = m * n *cos(m,n)2k=(1+k 2)*1 2 solution k=2+root number 3 or 2-root number 3 does not exist integer k satisfies the condition.

The Cartesian coordinate system is established with a and b as unit vectors on the x-axis and y-axis respectively.

tgm=1/k

tgn=ktg(m-n)=(1 k-k) (1+k*1 k)=1 2k-k2=root number 3 or negative root number 3

k = root 3 + root 2, root 3 - root 2 (two roots are rounded).

The Cartesian coordinate system is established with a and b on the x-axis and y-axis respectively.

tgm=1/k

tgn=ktg(m-n)=(1 k-k) (1+k*1 k)=1 2k-k2=root number 3 or negative root number 3

k = root 3 + root 2, root 3 - root 2, experiment k does not exist.

Let c=(x,y), because the modulus of a and b are equal, so a*c=b*c so x+2y=2x+y

And because |c|=1

So x 2 + y 2 = 1

So. x=y=plus or minus 2 root number 2

c = plus or minus (2 out of 2 root number, 2 out of 2 root number).

Because c is a unit vector.

So. c|=1

Vector b=(3,4).

a b so the vector a = (3k, 4k) == See "Vector parallel".

And because |a|=3=root[(3k) 2+(4k) 2]=5|k|k=±3/5

Vector a = (9 5, 12 5) or = (-9 5, -12 5) thank you

Solution: Let the vector mp = vector mn , then.

Vector om = 3 8 vector ob , vector on = n vector oa vector op = vector om + vector mp =3 8 (1- ) vector ob + n vector oa

Vector oc = 2 vector op

Vector op = 1 4 ( vector ob + vector oa ) 3 8 (1- ) = 1 4 , n = 1 4 = 1 3 , n = 3 4

1.Let op=ai

The modulo of PA+PB=OA-OP+OB-OP=(4-2A)I+J PA+PB is (4(A-2) 2+1 under the root number, and the minimum value is 1 P(2,0).

bp=(a-3)i+j

ap·bp=(a-2)^2-3

Minimum when a=2.

cos angle apb = -3 root number 10

Angular apb=-arccos3 root number 10

1) Seek |Vector pb|The minimum value is 1. The corresponding p-point coordinates are (2,0).

2) Find the vector AP and vector BP, the minimum is -3 cos, the angle APB = -3, 10 10

Because: a|=3，|b|=4

a+b) (a+3b)=a 2+4ab+3b 2 to bring in a|=3，|b|=4

9+4ab+48=33

4ab=—24

ab=—6 "The above ab is a vector".

cosc=|a|Multiply by |b|Divide by |a|multiply the modulo by |b|of the mold.

So cosc=—

That is, c (the angle between a and b) is 120°

The symbols are too hard to hit, hehe.

The answer upstairs is correct, doing vector problems pay attention to the algorithm, and pay attention to the direction, it's not too difficult, but it's very useful.

ka+b/=√3/a-kb/

The two sides are squared, where a 2 = 1 and b 2 = 1

a·b=(k 2+1) (4k).

k 2 + 1) (4k)=k 4 + 1 (4k) Since k >0, [k 4] *1 (4k)] = 1 16, if and only if k 4 = 1 (4k), i.e., k = 1, a·b is the smallest, which is 1 2a·b=|a||b|cosθ=1/2

Solution = 60°

If you want to prove that the points are in the same plane, you only need to prove that the vectors containing the points are in the same plane. If the proof vectors are in the same plane, it is only necessary to prove that one vector can be represented by the sum of the other vectors.

Proof: ab = (3, 4, 5), ac = (1, 2, 2), cd = (8, 12, 14).

Let AAB+BAC=CD

then 3a+b=8

4a+2b=12

5a+2b=14

It can be solved a=2 and b=2

The AB vector, AC vector, and CD vector are in the same plane.

So a, b, c, d are all in the same plane.