Recommend more difficult vector questions, questions about vectors

1. It is known that the 3 points a b c on the plane satisfy the modulus of vector ab = 3 the modulus of vector bc = 4 and the modulus of vector ca = 5

Find Vector AB·Vector BC+ Vector BC·Vector CA+ Vector CA·Vector AB=?

Answer: Obviously, the triangle ABC should be a right triangle with b as the right-angled vertex.

So the cosine of the angle between the vector ab and the vector bc is 0

The cosine of the angle between the vector ca and the vector bc is -4 5

The cosine of the angle between the vector ab and the vector ca is -3 5

So vector ab · vector bc + vector bc · vector ca + vector ca · vector ab = 3 * 4 * 0 + 4 * 5 * (-4 5) + 3 * 5 * (-3 5) = -25

2. The vector a=(2,-3), vector a is perpendicular to vector b, and b is not equal to 0, then the unit vector of b satisfying the condition is

Answer: Let b(x,y) a be perpendicular to b, then a*b=0.Get: 2x-3y=0

To make b a unit vector: x 2 + y 2 = 1

then x=3 sqrt(13) y=2 sqrt(13)

3. Knowing the points a(1,1),b(-1,5) and a = 1 2a b, a d = 2a b, a e = -1 2a b, find the coordinates of points c, d, e.

Answer: According to the problem, the vector ab=-2,4 Let the coordinates of point c be x, and y then the vector ac=x-1, y-1

And the vector ac=1 2ab, i.e., x-1, y-1=-1, 2 i.e., x-1=-1, y-1=2 is solved to x=0y=3

So the coordinates of c are 0,3

In the same way, the coordinates of d and e can also be found.

Is this a recommended so-called difficult vector problem? Generally.

1.Knowing that a and b are both unit vectors, and their angles are 60 degrees, then |a+3b|Amount?

Solution: Let a=(1,0); b=(cos60°，sin60°)=1/2，√3/2)

Therefore, a+3b=(1+3 2,0+3 3 2)=(5 2,3 3 2).

a+3b︱=√5/2)²+3√3/2)²]25/4+27/4)=√52/4)=√13

2。Known vectors a=(1,2),a b=5,|a-b|=2 5, then |b|Amount? Wang Zhishen.

Solution: Let b = (x, y), then a b = x + 2y = 5....1)；

a-b=(1-x，2-y)，|a-b|=√1-x)²+2-y)²]2√5；

There is (1-x) +2-y) =20....2)

From (1) we get x=5-2y, and substituting (2) gives (2y-4) +2-y) =4y -16y+16+4-4y+y =5y -20y+20=20

Therefore, 5y -20y = 5y(y-4) = 0, so y = 0, y = 4;Thus we get x = 5, x = -3

Therefore, b=(5,0) or b=(-3,4). So we get b =5

3.Let the plane vector a=(1,2),b=(-2,y), and if a b, then |3a+b|=？

Solution: a b, -2 1=y 2, i.e. y=-4, so b=(-2,-4).

So 3a+b=(3-2,6-4)=(1,2); 3a+b = 1 +2 loss) = 5

The mold caftan is 1

The vector of is a unit vector.

a=2e1+e2

The disadvantage is b=-3e1+2e2

Or Qingze. a·b

e1=e2e1^2=4,e1e2=2x2xcos60=2ab=(2e1+e2)(-3e1+2e2)-6e1^2+e1e2+2e2^2

6x4+2+2x4

1. a+b+c = under the root number (a 2 + b 2 + c 2 + 2a · b + 2b · c + 2c · a) =

a-b = (2cos - root number 3, 2sin -1) 2a-b 2 = 8-2 (root number 3cos + sin) = 8-8sin ( + 3).

When sin( + 3)=-1, the maximum value of 2a-b = root number 16=43, let a = (1,-2) b = (1,y).

The included angle is cos = a·b under the root number (5+5y 2) = 1-2y under the root number (5+5y 2) and greater than 0

then 1-2y>0

y<1/2

The format is a little irregular.。。。 Understand.

The sufficient and necessary condition for the perpendicular of the vector is that the inner product of the sail family of the other type is zero, then there are known conditions to get:

a+3b)(7a-5b)=0, 7a 2-15b 2+16ab=0 (1).

a-4b)(7a-2b)=0

7a 2+8b 2-30ab=0 (2) is obtained by (1) (2).

a 2 = b 2 = 2ab, so |a||b|=a^2=b^2=|a|^2=|b|^2=2ab

Then cos=ab |a||b|=ab 2ab = 1 2 pushes out a, the angle of b is 60 degrees.

a+3b)*(7a-5b)=0 7a^2+16ab-15b^2=0a-4b )*7a-2b )=0 7a^2-30ab+8b^2=0 b^2=2ab a^2=2ab a^2=|a|^2

a|=|b| ab=|a||b|cos2|a||b|cos jujube bureau =|a|2 cos rock excavation = 1 open core 2

The angle between A and B is 60 degrees.

From b=c, we can get a point product b = a point product c, but b = c does not prove a point product b = a point product c. So the product of a point b=a point product c is b=c.

The first one (8, 8, 12).

The second seems to be 2 third.