Vector Calculations in Mathematical Plane Geometry! High school one

Solution: (1).

bc vector = ac-ab = (-1, k-3).

If b is 90 degrees, -2+3k-9=0, k=11 3 If c is 90 degrees, -1+k 2-3k=0, k=(3+or-root number 13) 2 If a=90 degrees, k=-2 3

ab=(b-a,a-b)

Illustrate. 1. Prove x, y, z three points colinlinear, prove that the angle xyz = 180 °2, prove x, y, z three points collinear, choose a straight line pq over y, prove angle xyq = angle pyz3, prove x, y, z three points collinear, choose a ray xp through x, prove angle pxy = angle pxz4, prove x, y, z three-point collinear, prove xy + yz = xz5, x, y, z three-point collinear, prove xy, xz are parallel or perpendicular to a certain line 6, use the angle formula.

7. The inverse theorem using Menelaus' theorem.

8. Prove that x, y, z are three points collinear, prove that the area of "triangle" xyz is 09, and prove that one of the points is on the straight line determined by the other two points.

10. Use the same method.

bc=(c-b,b-c) means ab=(a-b) (b-c)bc means ab,bc are in the same direction.

And because AB and BC pass point B at the same time.

It can be obtained: ab and bc are the same straight line, i.e., a, b, c three points are collinear.

If the angle a=90 degrees,Then 2+3k=0 k=3/2 I didn't have a pen in the Internet café, so I didn't forget it.。。。 The collinear can be used in the textbook of your textbook. You can refer to my discussion one by one, such as calculating the bc vector to discuss the b angle.

C corner. It's the calculation of the 2th function... I'm old, and I haven't done a high school math problem in a long time.

Hello! The answer is:

The BAI diagram shows:

The answer du is 6x + 10y + 7z - 50 = 0 First find the vector zhi formed by the two points, the plane of the coplanar three internal points of dao, the normal volume vector n is the inner product of the two vectors, and then use the point equation to express the plane normal vector.

If there is anything you don't understand, you can ask at any time, I will try my best to answer, I wish you academic progress, thank you.

According to the three points, two vectors are arbitrarily rounded, and then the product of these two vector vectors is found, and the vector is the plane normal vector.

1. The law of triangles 2. The law of parallelograms.

Let a vector = (x1, y1) and b vector = (x2, y2), then: a vector + b vector = (x1 + x2, y1 + y2).

Subtractive triangle rule: let a vector = (x1 + y1) and b vector = (x2, y2), then: a vector + b vector = (x1-x2, y1-y2).

A vector * B vector = B vector * A vector.

Let a vector = (x1, y1) and b vector = (x2, y2), then: a vector + b vector = (x1 + x2, y1 + y2).

Establish. The spatial coordinate system is established with d as the origin.

Let oa=oc=op=2

Then set the f coordinate to (0,y,z).

Because the coordinates of point e are (0,1,1).

The p-point is (0,0,2).

Point b is (2,2,0).

So the EF vector is (0,y-1,z-1).

The pb coordinates are (2,2,-2).

And because of EF pb...

So the p-point coordinates are (0,0,0).

So you can figure it out. Point P coincides with point D.

Set the points a(x1,y1,z1),b(x2,y2,z2) on this plane, then the vector ab (the arrow can't be typed, sorry) = (x1-x2, y1-y2, z1-z2).

Substituting these two points into the plane equation yields a(x1-x2)+b(y1-y2)+c(z1-z2)=0

i.e. (a,b,c)(x1-x2,y1-y2,z1-z2)=0, and this equation is consistent for all points a and b on the plane.

So the normal vectors are (a, b, c).

Let : m(x,y) n(x0,y0).

Vector ma=(1-x,1-y) to the stool early reputation jujube segment an=(x0-1,y0-1).

Vector ma = 2 * vector an (1-x,1-y) = 2*(x0-1,y0-1).

x=2-2x0, y=2-2yo

m (x,y) on round opening type c: (x—3) 2+(y—3) 2=4.

x=2-2x0, y=2-2yo satisfies :(x—3) 2+(y—3) 2=4

Available: (2x0+1) 2+(2y0+1)=4