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Given the rectangle ABCD-A' B' C' D', simplify the following vector expressions and mark out the vector of the simplified result. {1} Small arrow above AB + small arrow above BB'; {2} AB plus small arrow + BC plus small arrow + BB' plus small arrow. Given the rectangle ABCD-A' B' C' D', the following vector expression is simplified and the vector of the simplified result is marked. {1} Small arrow above AB + small arrow above BB'; {2} AB plus small arrow + BC plus small arrow + BB' plus small arrow.
1) AB + BB'= AB'(2) AB + BC + BB'= AB'+ BC = AB'+ B' C'= AC'
Find the straight line and plane angle process. Judge my process what is wrong! First, set up the system, mark the coordinates of each point, second, find out the projection of the straight line AB on the plane, set it as AC third, use the formula, find the angle between the vector AB and the vector AC, this angle is the angle between the straight line AB and the plane! Look at my question. Why do the two methods work out different answers? Find the straight line and plane angle process. Judge my process what is wrong! First, establish the system, mark the coordinates of each point, second, find out the projection of the straight line AB on the plane, and set it as AC third, use the formula to find the included angle between the vector AB and the vector AC, and this included angle is the included angle between the straight line AB and the plane! Look at my question. Why do the two methods work out different answers?
If use space vector word, need not look for photography at all, look for projective also line, but the method is non-mainstream!
The non-mainstream method will lead to the non-mainstream error, after the establishment of the coordinate system, in the search for projective, analytic geometry method also have to prove a paragraph, projective is who? The method is right, the details will be wrong, look for yourself well ~ projective is not so easy to find, find also have to prove, in addition, the angle of vector is not line angle, the line angle is less than 90°, also have to turn, vector angle less than 180!
First, establish the system, mark the coordinates of each point, and second, calculate the normal vector AB of the plane, and find the direction vector MN of a straight line,
Thirdly, the angle a between vector AB and vector MN is obtained by using the formula,
Then, when a is less than 90°,90°-a is the line angle
When a is greater than 90°, a-90° is the line angle,
I.e.90°-a absolute
I. In the known cube ABCD—A1B1C1D1, E and F are the midpoint of D1D and BD respectively, G is on the edge CD, and CG=1/3GD, H is the midpoint of C1G. 1. Verification: EF⊥B1C; 2. Calculate the cosine value of the angle formed by EF and C1G; 3. Find the length of FH Second, in the rectangular parallelepiped ABCD-A1B1C1D1 with an edge length of a, is there a point P on the edge DD1 to make the B1D⊥ plane PAC? 3. In the quadrangular pyramid A-BCDE, the bottom surface BCDE is rectangular, the side surface ABC is perpendicular to the bottom surface BCDE, BC=2, CD=2, AB=AC. 1. Prove AD⊥CE 2. Let the side surface ABC be equilateral triangle and the dihedral angle C-AD-E. It is best to use space vectors.
0
If A, B∈a, C, D∈b, AD⊥b, BC⊥b, AB=2, CD=1, then the angle between a and b is?
0
How to judge whether the "speed" in the title is a vector or just a size? For example, a choice of questions say A speed of 5m/s, B 10m/s, two people may be in the same direction may be in the opposite direction, but in the big question said uniform deceleration straight line motion start speed is 10m/s, a few seconds later speed is 5m/s, seek acceleration why 5 and 10 must be in the same direction? Five can't be five? How to judge whether the "speed" in the title is a vector or just a size? For example, a choice of questions said that A speed 5m/s, B 10m/s, the two people may be in the same direction may be in the opposite direction, but in the big question said that uniform deceleration linear motion start speed 10m/s, a few seconds later speed 5m/s, find the acceleration why 5 and 10 must be in the same direction? Five can't be five?
1. In high school physics, velocity is a vector. Unless the topic only discusses the magnitude of velocity, don't say the direction of velocity.2. Choose one topic: Velocity 5m/s, B10m/s, both may be in the same direction and may be in the opposite direction.
1. In high school physics, velocity is a vector. Unless the topic only discusses the magnitude of velocity, don't say the direction of velocity.2. Choose one topic. Speed A is 5m/s, B is 10m/s. Both may be in the same direction and may be in the opposite direction.
Space vector fundamental theorem If any point O of a given space and three points A.B.C which are not collinear satisfy OP=xOA+yOB+zOC (x.y.z∈R), then the necessary and sufficient condition for "point P is in the plane ABC" is "x+y+z=1". Space vector fundamental theorem Given any point O of a space and a non-collinear three point A.B.C, satisfying OP=xOA+yOB+zOC (x.y.z∈R), then the necessary and sufficient condition for "point P is in the plane ABC" is "x+y+z=1".
This problem is not accurate enough to describe the basic theorem of space vector, and it is suggested to modify as follows:
Given any point O of a space and three points A.B.C which are not collinear, then point P is located in the plane ABC if and only if there exists x.y.z∈R and satisfies x+y+z=1 such that OP=xOA+yOB+zOC.
Certification:(Sufficiency)
X+y+z=1
Z =1-x-y
OP=xOA+yOB+zOC
OP=xOA+yOB+(1-x-y) OC
OP=x (OA-OC)+y (OB-OC)+OC
OP-OC = x (OA-OC)+ y (OB-OC)
CP=xCA+yCB
If the conditions A, B and C are not collinear, then CA and CB are not collinear vectors.
According to the fundamental theorem of the plane vector, the point P lies in the plane ABC
Sufficiency holds
(Necessity)
Point P is in plane ABC
If the conditions A, B and C are not collinear, then CA and CB are not collinear vectors.
According to the fundamental theorem of plane vector, there exists a real number x, y such that
CP=xCA+yCB
OP-OC = x (OA-OC)+ y (OB-OC)
OP=x (OA-OC)+y (OB-OC)+OC
OP=xOA+yOB+(1-x-y) OC
Let z=1-x-y
Then x+y+z=1 and OP=xOA+yOB+zOC
That is, there is a real number x, y, z satisfying x+y+z=1 such that OP=xOA+yOB+zOC
Necessity established
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