Senior 2 Mathematics Space Geometry, High School Mathematics Space Geometry

2*sqrt(34), which is the arithmetic square root of 34.

The pre-graduate school math space geometry in high school is:

1.Recognize the points, lines, and surfaces in space.

2.Then it is to understand the relationship and characteristics of points and lines, points and surfaces, lines and lines, lines and planes in space.

3.Regarding the lines in space, it is necessary to understand the relationships and conditions of parallelism, intersection, perpendicular, straight lines on different planes, and the distance between two straight lines. For the lines, planes, and surfaces in space, it is necessary to understand the conditions under which they intersect, parallel, especially perpendicular, and so on.

In high school mathematics, the students in spatial geometry learn to apply the contents of points 1, 2, and 3 mentioned above in spatial geometry.

Finally, I have to learn the analytic geometry of space.

My brother is holding a book all day long, with short hair and a pair of eyes on the bridge of his nose, he is very tall.

My parents took me to my brother's house, and my brother walked out of the study with a book in his hand, and greeted my parents and said, "Uncle, hello aunt." Mom and Dad also nodded again and again:

Good, good. After that, the elder brother returned to the study, sat on the sofa, crossed Erlang's legs, put the book on his lap, and began to read the book with relish again, and the guests were all under the care of the eldest mother. followed my brother into the study, "Wow!

The bookcase was filled with rows of books, and there were several boxes on the floor, which were also full of books. Classical classics, history, prose, **......As long as you can name the title, he will probably have it.

On Sunday, we went to my grandmother's house for dinner, and when he came, he carried a bag in his hand, what was in the bag? It's not going to be a book again. Sure enough, as I expected, it was a book, outside of class, and in class.

As soon as my brother put down the bag, he took out a copy, what "Can mosquito blood really make dinosaurs?" What a novelty book!

After eating, the table was full of big fish and meat, but these dishes did not attract my brother, my brother served the rice, picked up the chopsticks and ate quickly, stirred the food together, and pulled it into his mouth, as if he was afraid that the dishes would be eaten up later. After eating, I put the chopsticks on the table, and I put up my legs to read the book, which has been replaced by another book, called "Curious African Animals and Plants". I said:

Did you finish reading one just now? Potato shouted, "He didn't hear it, he didn't hear the good-looking TV series next to him, grandma gave us fruit, and he didn't notice it."

On the way to the car, he took out a textbook from his bag, leaned on the back of the car seat, and concentrated on the silver hand dust. I thought: Won't my brother have a headache? The car is rocking, and it is no wonder that the eyes are short-sighted. He is really a book lover, and he loves to read books to this extent.

This is my book lover brother, who takes up books every minute and every second.

Perpendicular to each other. pc , ab is included in , pc ab is the same as pd , ba is included in , pd ab and pc pd p are obtained

AB Plane PCD

ab⊥dc

(1) BF plane ace, BF AE can be obtained, because BC plane ABE, BC AE can be obtained, that is to say, AE BF, AE BC, BF and BC intersect and BC moreover, BF and BC are both plane BCE, BF plane BCE can be obtained, from which AE BE can be obtained

2) Do the auxiliary line, find the midpoint of AC, if it is H point, then MH is the median line of ABC, so MH BC, ABCD is rectangular, so MH AD, also use the median line to get NH AE, so that there are two days of intersecting straight lines are parallel to the plane DAE, so the plane MHN plane DAE, so any line in the plane MHN is parallel to the plane DAE, Mn plane MHN, so, MN plane DAE

(1) BC Vertical Plane ABE

BC Vertical AE

and bf perpendicular plane ace

BF Vertical AE

and BC to B

AE Vertical Plane BCE

AE Vertical BE

2) Take the q in the middle of be

mq‖ae , nq‖ad

And mq and nq intersect, and ae and ad intersect.

The two faces are parallel.

then we get MN parallel plane DAE

Connecting fg eh go1 oh eo1 of, it is found that the quadrilaterals eghf eo1fo oho1g are all parallelograms (the proof is exactly the same, they are all a group of opposite sides parallel and equal, for example, go1 is parallel and equal to oh), so eg=hf og=ho1 oe=fo1, the three sides correspond to equal, obviously the two triangles are congruent.

Set the three sides of the box to be equal in length.

Hello. The front, side, and top views are the three views of the geometry, with which the shape of the geometry can be roughly determined, and the segment can be obtained by vector addition of the projection that is decomposed into three faces, then the length is equal to.

a^2+b^2+(√6)^2]

That is, 7 a 2+b 2=1

a>0,b>0

From the mean inequality, obtained.

a^2+b^2≥2ab

a+b)^2=a^2+b^2+2ab=1+2ab≤1+（a^2+b^2)=2

Mostly A+B 2

I think the answer is wrong. Thank you.

Answer 4 is false, and I agree with the view that the answer is root number 2.

If you are strong enough, your spatial thinking is to build a model in your brain, the relationship between the construction line and the line, if you don't have this tool to create a model, for example, the pen like the old world.

e and h are the midpoints of a1b1 and d1c1, respectively.

eh//a1d1，b1c1//a1d1

eh//b1c1

EH is not in the surface BCC1B1.

B1C1 is within the surface Bcc1B1.

EH surface BCC1B1

eh is inside the surface efgh, face efgh surface bcc1b1=fg eh fg

eh//a1d1//ad

fg//ad

fg is not in the add1a1 face.

ad is inside the surface add1a1.

fg surface add1a1

2 (1 2) 2 is the root number of the two.

The four sides are all one, and the diagonal is also one, then these four points constitute a regular tetrahedron, the shortest distance between AB and CD is the distance between two straight lines on different planes, which can be proved to be the length of the line segment between the midpoint of AB and the midpoint of CD.

Hint, ABCD is a regular tetrahedron, and I am really not in the mood to do anything else.

Sit back and wait for the strongman.

This space quadrilateral and its two diagonals form a regular tetrahedron, and the two moving points are on a bottom edge and its diagonal edge, so their shortest distance is the distance between the midpoints of the two sides, when p and q are the midpoints of ab and cd respectively, you can find cp=dp= 3 2, so the triangle cdp is an isosceles triangle, and pq is both the middle line of the triangle cdp and its height, i.e., pq cd, you can find pq 2 2