There are three ABC points on the plane. Draw a straight line through each two points. You can draw a straight line at most There are three ABC points on the plane. Draw a straight line through each two points. You can draw () lines at most and () lines at least It takes at least () nails to fix the wood to the wall Why? The following statement is correct: there are two kinds of positional relations between () 1, a and line L 2. There are three intersections when three lines intersect

3 at most and 1 at least

Five points ABCDE and ABC are on the same line. One of the three points is not on the same line. How many different lines can be drawn when two points draw a line
If it is allowed to change the position of five points arbitrarily, how many lines can be drawn at least and how many at most

First of all, any two points can draw C (5,2) = 10 lines, because ABC is on the same line, so there are two less lines, and because any other three points are not collinear, so there are no more duplicate lines. Therefore, there are 10-2 = 8 lines
If the position is arbitrarily changed, referring to the above analysis, it is obvious that C (5,2) = 10 at most and 1 at least (at this time, 5 points are collinear)

How many lines can be drawn through three points which are not on the same line in the same plane?

There are countless words after one point, three words after two points, and no words after three

If the distance from three points a, B and C not on the same straight line to plane a is equal, and ABC does not belong to plane a, then the position relationship between plane ABC and plane a

ABC is an equilateral triangle and a is in the middle of the triangle

Given that the radius of the ball o is 1, a, B and C are all on the spherical surface, and the spherical distance between each two points is π 2, then the distance from the spherical center O to the plane ABC is ()
A. 13B. 33C. 23D. 63

Obviously, OA, OB and OC are perpendicular. As shown in the figure, let O1 be the center of the circle obtained by intercepting the ball on the plane where ABC is located, ∵ OA = ob = OC = 1, and OA ⊥ ob ⊥ OC, ≁ AB = BC = CA = 2.. O1 is the center of △ ABC.. O1A = 63. From oo12 + o1a2 = oa2, oo1 = 33 can be obtained

There are three points a, B and C on the sphere, ab = 18bc = 24AC = 30, and the distance from the center of the sphere to the plane ABC is 1 / 2 of the radius of the sphere
Well, the radius of the ball is
The known answer is ten times three,

From the relationship of the three sides, we can see that the triangle ABC is a right triangle. The center of its circumscribed circle is the midpoint of the hypotenuse AC. connect the midpoint of AC with the center of the ball, and then connect the center of the ball with point A. suppose that the center of the ball is point O and the midpoint of AC is point D, then the triangle ODA is a right triangle

The distance from a point I to three times in the triangle ABC is 3cm, and the lengths of three sides a, B and C are 2.1cm, 3.7cm and 4.2cm respectively

The area of a triangle is equal to the sum of the areas of IAB, IBC and ICA,
That is s = (2.1 + 3.7 + 4.2) × 3 / 2 = 15

Let a (2,0,1), B (0,1,2), C (2,1,4), D (- 1,2,6), then what is the distance from D to plane ABC

According to the coordinates of a, B and C, AB, AC and BC can be obtained
Find the normal vector n of the plane
Then find the projection of Da, dB or DC on the normal vector, which is the distance from D to plane ABC
The rest is the calculation process
I believe you can handle it

In △ ABC, BC = 21, ∠ BAC = 120 ° and the distance from a point p outside the plane of △ ABC to a, B, C and a is 14, then the distance from P to plane ABC is?

The distance from P to a, B, C and C is 14
So the projection of P on the bottom is the outer center of the bottom
Sine theorem a / Sina = B / SINB = C / sinc = 2R (where R is the radius of triangle circumcircle)
21/sin120°=2R
R=21/√3=7√3
The distance from P to plane ABC is h
Pythagorean theorem has 14 & # 178; = H & # 178; + R & # 178;
h=7……
I don't know what to say or ask

As shown in the figure △ ABC in the plane, find the point with equal distance from the three sides of the line

(1) Take B as the center of the circle, draw a circle with any length as the radius, intersect AB and BC at two points D and e respectively, (2) then draw a circle with D and E as the center of the circle, and draw a circle with a radius greater than 12de, the two circles intersect at f and connect BF, then BF is the bisector of ∠ B; similarly, make the bisector of ∠ a, and the two bisectors intersect at point G1, then point G1 is the bisector of ∠ B

Five points ABCDE and ABC are on the same line. One of the three points is not on the same line. How many different lines can be drawn when two points draw a line If it is allowed to change the position of five points arbitrarily, how many lines can be drawn at least and how many at most