What's the difference between tangent vector and normal vector For example, the tangent vector n = (1,2,1), what is the normal vector, and the difference between the inner normal and the outer normal,

What's the difference between tangent vector and normal vector For example, the tangent vector n = (1,2,1), what is the normal vector, and the difference between the inner normal and the outer normal,

If it's a space curve, then the points on the curve should have tangent vectors and normal planes. Similarly, if it's a space surface, then there are normal vectors and tangent planes. Tangent vectors and normal vectors will be discussed only on a plane smooth curve

What's the difference between unit normal vector and normal vector?

The unit vector is a vector whose module length is one
The normal vector is a vector perpendicular to a known plane

Xiaolan's room is 3.5 meters long, 3 meters wide and 3 meters high. Except for the 4.5 meters of doors and windows, the walls and roofs of the room are pasted with wallpaper. How much wallpaper does this room need?

3.5 × 3 + 3.5 × 3 × 2 + 3 × 3 × 2-4.5 = 10.5 + 21 + 18-4.5 = 45 square meters; a: this room needs at least 45 square meters of wallpaper

Let f (1.0) e = 1 / 2 be a focal point of the ellipse CX2 / A2 + Y2 / B2 = 1. Let a straight line passing through f intersect the ellipse at M N, and a vertical line in Mn intersects the Y axis at P (0, Y0) to find the Yo range

Obviously, C = 1, a = 2, the elliptic equation x ^ 2 / 4 + y ^ 2 / 3 = 1
Let m (x1, Y1), n (X2, Y2), midpoint K (x0, Y0), and straight line Mn: My + 1 = x be substituted into elliptic equation (slope 1 / M)
3(my+1)^2 +4x^2-12=0
(4+3m^2)y^2 +6my-9 = 0
So Y0 = (Y2 + Y1) / 2 = - 3m / (4 + 3M ^ 2); x0 = my0 + 1 = 4 / (4 + 3M ^ 2)
Middle vertical KP: - M (x-4 / (4 + 3M ^ 2) = (y + 3m / (4 + 3M ^ 2)
X = 0; y = m / (4 + 3M ^ 2) = 1 / (4 / M + 3M) m ∈ R (because no matter what value m takes, it has two intersections with the ellipse)
4/m+3m ∈(-∞,-4√3 ]∪[4√3,+∞)
So y ∈ [- √ 3 / 12, √ 3 / 12]

If the height of a triangle is 30 meters, then the height of a parallelogram is () meters

If the height of a triangle is 30 meters, then the height of a parallelogram is (15) meters

The eccentricity of the ellipse x 2A2 + y 2B2 = 1 (a > b > 0) is 32. The ellipse and the straight line x + 2Y + 8 = 0 intersect at P, Q, and | PQ | = 10. The equation of the ellipse is obtained

From C2 = A2-B2, A2 = 4B2. From x24b2 + y2b2 = 1x + 2Y + 8 = 0, eliminating x, 2Y2 + 8y + 16-b2 = 0. From the relationship between root and coefficient, Y1 + y2 = - 4, y1y2 = 16 − B22. | PQ | 2 = (x2-x1) 2 + (y2-y1) 2 = 5 (y1-y2) 2 = 5 [(Y1 + Y2) 2-4y1y2] = 10

If the perimeter of a rectangle is 4a-b and one side is 2a-b, the other side is 2a-b______ ．

The other side is 12 [(4a-b) - 2 (2a-b)] = 12 (4a-b-4a + 2b) = 12b

As shown in the figure, PA, Pb are the tangent of circle O, AB is the stealing point, ∠ P = 60 °, ab = 4 root sign 3, find the radius of circle o

PA and Pb are circular tangents, so ∠ PbO = ∠ Pao = 90
∠AOB=360-∠PAO-∠PBO-∠P=120
According to the vertical diameter theorem, AC = AB / 2 = 2 √ 3
Therefore, AC: Ao = √ 3:2
AO=4

It is known that, as shown in the figure, in △ MPN, h is the intersection of high MQ and NR, and PQ = HQ, find the degree of: ∠ QMN

It is easy to know from the condition that RMQ = rnq
And in right triangle PQ = HQ
So right triangle PQM is equal to hQN
QM=QN
So the angle QMN = qnm = 45 degrees

If AB = 2cm, BC = 4cm, what is the area of quadrilateral aecf

∵AB=2CM,BC=4CM,∴AC=2√5CM,∴OA=√5CM
∵OF/CD=OA/AD,∴OF=√5/2CM
∴AF=5/2CM
It is also known that EF ⊥ AC, OA = OC, aecf is rhombic
S quadrilateral aecf = AF × AB = 5 / 2 × 2 = 5cm ∧ 2