P is the point outside the plane of the RT triangle ABC. The distance from P to the right angle vertex C is 24 cm, and the distance to the two right angle sides is 6 √ 10, Then the distance from P to plane ABC?

Let PD ⊥ AC in D, PE ⊥ BC in E, Po ⊥ plane, ABC in O. connect PC, OD, OE, OC. Then PC = 24, PD = = PE = 6 √ 10. From the three perpendicular theorem, we can get OD ⊥ AC, OE ⊥ BC

If the distance from a point P out of the plane where RT △ ABC is located to the vertex of the right angle is 24cm, and the distance to the two right sides is 6 √ 10cm, then what is the distance between P and plane ABC?

The process of solving this question is as follows:

P is the point outside the plane of the RT triangle ABC. The distance from P to the right angle vertex C is 24 cm, and the distance to the two right angle sides is 6 √ 10. Find the angle between PC and the plane

Make PD ⊥ AC in D, PE ⊥ BC in E, Po ⊥ plane, ABC in O. connect PC, OD, OE, OC. Then PC = 24, PD = = PE = 6 √ 10, ∠ PCO is the angle formed by PC and plane
Od ⊥ AC, OE ⊥ BC can be obtained from the three perpendicular theorem
It is easy to prove that △ PDO ≌ △ PEO, so od = OE, so OC bisection ∠ ACB, ∠ BCO = 45 °
In RT △ PEC, from Pythagorean theorem, we can get CE = ＝6√6
In RT △ Coe, OC = CE / cos ∠ OCE = (6 √ 6) / (√ 2 / 2) = 12 √ 3
In RT △ POC, cos ∠ PCO = OC / PC = (12 √ 3) / 24 = √ 3 / 2
So ∠ PCO = 30 ° is what we want
Please check the figures

In △ ABC, ab = 9, AC = 15, ∠ BAC = 120 ° and the distance from a point p outside the plane to the three vertices of △ ABC is 14, then the distance from point P to plane ABC is:______ .

Analysis: note that the projection of P on plane ABC is O, ∵ PA = Pb = PC
/ / OA = ob = OC, that is, O is the outer center of △ ABC, and only OA (radius of circumscribed circle of △ ABC) is required,
In △ ABC, it is known by cosine theorem that:
BC = 21, according to the sine theorem, 2R = 21
sin120°=14
3，∴OA=7
3, Po = 7
So the answer is: 7

In the triangle ABC, ab = 9ac = 5, angle BAC = 120, the distance from the point P out of the plane to the three vertices is 14P, and the distance from the plane ABC is

The distance from point P out of plane to plane ABC, then the projection of point P in plane ABC is
Let the radius be r, then AQ = 2R / radical 3
The inner Q point of triangle ABC is a point on the bisector of angle BAC
I'll just mention that it's not easy to say that there's no diagram. We'll make QE vertical AB through Q point
Triangle AQP, triangle AQE, triangle AEP are right triangle
The Pythagorean theorem can be used
If you don't know, I'll be more specific next time

If O is any point in the plane and satisfies (OB + oc-2oa). (AB-AC) = O., then what triangle ABC must be? OA, ob, OC are vectors

Because (OB + oc-2oa) * (AB-AC)
＝[(OB-OA)+(OC-OA)]*(AB-AC）
=(AB+AC)*(AB-AC）
=|AB|²-|AC|²
=0
So | ab | = | AC |
That is, A.C|
So the triangle ABC must be an isosceles triangle

O is a moving point in the plane where the triangle ABC is located, connecting OB and OC, and connecting the midpoint D, e, F and g of AB, ob, OC and AC in turn If defg is a quadrilateral, what is the reason for the point

If the quadrilateral defg is a rectangle, the point O should pass through point a and be perpendicular to the straight line BC. From (1), we can see that defg is a parallelogram

If OA * ob = ob * OC = OC * OA, then what is the center of the triangle ABC? Proof process (all OA above are vectors)

OA *OB=OB*OC
0 = ob * (oa-oc) = ob * Ca, ob ⊥ CA is the same as OA ⊥ BC OC ⊥ ab
O is the vertical center of ⊿ ABC
Please note that this can get a triangle three high intersection of a vector proof method
(from OA ⊥ BC, ob ⊥ AC, OC ⊥ AB!)

In the RT triangle ABC, the angle c = 90 degrees, AC = 15, BC = 20, CD vertical plane ABC, CD = 5 Hope there is a process, easy to understand

Through point d to AB, make a vertical intersection with point E; connect CE; because CD is perpendicular to AB; De is perpendicular to AB, CE is perpendicular to AB; ab = 25; ab * CB = CE * AB is CE = 12; from RT triangle DCE, de = 13; that is, the distance from D to AB is 13

In the right triangle ABC, ∠ ACB = 90 °, CD is the height of AB side, ab = 13cm, BC = 12cm, AC = 5cm, find; 1. The area of △ ABC; 2. The length of CD I'll give you 100

The area of △ ABC = BC × AC = 125 △ 2 = 30cm
CD = △ ABC area × 2 △ AB = 30 × 2 △ 13 = 60 / 13 (CM)

P is the point outside the plane of the RT triangle ABC. The distance from P to the right angle vertex C is 24 cm, and the distance to the two right angle sides is 6 √ 10, Then the distance from P to plane ABC?