As shown in the figure, in the tetrahedral aboc, OC ⊥ OA, OC ⊥ ob, AOB = 120 degrees, OA = ob = OC = 1, let p be the midpoint of AC, Q be on AB, and ab = 3aq, prove PQ ⊥ OA How do you get the point vector o (0,0,0), a (1,0,0), B (- 1 / 2, √ 3 / 2,0), C (0,0,1) B

As shown in the figure, in the tetrahedral aboc, OC ⊥ OA, OC ⊥ ob, AOB = 120 degrees, OA = ob = OC = 1, let p be the midpoint of AC, Q be on AB, and ab = 3aq, prove PQ ⊥ OA How do you get the point vector o (0,0,0), a (1,0,0), B (- 1 / 2, √ 3 / 2,0), C (0,0,1) B

Let B (x, y, z), OC be perpendicular to ob, so XYZ is multiplied by 001 = 0 to get z = 0
AB = ob + OA - 2 times ob times OA times cos120 We get 3x = y, and X must be negative
OB = 1, x + y = 1, just solve the equation

Ask a high school geometry math problem
How long is the diagonal of a cuboid if the area of three sides of the same vertex of the cuboid is root 3, root 5 and root 15 respectively?
Please try your best to write out the detailed process of problem solving. Thank you very much

Let trigonometric length a, B, C
AB = root 3
AC = root 5
BC = root 15
After multiplication, we get ABC = root 15,
So a = 1
B = root 3
C = root 5
Diagonal length = 1 + 3 + 5 under radical = 3

Online, etc. fast. Mathematical problems. Geometry. Proof
Points E and D are on BC, BD = CE, ∠ 1 = 2, ad = AE
There are pictures
___________________ A
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___ B_______ E ____________ D _______ C
The underline in the picture is not in the title.
You can copy it elsewhere and remove the underline. Look at the picture again [except for the three sides of the triangle. 】
Note: at present, it belongs to the category of congruent triangle.

In △ abd and △ ace,
Ad = AE (known)
∠ 1 = ∠ 2 (known)
BD = Ce (known)
∴△ABD≌△ACE（SAS）
‖ AB = AC (corresponding edges are equal)

As shown in the figure, in △ ABC, ad bisects ∠ BAC, ad = AB, cm ⊥ ad intersects ad extension line at point M. verification: am = 12 (AB + AC)

It is proved that: lengthen am to N, make DM = Mn, connect CN, ∵ cm ⊥ ad, DM = Mn, ∵ CN = CD, ∵ CDN = DNC, ∵ DNC = DAB, ∵ ad = AB, ∵ B = ADB, ∵ B = ANC, ? bad = CAD, ? ADB = ACN, ∵ ANC = ACN, ? an = AC, ? AB + AC = AD + an = AD + am + Mn = AD + am + DM = 2am, ∵ am = 12 (AB + AC)