As shown in the figure, it is known that the edges OA, OB and OC of o-abc are perpendicular, and OA = 1, OB = OC = 2, e is the midpoint of OC Find the sine value of the angle between the line be and the plane ABC. ② find the distance between the point E and the plane ABC

As shown in the figure, it is known that the edges OA, OB and OC of o-abc are perpendicular, and OA = 1, OB = OC = 2, e is the midpoint of OC Find the sine value of the angle between the line be and the plane ABC. ② find the distance between the point E and the plane ABC

Let the distance between E and plane ABC be d,
S△OBC=2*2/2=2,
S△BEC=S△OBC/2=1,
OA ⊥ plane BEC,
VA-BEC=S△BEC*AO/3=1/3,
AC=√5,AB=√5,BC=2√2,
Take the midpoint m of BC, BM = √ 2
AM=√（5-2）=√3,
S△ABC=BC*AM/2=√3*2√2/2=√6,
VE-ABC=S△ABC*d/3=√6d/3,
VA-BEC=VE-ABC,
√6d/3=1/3,
d=√6/6,
BE=√5,
Let be and plane ABC form an angle,
∴sinφ=d/BE=(√6/6)/√5=√30/30.
2. The distance between point E and plane ABC is d = √ 6 / 6
Analytic geometry solution in space
Take o as the origin and OA, OB and OC as the X, y and X axes to establish the null
A rectangular coordinate system between the two,
A（1,0,0）,B（0,2,0）,C（0,0,2）,
The plane ABC equation is: X / 1 + Y / 2 + Z / 2 = 1,
2x+y+z-2=0,
The coordinates of point e are (0,0,1),
Vector be = (0, - 2,1),
The linear equation is: X / 0 = (Y-2) / (- 2) = (Z-1) / 1,
sinφ=|0-2+1|/（√（4+1）*√（4+1+1）=√30/30.
According to the space point line distance formula,
The distance between E and plane ABC is d = | 0 * 2 + 0 * 1 + 1 * 1-2 | / √ (4 + 1 + 1) = √ 6 / 6,

In the triangular pyramid o-abc, the three edges OA, OB and OC are perpendicular to each other
OA = ob = OC, M is the midpoint of AB side, then the tangent of the angle between OM and plane ABC

According to the condition, △ ABC is an equilateral triangle, M is the midpoint of AB, so om ⊥ AB, cm ⊥ AB, and the projection of O on plane ABC, O 'is on cm, so ∠ OMC is the angle between OM and plane ABC. Let OA = ob = OC = a, then AB = √ 2a, OM = 1 / 2Ab = √ 2A / 2, CM = √ 3 / 2 * (√ 2a) = √ 6A / 2, so o'm = 1 / 3cm = √ 6A / 6

It is known that the side edges OA, OB and OC of the triangular pyramid o-abc are perpendicular, and OA = 1, OB = OC = 2, e is the midpoint of OC, and the distance from C to Abe is calculated

You first draw a picture and use the method of volume conversion (triangular pyramid e-abc)
If the area of Abe is 1.5, the area of BEC is 1 (AO is perpendicular to the plane BEC)
Let the distance be h, then s △ Abe * H = AO * s △ bec
It should be equal to 2 / 3. Check it by yourself. The method is right