3D trigo - Ans.fyi 參考答案

3D trigo

[匿名]

2014-12-04 1:38 am

Please look at the picture below.
For part (c), according to the solution, it let P be a point lying on BC such that NP is perpendicular to CB. Then the required angle is angle VPN.

My question is why in this case, VP must be perpendicular to BC.

My teacher said we can use M2 vector to prove, but only 3 marks, no need to do so. And, I did not study M2.

So, can anyone explain? Thanks.

https://farm8.staticflickr.com/7540/15750867817_831fbe 1c 80_o.jpg

圖片參考：https://s.yimg.com/lo/api/res/1.2/yosZi4IgHEde1Rl0heMqeg--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/https://farm8.staticflickr.com/7540/15750867817_831fbe1c80_o.jpg

更新1:

Re wy: You mean "therefore VP (NOT NP) is also perpendicular to BC. "?

更新2:

only NP is selected to be perpendicular to CB... but not VP??

回答 (1)

用戶10012

2014-12-04 3:41 pm

✔ 最佳答案

Since P is the point on BC such that VP is perpendicular to BC, notice that NP will be the projection of VP on plane ABC or NP will be the "shadow" of VP on ABC, therefore NP is also perpendicular to BC. Also notice that ANP is NOT a straight line, AN produce will meet BC at another point and it will not be perpendicular to BC.
(c)
Assume you calculated CM ( let it be k) and x, so NM is known ( = a)
CN = CM - NM = k - a
Angle MCB = 30 degree (equilateral triangle)
NP/CN = sin 30 = 1/2
so NP = CN/2 = (k - a)/2
So angle between plane = arctan (VN/NP) = arctan ( 2x/(k - a)].

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