Set by the question: vector AB+ vector BC+ vector CD+ vector DA=0 (vector).
Vector DA=- (vector AB+ vector BC+ vector CD), and the word "vector" is omitted below.
That is DA=-(AB+BC+CD).
|DA|^2=[-(AB+BC+CD)]^2.
=(ab^2+bc^2+cd^2+2ab*bc+2bc*cd+2cd*ab)
【|ab|^2+|bc|^2+|cd|^2+2|ab||bc|cos<； AB，BC & gt+2 | BC | CD | cos & lt； BC，CD & gt+2 | CD | | AB | cos & lt； CD，AB & gt。
∵|AB|=200，|AB|^2=40000，； |BC|= 100√2，|bc|^2=20000； |CD|=50√2，|CD|^2=5000。
Vector included angle:
& ltBC，CD " =(90-60)= 30；
& ltCD，AB & gt=Θ .
|da|^2=40000+20000+5000+2*200* 100√2cos 135+2 * 100√2 * 50√2 * cos 30+2 * 50√2 * 200 cosθ。 = 65000-40000+ 10000√3+20000√2 cosθ
= 25000+ 17320+28284 cosθ( 1)
According to the equation (1), |DA| and θ are both unknowns, and one of them needs to be solved by other methods.
According to the topic, A and C are on the positive semi-axis of Y coordinate, and ABC forms an isosceles right triangle.
In the triangle Rt△ABC, ∠ C = 90, ∠ ABC = ∠ BAC = 45.
∴| AC | = | BC | = 100 √ 2, | CD | = 50 √ 2, and the included angle between CD and AC ∠ ACD = 60.
DE is perpendicular to AC, and the vertical foot is E.
In Rt△CED, ce = | CD | cos60. de = | CD | sin60。
∴CE=( 1/2)|CD|=25√2。
DE=50√2*√3/2=25√6。
AE=AC-CE= 100√-2-25√2。
=75√2.
At Rt△AED. |DA|^2=AE^2+DE^2.
|DA|^2=(75√2)^2+(25√6)^2.
= 15000.
By simplifying generation | da | 2 (1), we get:
cosθ=-27.32/28.284
θ=π-arccos(0.9659)= 180- 15
∴Θ≈ 165 ,
A: The plane takes off from point A, passes through B, C and D, and flies back to point A from D with a displacement of 122.47 km, and the angle between the direction and the original takeoff direction is about 165.