public String removeType(String sou){ String result=sou; if( sou.indexOf ("^^")>0){ result= sou.substring (0, sou.indexOf ("^^")); } return result; }

public String removeType(String sou){ String result=sou; if( sou.indexOf ("^^")>0){ result= sou.substring (0, sou.indexOf ("^^")); } return result; }

if( sou.indexOf ("^ ^") > 0) ---- -- that is to say, return the position where the string sou first appears ^ ^: for example:
"Sfjkd ^ ^ FJK". Indexof ("^ ^") is 5. If there is no matching such as "sfjkdfjk". Indexof ("^ ^"), it will return - 1. This condition does not hold and cannot be executed below;
String result = sou; -- "is to assign a formal parameter to an actual parameter result;
result= sou.substring (0, sou.indexOf ("^ ^"); -- "is that when the condition is established, the string of Sou intercepts characters from 0 to N and assigns the resulting string to result;
Finally, it returns a value

Given the vector a = (K2 + K-3) I + (1-k) J, B = - 3I + (k-1) J, if the vector a is parallel to B, then K=______ ．

∵ a ∥ B ∥ there exists λ∈ r such that [(K2 + K − 3) I + (1 − K) J] = λ [− 3I + (K − 1) J] ∥ K2 + K − 3 = − 3 λ ① 1 − k = λ (K − 1) ② gets k = 1 from ② or λ = - 1 replaces ① gets k = 1, 2, - 3, so the answer is: 1, 2, - 3

In the triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively. If the vector AB multiplied by the vector AC = the vector Ba multiplied by the vector BC = k, K belongs to R
1. Judge the shape of triangle ABC
2. If C = √ 2, find K

(1) According to the meaning of the question, we can get: C * b * cosa = C * a * CoSb, then we can get b * cosa = a * CoSb, and B / SINB = A / Sina (sine theorem). We can get along with the two expressions, eliminate B and a, and get: cosa * SINB = Sina * CoSb, transfer term to get: cosa * SINB Sina * CoSb = 0, that is: cos (a + b) = 0, get a + B = 90 degrees, that is

Given △ ABC, vector AB = a, vector AC = B, for any point o on the plane ABC, the moving point P satisfies the vector OP = vector OA + X (a + b), X is greater than or equal to zero, ask whether the trajectory of the moving point P passes a certain point
The answer is the center of gravity. Since it's a moving point, it's a fixed point

When any point O in the plane ABC is fixed, for different values of X, point P will be in different positions. Therefore, point P is a moving point, and there is a track in the plane with the change of X. but for different points o, the track is generally different. Now the question is about the