High one vector addition operation It is known that △ ABC is a right triangle, ∠ BAC = 90 ° and ad ⊥ BC is equal to d Prove: | vector BC | ^ 2 = | vector DB + vector Da | ^ 2 + | vector DC + vector Da | ^ 2

High one vector addition operation It is known that △ ABC is a right triangle, ∠ BAC = 90 ° and ad ⊥ BC is equal to d Prove: | vector BC | ^ 2 = | vector DB + vector Da | ^ 2 + | vector DC + vector Da | ^ 2

Because ad ⊥ BC
|Vector DB + vector Da | ^ 2 + | vector DC + vector Da | ^ 2
=IBDI^2+IDAI^2+2IBDIIDAICOSπ/2+IDCI^2+IDAI^2+2IDCIIDAICOSπ/2
=IBDI^2+IDAI^2+IDCI^2+IDAI^2
=IABI^2+IACI^2
=IBCI^2

Given that E and F are the midpoint of AD and BC respectively in any quadrilateral ABCD, we prove that the vector EF = 1 / 2 (vector AB + vector DC)

EF=EA+AB+BF
EF=ED+DC+CF
Where EA and ED modules are equal and opposite, so EA + ed = vector 0
BF and CF are the same, BF + CF = vector 0
Add the above two expressions,
2EF=AB+CD
EF=1/2(AB+CD)
Note: all letters denote vectors

High one vector addition
If the speed of a ship in still water is 6km / h and the current speed is 3km / h, which direction must it go in order to ensure the ship to move in the direction perpendicular to the current? What is the actual speed of the ship?
The actual speed is 3km / h

It's time to draw a picture
First, draw a right triangle. Let the long one be the vector AB, which is the vertical direction. Let the short one be the vector CB, which is the direction of the current. Let the vector AC be the direction of the ship
Absolute value vector AB = radical 36 + 9 = 3 radical 3
So the angle BCA is 60 degrees