Pre-Calculus Vector Problem. In this question vector i represents a vector due east and vector j represents a vector 1 km due north. 

An aircraft flies (at a constant height) with a speed of $800$ km/h. It flies in a fixed direction from $A$ to $B$ for $30$ minutes, and then, in another fixed direction (diagonally), from $B$ to $C$ for $45$ minutes (which forms a triangle ^_^). 
(a) Find the distances $AB, BC$. 
  The direction from $A$ to $B$ is due east. The direction from $B$ to $C$ is parallel to $24i + 7j$. 
(b) Use this information to find the angle theta that the aircraft turns through at $B$. Give your answer in degrees to the nearest tenth of a degree. 
(c) Find the Angle $ABC$. Hence find the direct distance from a to $C$. 
(d) Another aircraft flies directly from $A$ to $C$. 
(i) Show that vector $AC= 976 i = 168 j$. 
(ii) Find the angle $CAB$. 
(iii) When this aircraft is closest to $B$, how far is it from $A$?

Okay I think I got part A. $AB= 400$ km. $BC= 600$ km. 
I am sure that "the direction from $B$ to $C$ is parallel to $24i+7j$" is supposed to help me solve the rest, but I do not know what it means. 
Could someone explain that to me and how to solve the rest?
 A: If the airplane flies from $A$ to $C$ for 30 minutes, then we can claim that the magnitude of the vector would be $400$ as the speed of the airplane is $800 ~ \rm km/h$.  Therefore, the vector would be $400\hat i$ as the the direction is due east. 
It is a known fact that for a vector, $|\vec F|\hat F = \vec F $ where $\hat F $ is the unit vector in the direction of $\vec F$.  What is the unit vector in the direction of $\vec{BC}$? Since $\vec {BC}$ is parallel to $24\hat i + 7 \hat j $, it is obvious that the unit vector is $\frac{24\hat i + 7 \hat j}{|24\hat i + 7 \hat j |} = \frac{24}{25}\hat i + \frac{7}{25}\hat j$. Also, the magnitude of $\vec {BC}$ is $800 \times \frac{3}{4} = 600$.
Now, using the relation $|\vec F| \hat F = \vec F$, we get $600 \times \left( \frac{24}{25} \hat i + \frac{7}{25}\hat j\right) = 576 \hat i + 168 \hat j $.
Also, note that $\vec {AC} = \vec{AB} + \vec{BC} = 400 \hat i + 576 \hat i + 168 \hat j = 976 \hat i  + 168 \hat j$.
A: If you choose the origin to be at $A$, point $B$ is $(400,0)$.  You are correct that the length of $BC$ is $600$.  $24i+7j$ is a vector from $(0,0)$ to $(24,7)$.  If you move the starting point to $B$, it is a vector from $(400,0)$ to $(424,7)$.  You need to scale up the vector $(24,7)$ to make the length $600$.  Can you find the length of $(24,7)$?  This should let you find the coordinates of $C$.  Draw a picture.  The turning angle is between the east ($x$) axis and $BC$.  You will need an inverse trig function to find the angle.
Under i, it is not true that $AC=976i=168j$.  The numbers are right, but what you have written says that three vectors are equal and they are not.
