Question 90501
Hi,<br />

The planes' flightpath can be modelled by a straight line. The position of plane i after time t is given by:<br />

*[tex P_i = S_i + tD_i]<br />

Where *[tex P_I] is the position of plane *[tex i], *[tex S_i] is the starting position of plane *[tex i], and *[tex D_i] is the direction(vector) in which plane *[tex i] flys in a time unit.<br />

The disatnce between the planes is obviously *[tex |P_1-P_2|=\sqrt{(P_1-P_2)\cdot(P_1-P_2)}].<br />

Since both planes start at the airport *[tex S_1 = S_2]. Substituting and tidying up gives the distance as *[tex t\sqrt{|D_1|^2 + |D_2|^2 -2D_1\cdot D_2}].<br />

Now in an hour the first plane moves 190miles, so *[tex |D_1|=190] and similarly *[tex |D_2|=160]. That's about as simple as we can get, so now we have to actually deal with some vectors to calculate *[tex D_1\cdot D_2].<br />

By definition *[tex D_1=190\left(\begin{array}{c}\cos(86)\\ \sin(86)\end{array}\right)] and similarly *[tex D_2=160\left(\begin{array}{c}\cos(176)\\ \sin(176)\end{array}\right)]. So their dot product is:<br />

*[tex 190*160\cos(86)\cos(176) + 190*160\sin(86)\sin(176)]<br />

At this point you realise all your hard work is for nothing because the angles differ by 90 degrees so you could have just used Pythagorous theorem, but if they didn't differ by 90 degrees, this is how you would do it. Anyway, because they differ by 90 degress *[tex D_1\cdot D_2=0].<br />

The planes have been in the air for 1.5 hours so *[tex t=1.5]. That gives the answer as:<br />

*[tex 1.5\sqrt{190^2 + 160^2}]<br />

I'll leaave you to do the math.<br />

Kev