Let the plane going northweat be a and the plane going east be b. 

After 3 hours, a has travelled:

403 mph x 3 = 1209m

While b has travelled:

344 mph x 2 = 1032m

This result in a triangle with a given angle and two sides/SAS. We can therefore use the law of cosine to find the distance between the two planes after 3 hours. 

Let the distance be c:

c^2 = a^2 + b^2 - 2abcosc

c^2 = (1209)^2 + (1032)^2 - 2(1209)(1032)cos135

c^2 = 1461681 + 1065024 - 2495376cos135

c = 2071.5 (correct to 1 decimal place)