The thrust of an airplane\'s engine is 700 mph. It is aimed in the direction <6,

Question

The thrust of an airplane's engine is 700 mph. It is aimed in the direction <6,-3,2> but its velocity with respect to the ground is <580,-330,160> mph. What is the wind's velocity?

I solved this by finding the direction angles (a, b, and c) using the position vector, then finding the plane's thrust vector <700cosa,700cosb,700cosc>. Then I found the difference between this vector's components and <580,-330,160>. The magnitude of that vector is then the wind's velocity.

So, did I do this correctly? And, is there an easier way?

Explanation / Answer

unit vector in the direction of <6,-3,2> is a = [6i-3j+2k ] / [6^ 2+(-3)^ 2+2^ 2]                                                                   = [6i-3j+2k] / 7 thrust of an airplane's engine V = 700 mph So, resultant speed vector V = 700 mph { [6i-3j+2k ] / 7 }                                              = (600 i -300 j +200 k ) mph velocity with respect to ground V ' = 580 i -330 j +160 k wind's velocity = V - V '                             = ( 600-580 ) i + (-300+330 ) j +(200-160) k                            = 20 i +30 j + 40 k magnitude of wind's velocity = [ 20^ 2+ 30^ 2+ 40 ^ 2]                                             = 53.85 mph

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