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Gradient of line and vectors

Posted by White Group Mathematics on August 26, 2016 at 2:25 AM



Hi Mr Koh,

                


                     I can't seem to get around this problem, could you please help? :/


                 

  




Student X

          







Gradient of the line = -4/3 by rewriting its equation in the form y=mx+c. As such, it is as good as saying that for every 1 unit one moves in the x direction, it also has to move -4/3 units in the y direction. This general movement is hence represented by i - 4/3j.



Since OA is parallel to the line, then we say vector OA= k (i - 4/3j) , where k is a real valued constant. Given the magnitude of vector OA is 20 units, we can further surmise that k √ [ 1² + (-4/3)² ] = 20, solving this gives k=12 or -12. Thus vector OA= 12 (i - 4/3j) =12i -16j or -12 (i - 4/3j) = -12i+16j (shown)





Hope this helps. Peace.








Best Regards,

Mr Koh





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