mathematics HELP!!!! @ bxzbthf的部落格

標題:

mathematics HELP!!!!

此文章來自奇摩知識+如有不便請留言告知

發問:

A tracking station observes an aeroplane at two successive time to be (-500, 0, 1000) and (400, 400, 1050) relative to axes x in an easterly direction, y in a northerly direction, and z vertically upwards, with distances in metres.a) Find the equation of the path of the aeroplane in vector form.b) Controls... 顯示更多 A tracking station observes an aeroplane at two successive time to be (-500, 0, 1000) and (400, 400, 1050) relative to axes x in an easterly direction, y in a northerly direction, and z vertically upwards, with distances in metres. a) Find the equation of the path of the aeroplane in vector form. b) Controls advises the aeroplane to change course from its present position, (400, 400, 1050), to level flight at the current height and turn easterly through an angle of 90°; what is the equation of the newpath in both vector form and Cartesian coordinates? 更新: C) Draw a diagram to illustrate the situation.

最佳解答:

a) (-500, 0, 1000) and (400, 400, 1050) let vector r be the equation of the path (note: r should have an arrow on its head) r = <-500, 0, 1000> + t*<400-(-500), 400-0, 1050-1000> r = <-500, 0, 1000> +t*<900, 400, 50> r = <-500+900t, 400t, 1000+50t> Check: sub t=1, we have r(1) = <400, 400, 1050> b) This is actually finding the equation for the line that goes through (400, 400, 1050) and perpendicular to <900, 400, 0> perpendicular: dot product = 0 <900, 400> dot = 0 so, the vector is either in the direction <400,-900> or <-400,900> Using the graph in part c, we know that the aeroplane in the direction of <400, -900, 0> let vector v be the equation of the new path (note: v should have an arrow on its head) v = <400, 400, 1050> + t*<400, -900, 0> v = <400+400t, 400-900t, 1050> or in Cartesian coordinates, x = 400+400t, y = 400-900t, z = 1050 c) graph http://image82.webshots.com/482/5/17/65/2691517650104626724faCylX_ph.jpg

其他解答:8758B59A7FA1EEA7