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Math Help - find magnitude of a vector

  1. #1
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    find magnitude of a vector

    Hi folks!

    I joined today to find the answer to a vector question that has been driving me nuts. You know how it is! So here goes:

    The two vectors a and b are of equal magnitude k (k not 0) and the angle between a and b is 60 degrees. If c = 3a - b and d = 2a - 10b

    i) show that c and d are perpendicular
    ii) find magnitudes of c and d in terms of k.

    i) is OK. c.d = 0 for perpendicular vectors so evaluate (3a - b).(2a - 10b). This comes to 0 since a.b = a.b. cos 60. = k squared / 2 and a squared = b squared = k squared. (Sorry about the notation folks! - I'll figure out the superscripts later)

    My problem is the second part. Can anyone help?

    regards and thanks
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  2. #2
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    Quote Originally Posted by s_ingram View Post
    Hi folks!

    I joined today to find the answer to a vector question that has been driving me nuts. You know how it is! So here goes:

    The two vectors a and b are of equal magnitude k (k not 0) and the angle between a and b is 60 degrees. If c = 3a - b and d = 2a - 10b

    i) show that c and d are perpendicular
    ii) find magnitudes of c and d in terms of k.

    i) is OK. c.d = 0 for perpendicular vectors so evaluate (3a - b).(2a - 10b). This comes to 0 since a.b = a.b. cos 60. = k squared / 2 and a squared = b squared = k squared. (Sorry about the notation folks! - I'll figure out the superscripts later)

    My problem is the second part. Can anyone help?

    regards and thanks
    The magnitude of c is \sqrt{c\cdot c}. Now, c\cdot c= (3a- b)\cdot(3a- b)= 9 a\cdot a- 6 a\cdot b+ b\cdot b. Of course, a\cdot a= b\cdot b= k^2 any you calculated a\cdot b before. Do the same with the magnitude of d.
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  3. #3
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    It's always easy when someone else gives you the answer! I was faffing about looking at the equalateral triangle made by vectors a and b, then I tried expressing c and d in terms of i and j (vectors at right angles) when all I had to do was mod c = SQRT c.c! Thanks to you HallsofIvy.
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