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    find the angle that the vector 9i - 5j +3k makes with
    a) the positive x axis
    b) the positive y axis


    how do i solve this?

    do i use the a.b = |a||b|cosX formula?
    if so, how?
    cos it seems like it would have a lot of zeros in it... :confused:

    ta,

    JB
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    (Original post by jumblebumble)
    find the angle that the vector 9i - 5j +3k makes with
    a) the positive x axis
    b) the positive y axis


    how do i solve this?

    do i use the a.b = |a||b|cosX formula?
    if so, how?
    cos it seems like it would have a lot of zeros in it... :confused:

    ta,

    JB
    Well, any vector which has a direction vector in the direction of the positive x-axis, for example, 5x +0y+0z or 854x + 0y+0z.
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    (Original post by Clarity Incognito)
    Well, any vector which has a direction vector in the direction of the positive x-axis, for example, 5x +0y+0z or 854x + 0y+0z.
    what do you mean?
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    (Original post by jumblebumble)
    what do you mean?
    There are infinitely many vectors that can describe the direction of the positive x axis, but you don't want them moving into the y or z plane because then it wouldn't be describing the direction of the +ve x axis.
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    For the first one.

    I'd forget about the dot product. Just drop a perpendicular from the point specified by the vector down to the x-axis. You now have a right angled triangle.

    What's the length of the hypotenuse?

    What's the length along the x axis?

    Hence cosine of your angle between the vector and the x-axis is?

    Hence....


    In this case the x co-ordanate is positive so you have found the angle with the postive x-axis.

    In the second one you'll have to work a little harder.
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    (Original post by ghostwalker)
    For the first one.

    I'd forget about the dot product. Just drop a perpendicular from the point specified by the vector down to the x-axis. You now have a right angled triangle.

    What's the length of the hypotenuse?

    What's the length along the x axis?

    Hence cosine of your angle between the vector and the x-axis is?

    Hence....


    In this case the x co-ordanate is positive so you have found the angle with the postive x-axis.

    In the second one you'll have to work a little harder.
    Thanks. I've worked out both parts now
 
 
 
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