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    Can someone help me with the second part??
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    Zacken
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    (Original post by Mystery.)
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    Can someone help me with the second part??
    If you've evaluated p.(q+r) then you know that:

    p \cdot (q+r)  = |p| |q+r| \cos 0, since (q+r) is just moving in the direction q up to A and then moving A down which gets you on the direction P. So the angle between p and (q+r) is 0.

    Can you do something similar with the second part? Remember that you've worked out r.(p-q), etc...
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    For |q+r|, recognize the question is merely requiring you to discover the magnitude of the resultant vector q+r. Let the foot of perpendicular from A to the line CD be E. Can you therefore see that vector BC= r = vector AE?

    If you can, the resultant of q+r is simply given by the vector DE. |q+r| therefore is about computing the length of the base of the upright right-angled triangle ADE.


    To discover |p-q|, consider the following:

    |p-q|^2=(p-q)•(p-q)
    = p•p -2 p•q + q•q
    = |p|^2 - 2 p•q +|q|^2 --------------(1)

    Each of the 3 components of the RHS of the above expression (1) can be computed fairly quickly.
    Obviously, |p|^2= 16 and |q|^2= 9, while 2p•q =2 |p||q| cos (30°) = 12√3

    Assemble these 3 quantities to calculate the overall RHS value, square root it, and voila, the answer is right in your face.

    Hope what I have articulated was useful for your understanding. Peace.
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    (Original post by Zacken)
    If you've evaluated p.(q+r) then you know that:

    p \cdot (q+r)  = |p| |q+r| \cos 0, since (q+r) is just moving in the direction q up to A and then moving A down which gets you on the direction P. So the angle between p and (q+r) is 0.

    Can you do something similar with the second part? Remember that you've worked out r.(p-q), etc...
    Yeah, kinda get it now. Thanks!
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    (Original post by WhiteGroupMaths)
    For |q+r|, recognize the question is merely requiring you to discover the magnitude of the resultant vector q+r. Let the foot of perpendicular from A to the line CD be E. Can you therefore see that vector BC= r = vector AE?

    If you can, the resultant of q+r is simply given by the vector DE. |q+r| therefore is about computing the length of the base of the upright right-angled triangle ADE.


    To discover |p-q|, consider the following:

    |p-q|^2=(p-q)•(p-q)
    = p•p -2 p•q + q•q
    = |p|^2 - 2 p•q +|q|^2 --------------(1)

    Each of the 3 components of the RHS of the above expression (1) can be computed fairly quickly.
    Obviously, |p|^2= 16 and |q|^2= 9, while 2p•q =2 |p||q| cos (30°) = 12√3

    Assemble these 3 quantities to calculate the overall RHS value, square root it, and voila, the answer is right in your face.

    Hope what I have articulated was useful for your understanding. Peace.
    Thanks for such a detailed answer! Much appreciated.
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    No problem. Peace.
 
 
 
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