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    (Original post by BabyMaths)
    OK \vec{OM}= 4 i -11j +4k + lambda(i + 4j + k)

    The dot product of this with i+4j+k (the direction of AB) is 0.
    Why is OM equal to this ? Only M can be a point not the whole equation (OM). Then it can't give 0 with cross product.
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    (Original post by zaback21)
    Why is OM equal to this ? Only M can be a point not the whole equation (OM). Then it can't give 0 with cross product.
    Do the first part of the equation and you get
    r=(4,-11,4)+lambda(1,4,1)

    M is some point on this line, between A and B. Remember the direction of AB is (1,4,1). So when we are at A we must move a certain multiple of (1,4,1) vector to get to this point M. We let this multiple=lambda. To break this down, the journey from O to M, which gives the position vector of M, is like this

    1. Go to A from O, so vector (4, -11,4)
    2. Move a certain multiple of (1,4,1) to go from A to M
    Overall, you went from O to M

    A sketch really really helps a lot to visualise things .
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    (Original post by krisshP)
    Do the first part of the equation and you get
    r=(4,-11,4)+lambda(1,4,1)

    M is some point on this line, between A and B. Remember the direction of AB is (1,4,1). So when we are at A we must move a certain multiple of (1,4,1) vector to get to this point M. We let this multiple=lambda. To break this down, the journey from O to M, which gives the position vector of M, is like this

    1. Go to A from O, so vector (4, -11,4)
    2. Move a certain multiple of (1,4,1) to go from A to M
    Overall, you went from O to M

    A sketch really really helps a lot to visualise things .
    I know what O to M is. I know what OM is. But the equation of line OM is not what he wrote. It is only equal at the point M and O has nothing to do with it. Its the equation AB which should be (4,-11,4)+lambda(1,4,1) not OM=(4,-11,4)+lambda(1,4,1).

    Equation of OM is = t(ai+bj+ck)
    where O=(0,0,0) M=(a,b,c) and t is a parameter.
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    (Original post by zaback21)
    I know what O to M is. I know what OM is. But the equation of line OM is not what he wrote. It is only equal at the point M and O has nothing to do with it. Its the equation AB which should be (4,-11,4)+lambda(1,4,1) not OM=(4,-11,4)+lambda(1,4,1).

    Equation of OM is = t(ai+bj+ck)
    where O=(0,0,0) M=(a,b,c) and t is a parameter.
    You are correct. If I was not being lazy I would have said \vec{OM}=......... for some value of \lambda.

    Sorry for any confusion caused.

    I must be less sloppy.
    I must be less sloppy.
    Only another 98 lines to do..
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    (Original post by BabyMaths)
    You are correct. If I was not being lazy I would have said \vec{OM}=......... for some value of \lambda.

    Sorry for any confusion caused.

    I must be less sloppy.
    I must be less sloppy.
    Only another 98 lines to do..
    Ha ha lol !
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    (Original post by BabyMaths)
    X
    a=3i+5j+7k find the angle a makes with the x-axis?
    Name:  1377108456863.jpg
Views: 49
Size:  19.7 KB

    With |a|=√83

    Cos∅=3/√83
    ∅=71° <-- correct

    Tan∅=5/3
    ∅=59° <-- wrong, but WHY?

    Thank you very much
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    (Original post by krisshP)
    a=3i+5j+7k find the angle a makes with the x-axis?
    Name:  1377108456863.jpg
Views: 49
Size:  19.7 KB

    With |a|=√83

    Cos∅=3/√83
    ∅=71° <-- correct

    Tan∅=5/3
    ∅=59° <-- wrong, but WHY?

    Thank you very much
    Well, that triangle is all wrong. You can't even make a triangle with those side lengths.

    I might have a go at a diagram in a minute. But mainly I should say, stick to using the dot product in these questions, even though it's good to try out your own ideas too.
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    Thanks a lot. The diagram did help me to visualise the situation far better.
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    (Original post by krisshP)
    As previously said, here's the correct working out
    With |a|=√83

    Cos∅=3/√83
    ∅=71° <-- correct

    But the thing is the working out implies that 3 is the length of the adjacent side:confused:. But the diagram shows that 3 is NOT the length of a side of the triangle. i hope i am making some sense here. But i agree though that the triangle's hypotenuse is sqrt34.


    From the diagram only,

    For triangle's adjacent side (a)
    3^2 + 5^2=a^2
    a=sqrt34

    Tanø=7/sqrt34
    Ø=47.7

    Sorry, but the diagram did help me to visualise the situation far better, but I now question the "correct" working out.
    The angle you found, \displaystyle \arctan \left( \frac{7}{\sqrt{34}}\right), is the angle between the vector and the xy plane. To find the angle between the vector and the x axis you could use the cosine rule but you should just use the dot product.
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    (Original post by BabyMaths)
    The angle you found, \displaystyle \arctan \left( \frac{7}{\sqrt{34}}\right), is the angle between the vector and the xy plane. To find the angle between the vector and the x axis you could use the cosine rule but you should just use the dot product.
    I get it now, what I Previously said doesn't give the required angles that is needed. I looked at everything again and then realised this, so I edited my post. Yes I understand why you suggest using the dot product rather than basic trig. Looking at that angle makes it be difficult to define what really even is the opposite/adjacent/hypotenuse side. So it's better to just stick to the dot product and keep things simple. I guess a "fresh eye" helps a lot in Maths if you are stuck on something


    Thank you
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    (Original post by BabyMaths)
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    Name:  IMG_20130822_111528-1.jpg
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    I'm stuck on the second part and was wondering during if you can help me out please.

    Part a.
    AB=(4,-8)

    Line 1
    r= (3,4) + lambda(4,-8)
    r=(3,4) + lambda(1,-2)

    Part b.

    Let the direction of line 2 be (x,y), so for line 2
    r=(14,7) + mu(x,y)

    Since line 2 is perpendicular to line 1,

    (1,-2).(x,y)=0

    x-2y=0

    I suspect that simultaneous equations are involved to get x and y, but I'm unsure how to get the second equation:confused:. I feel like I have a brain freeze.

    Thank you
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    (Original post by krisshP)
    Name:  IMG_20130822_111528-1.jpg
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    x-2y=0

    I suspect that simultaneous equations are involved to get x and y, but I'm unsure how to get the second equation:confused:. I feel like I have a brain freeze.

    Thank you
    You don't need simultaneous equations.

    Recall that the direction vector of a line is only unique up to scalar multiples, i.e. (1,2), (2,4) specify the same direction.

    So you have 2y=x.

    And your direction vector (x,y) becomes (2y,y). Can you take it from there?
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    (Original post by ghostwalker)
    You don't need simultaneous equations.

    Recall that the direction vector of a line is only unique up to scalar multiples, i.e. (1,2), (2,4) specify the same direction.

    So you have 2y=x.

    And your direction vector (x,y) becomes (2y,y). Can you take it from there?
    r=(14,7) + mu(2,1)

    This is right according to the textbook answer of
    r=(14,7) + mu(8,4)

    But where did the 8 and 4 one from? Surely that means that perhaps there's another method getting exactly these numbers?
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    (Original post by krisshP)
    r=(14,7) + mu(2,1)

    This is right according to the textbook answer of
    r=(14,7) + mu(8,4)

    But where did the 8 and 4 one from? Surely that means that perhaps there's another method getting exactly these numbers?
    If you recall if two lines are perpendicular then the product of their gradients is -1.

    The gradient of a line with direction vector (4,-8) is -8/4, So the gradient of a prependicular lines is 4/8, which gives a direction vector of (8,4).
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    (Original post by ghostwalker)
    If you recall if two lines are perpendicular then the product of their gradients is -1.

    The gradient of a line with direction vector (4,-8) is -8/4, So the gradient of a prependicular lines is 4/8, which gives a direction vector of (8,4).
    Makes perfect sense

    Thank you for helping me out
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    thank you for the help
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    can you explain it more clearly by drawing the square and point o in it.
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    (Original post by davros)
    OD is just the vector from O to D. You can get there in a number of ways e.g. O to A then A to D; or O to B and then B to D; or O to C and then C to D.

    Using the fact that ABCD is a square, you should be able to express vectors like AD, BD and CD in terms of a, b and c.
    i can't t understand it clearly can you draw the square and show it.
 
 
 
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