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    (Original post by zaback21)
    They simplified (-4,2,2) into the simplest ratio of (-2,1,1). (-4,2,2) = 2(-2,1,1). 2 is a scaler parameter that is included as multiple of u.

    So, lets say r=(1,5,2) + t(-4,2,2) = (1,5,2) + u(-2,1,1)

    u=2t
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    How would I go about doing the last question there? I know it starts off as r=(3,2,-5).. But then what? The unknown there really puts me off and confuses me .
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    (Original post by krisshP)
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    How would I go about doing the last question there? I know it starts off as r=(3,2,-5).. But then what?
    Look at the equation of the line that it has to be parallel to. This is r=i+3j+lambda*(......).

    The bit in brackets gives you the direction.
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    (Original post by krisshP)
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    How would I go about doing the last question there? I know it starts off as r=(3,2,-5).. But then what? The unknown there really puts me off and confuses me .
    Rewrite the equation as:

    r= i+ 3j + lambda ( 2i -4j + 5k)
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    (Original post by BabyMaths)
    Look at the equation of the line that it has to be parallel to. This is r=i+3j+lambda*(......).

    The bit in brackets gives you the direction.

    (Original post by zaback21)
    Rewrite the equation as:

    r= i+ 3j + lambda ( 2i -4j + 5k)
    That made me feel dumb

    Now I got r=(3,2,-5) +lambda(2,-4,5)

    Thanks
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    (Original post by krisshP)
    That made me feel dumb

    Now I got r=(3,2,-5) +lambda(2,-4,5)

    Thanks
    Asking never hurts. Now you will never make this mistake again. We all did when we first started learning. You just get better with practice.
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    (Original post by zaback21)
    Asking never hurts. Now you will never make this mistake again. We all did when we first started learning. You just get better with practice.
    Well said, it does feel a bit mind boggling with the vector forms of i, j and k stuff and lambda coming all at you and I couldn't see a way forward I was clueless. I guess I'll get much better with a lot of practice.

    One must fail to learn I believe :rolleyes:
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    (Original post by krisshP)
    Well said, it does feel a bit mind boggling with the vector forms of i, j and k stuff and lambda coming all at you and I couldn't see a way forward I was clueless. I guess I'll get much better with a lot of practice.

    One must fail to learn I believe :rolleyes:
    Haha take the Further Maths and you will see how interesting it gets in 3D Vector Space ( a must if you plan to study science, maths or engineering).
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    (Original post by zaback21)
    Haha take the Further Maths and you will see how interesting it gets in 3D Vector Space ( a must if you plan to study science, maths or engineering).
    Is that in A2 or at AS?
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    (Original post by krisshP)
    Is that in A2 or at AS?
    Its in FP1-FP3. Check the content. Most likely A2. But you can take them in AS too if you want I guess. Well I did. I took two Maths at A Level.
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    (Original post by zaback21)
    X
    find the angle in degrees between the vectors a=-8i+j+4k and b=6i+3j+6k

    I got 105°. What did you get? My textbook says it is 75°, why?:confused:

    Thanks
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    (Original post by krisshP)
    find the angle in degrees between the vectors a=-8i+j+4k and b=6i+3j+6k

    I got 105°. What did you get? My textbook says it is 75°, why?:confused:

    Thanks
    \arccos \left(-\frac{7}{27}\right)=105.026.. degrees.

    Did the question ask for the acute angle?
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    (Original post by BabyMaths)
    \arccos \left(-\frac{7}{27}\right)=105.026.. degrees.

    Did the question ask for the acute angle?
    Nope, a bit annoying

    Anyway thanks
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    (Original post by BabyMaths)
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    I feel so confused on this and I check it like 3 times

    AB.AC=12z+26

    |AB|=13

    |AC|=√(z^2 + 40)

    Cos60=1/2 from special triangles

    (12z+26)/[13√(z^2 + 40)] = 1/2

    2(12z+26)=13√(z^2 +40)

    24z+52=√(169z^2 +6760)

    576z^2 + 2704=169z^2 +6760

    407z^2 - 4956=0

    Using the quadratic formula gives me z=3.2 or z=-3.2 both to 2SF

    Where did I go wrong? Textbook says z=1.33 or z=-7.47 :confused: WHY?

    Thank you VERY much
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    (Original post by krisshP)
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    I feel so confused on this and I check it like 3 times

    AB.AC=12z+26

    |AB|=13

    |AC|=√(z^2 + 40)

    Cos60=1/2 from special triangles

    (12z+26)/[13√(z^2 + 40)] = 1/2

    2(12z+26)=13√(z^2 +40)

    24z+52=√(169z^2 +6760)

    576z^2 + 2704=169z^2 +6760

    407z^2 - 4956=0

    Using the quadratic formula gives me z=3.2 or z=-3.2 both to 2SF

    Where did I go wrong? Textbook says z=1.33 or z=-7.47 :confused: WHY?

    Thank you VERY much
    You made a mistake in the bold line. When you square (24z+52)^2, use (a+b)^2 formula=a^2+2ab+b^2

    So, it should be 576z^2 +2496z+ 2704=169z^2 +6760
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    (Original post by zaback21)
    You made a mistake in the bold line. When you square (24z+52)^2, use (a+b)^2 formula=a^2+2ab+b^2

    So, it should be 576z^2 +2496z+ 2704=169z^2 +6760
    Okay, thanks. I feel really dumb now
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    (Original post by BabyMaths)
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    Part a was fine for me and I got r=(4,-11,4)+lambda(1,4,1)

    In part b I hit a brick wall. The dot product of MA and MO must be 0 as both vectors are perpendicular. So

    x(x-4)+y(11+y)+z(z-4)=0

    I only have 1 equation with 3 unknowns. It would be okay if I had to equations with 3 unknowns that an be solved simultaneously, but this isn't the case. :confused: help me out please


    Thanks
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    (Original post by krisshP)
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    Part a was fine for me and I got r=(4,-11,4)+lambda(1,4,1)

    In part b I hit a brick wall. The dot product of MA and MO must be 0 as both vectors are perpendicular. So

    x(x-4)+y(11+y)+z(z-4)=0

    I only have 1 equation with 3 unknowns. It would be okay if I had to equations with 3 unknowns that an be solved simultaneously, but this isn't the case. :confused: help me out please


    Thanks
    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.
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    (Original post by krisshP)
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    Part a was fine for me and I got r=(4,-11,4)+lambda(1,4,1)

    In part b I hit a brick wall. The dot product of MA and MO must be 0 as both vectors are perpendicular. So

    x(x-4)+y(11+y)+z(z-4)=0

    I only have 1 equation with 3 unknowns. It would be okay if I had to equations with 3 unknowns that an be solved simultaneously, but this isn't the case. :confused: help me out please


    Thanks
    Is the answer (6i,-3j,6k) ?
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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.
    Wow, that makes perfect sense. I didn't think of it like that. I got lambda=2 and then I got OM vector =(6, -3, 6) which is correct.


    Thanks a lot!
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    (Original post by zaback21)
    Is the answer (6i,-3j,6k) ?
    Yep, I just got after using BabyMaths' advice.
 
 
 
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