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    I let C = x, y, x

    then did

    (5, 2, 6) dot [(2, -1, 3) x (x, y, z)] = 30

    -5z + 15y - 6x - 4z + 12y + 6x = 30

    27y - 9z = 30
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    (Original post by creativebuzz)
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    I let C = x, y, x

    then did

    (5, 2, 6) dot [(2, -1, 3) x (x, y, z)] = 30

    -5z + 15y - 6x - 4z + 12y + 6x = 30

    27y - 9z = 30
    1. (5, 2, 6)? It should be (5, 2, 0).

    2. Your cross product is wrong as well. It should be i(-z-3y) - j(2z-3x) + k(2y+x), the k component will disappear after dot product.
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    (Original post by Zacken)
    1. (5, 2, 6)? It should be (5, 2, 0).

    2. Your cross product is wrong as well. It should be i(-z-3y) - j(2z-3x) + k(2y+x), the k component will disappear after dot product.
    What a clumsy mistake *facepalm* Thank you Zacken! I redid the question with 0 instead of 6 and got the right answer
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    (Original post by creativebuzz)
    What a clumsy mistake *facepalm* Thank you Zacken! I redid the question with 0 instead of 6 and got the right answer
    Awesome - first class work.
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    (Original post by Zacken)
    Awesome - first class work.
    Thank you! (I would rep but SR won't let me) :P

    Last question,

    where did I go wrong in part d?

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    As P and M are perpendicular to each other I made P into a plane that is parallel to M and then put it in the form r.n(unit vector) = d(unit vector)

    for M: r . 1/root14(2, 1, 3) = 21/root14

    for P: r . 1/root14(2, 1, 3) = (1, 2, 1) . (2, 1, 3)

    = 7/root14

    so it's 21-7/root14 = root14
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    (Original post by creativebuzz)
    Thank you! (I would rep but SR won't let me) :P

    Last question,

    where did I go wrong in part d?

    Name:  Untitled.png
Views: 54
Size:  36.3 KB

    As P and M are perpendicular to each other I made P into a plane that is parallel to M and then put it in the form r.n(unit vector) = d(unit vector)

    for M: r . 1/root14(2, 1, 3) = 21/root14

    for P: r . 1/root14(2, 1, 3) = (1, 2, 1) . (2, 1, 3)

    = 7/root14

    so it's 21-7/root14 = root14
    I'm not sure why you think "P and M are perpendicular to one another" - it doesn't even really make sense to say that two points are perpendicular to one another nor to say that a plane is parallel to a point. I understand what you're saying but it's not the case that P and M are "perpendicular" to one another. You'll need to draw a triangle OMP and use the area and height (which will be the perpendicular distance) to find the perpendicular distance since you know the base and the area.
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    (Original post by Zacken)
    I'm not sure why you think "P and M are perpendicular to one another" - it doesn't even really make sense to say that two points are perpendicular to one another nor to say that a plane is parallel to a point. I understand what you're saying but it's not the case that P and M are "perpendicular" to one another. You'll need to draw a triangle OMP and use the area and height (which will be the perpendicular distance) to find the perpendicular distance since you know the base and the area.
    The reason why I thought P and M were perpendicular was because the line l (which contains P) is perpendicular to the plane (which contains M)
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    (Original post by creativebuzz)
    The reason why I thought P and M were perpendicular was because the line l (which contains P) is perpendicular to the plane (which contains M)
    Yes, but it doesn't make sense to talk about points being perpendicular. I can't give you a point on the y-axis and a point on the x-axis and say they are perpendicular, that doesn't make sense. I can say that the y-axis is perpendicular to the x-axis, though.
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    (Original post by Zacken)
    Yes, but it doesn't make sense to talk about points being perpendicular. I can't give you a point on the y-axis and a point on the x-axis and say they are perpendicular, that doesn't make sense. I can say that the y-axis is perpendicular to the x-axis, though.
    Yup, that's a very valid point! So how would you advise that I draw vector situations like this one? Because I would have use the whole area of a triangle if I had drawn it right in the first place to see the triangle clearly
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    (Original post by creativebuzz)
    Yup, that's a very valid point! So how would you advise that I draw vector situations like this one? Because I would have use the whole area of a triangle if I had drawn it right in the first place to see the triangle clearly
    The solution bank (for this question) has a nice drawing that showcases how to do this - I will admit that this part of the question was very difficult and required some insight.
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    (Original post by Zacken)
    The solution bank (for this question) has a nice drawing that showcases how to do this - I will admit that this part of the question was very difficult and required some insight.
    Yeah that's what I was checking for my answer! I tried drawing it again but it still doesn't work out as nicely as the one in the solution bank.

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    But I don't see how my drawing is technically wrong? (I know I could just copy the one in the solution bank but I'm more concerned about whether my method for drawing vectors is wrong because I can't seem to get what they go on my own)
 
 
 
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