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Math Help - Important Vector question!!

  1. #1
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    Important Vector question!!

    Hi guys, I have this question which I really need help ...

    The equation of a plane P1 in a cartesian coordinate system is given by 3x+y-z-6=0. THe cordinates of two points A and B are [1,1,3] and [2,1,2] respectively.

    i)Find the unit vector normal to the plane P1.
    ii)Find the vector AB.
    iii)Find the equation of a plane P2 that passes thru the points A and B and is perpendicular to the plane P1.

    so far i only got AB=OB-OA ..i think so which is equals to i-k.
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  2. #2
    Moo
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    Hello,

    If a plane has ax+by+cz=d as equation, a normal vector of it has coordinates (a,b,c)
    To find the unit vector, divide each coordinate by the norm of the vector, given by \sqrt{a^2+b^2+c^2}

    A vector \vec{MN} joining 2 points M(x_M, y_M, z_M) and N(x_N, y_N, z_N) has its coordinates as follow :

    \vec{MN} \begin{pmatrix} x_N-x_M \\ y_N-y_M \\ z_N-z_M \end{pmatrix}
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  3. #3
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    That means the vector normal to the plane is 3i+j-k andi divide it by its modulus so i will get its unit vector?
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  4. #4
    Moo
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    Yes !

    By "norm", i meant modulus yeah
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  5. #5
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    ok so i got 2 parts done..thanks a lot..What about the last part..?im clueless..
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  6. #6
    Moo
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    Hm well you can start by stating that the equation of P2 is in the form ax+by+cz+d=0 then replacing the coordinates of A and B in it.
    It's already 2 equations ^^

    Then...i have to think about it >_<
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  7. #7
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    Moo i dont get you..you mean just put in the cordinates of A and B..that means i do a scalar product of the 2?and the result?
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  8. #8
    Moo
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    No, you have the hypothetic equation of P2. You will have to determine a, b, c and d
    As A and B are in P2, the coordinates of A verify the equation of P2 and the coordinates of B verify the equation of P2.

    This will give you two equations in the aim to solve for a, b, c and d. But you need 2 others to find the complete equation.



    Oh, there can be a way of doing it... even 2. But i don't really like it because it doesn't use the previous questions...

    I don't remember if there are other properties we can get from the equation of a plane... With orthogonality.

    Perhaps you can try a thing :

    The vector (a,b,c) is orthogonal (perpendicular) to P2 (same rule as for the 1st question).
    As \vec{AB} is in P2, you can say that the scalar product of \vec{AB} with the vector (a,b,c) is 0...

    I'm not sure... can you try ?
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  9. #9
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    I just learned about orthogonal matrices.but cant seem to figure out equations of perpendicular planes.would the plane by any chancebe parallel to to the vector normal to the plane?
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  10. #10
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    Quote Originally Posted by ashes View Post
    Hi guys, I have this question which I really need help ...

    The equation of a plane P1 in a cartesian coordinate system is given by 3x+y-z-6=0. THe cordinates of two points A and B are [1,1,3] and [2,1,2] respectively.

    i)Find the unit vector normal to the plane P1.
    ii)Find the vector AB.
    iii)Find the equation of a plane P2 that passes thru the points A and B and is perpendicular to the plane P1.

    so far i only got AB=OB-OA ..i think so which is equals to i-k.
    1. The normal vector of P1 is \vec n = (3, 1, -1) as Moo has posted.

    2. The vector \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1, 0, -1)

    3. That means the plane P2 contains the point A and the 2 vectors which span the plane:

    \vec r = (1, 1, 3) + s \cdot (1, 0, -1) + t \cdot (3, 1, -1) which yield the equation of the plane in parametric form:

    \left|\begin{array}{lcr}x&=&1+s+3t \\ y&=&1 + t \\ z&=& 3-s-t\end{array}\right.

    4. To get this equation in coordinate form(?) calculate the normal vector of the 2 direction vectors which span the plane:

    (1,0,-1) \times (3,1,-1) = (1,-2,1)

    The plane contains point A. Calculate the dot-product to get the constant summand. Therefore the equation of the plane P2 is:

    x-2y+z-2=0
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  11. #11
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    A different, but longer way

    Quote Originally Posted by ashes View Post
    need the maths experts' help...
    let ax+by+cz=d be the perpendicular plane  3x+y-z=6

    then its normal vector is

    <a,b,c>

    so since the planes are perpendicular the dot product of their normals must be zero. so

    <a,b,c> \cdot <3,1,-1> =0 \iff 3a+b-c=0

    Now if we evaluate the plane at the two given points we get
    (1,1,3) and (2,1,2)

    a+b+3c=d \mbox{ and } 2a+b+2c=d

    This is an underdetermined system

    solving the system
    \begin{bmatrix}<br />
1 && 1 && 3 && -1 && 0 \\<br />
2 && 1 && 2 && -1 && 0 \\<br />
3 && 1 && -1 && 0 && 0 \\<br />
\end{bmatrix}

    \begin{bmatrix}<br />
1 && 1 && 3 && -1 && 0 \\<br />
0 && -1 && -4 && 1 && 0 \\<br />
0 && 0 && 2 && -1 && 0 \\<br />
\end{bmatrix}

    so d=2c c=c b=-2c a=c

    plugging into our equation we get

    ax+by+cz=d \iff cx-2cy+cz=2c divide by c and we get

    x-2y+z=2
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  12. #12
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    just check with you, so the unit vector normal to the plane, \hat{n}= [\frac{3}{\sqrt{11}},\frac{1}{\sqrt{11}},\frac{-1}{\sqrt{11}}] where \sqrt{11} is the modulus of the normal vector
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  13. #13
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    Quote Originally Posted by ashes View Post
    just check with you, so the unit vector normal to the plane, \hat{n}= [\frac{3}{\sqrt{11}},\frac{1}{\sqrt{11}},\frac{-1}{\sqrt{11}}] where \sqrt{11} is the modulus of the normal vector
    Yes.
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