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Math Help - Two short questions on vectors

  1. #1
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    Two short questions on vectors

    Hi
    The following question i am having problems with:

    1) Find the equation of the plane normal to the vector 2i+j-3k

    2x+y-3z = d

    |d|=5
    d= 5 or -5

    so

    P = \frac{5}{\sqrt{14}} or \frac{-5}{\sqrt{14}}

    so 2x+y-3z = \frac{5}{\sqrt{14}} or 2x+y-3z = \frac{-5}{\sqrt{14}}

    book's answer says 2x+y-3z =5\sqrt{14} or 2x+y-3z =-5\sqrt{14}

    2) Find the equation of the plane which passes through the points (1,6,1), (3,5,2) and (3,2,1)

    1,6,1), B3,5,2) and C3,2,1)" alt="A1,6,1), B3,5,2) and C3,2,1)" />

    n=AB*AC
    n=(OB-OA) * (OC-OA)
    n=((3i+5j+2k)-(i+6j+k)) * ((3i+2j+k)-(i+6j+k))
    n=(2i-j+k)*(2i-4j)

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


    n=i(0-4)-j(0-2)+k(-8+2)
    n=4i+2j+6k

    At plane (1,6,1)
    4(1)+2(6)-6(1)=d
    4+12-6=d
    d=10

    therefore equation is 4x+2y+6z=10<br />

    however book's answer is 2x+y-3z=5
    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    The following question i am having problems with:

    1) Find the equation of the plane normal to the vector 2i+j-3k

    2x+y-3z = d

    |d|=5
    d= 5 or -5

    so

    P = \frac{5}{\sqrt{14}} or \frac{-5}{\sqrt{14}}

    so 2x+y-3z = \frac{5}{\sqrt{14}} or 2x+y-3z = \frac{-5}{\sqrt{14}}

    book's answer says 2x+y-3z =5\sqrt{14} or 2x+y-3z =-5\sqrt{14}

    2) Find the equation of the plane which passes through the points (1,6,1), (3,5,2) and (3,2,1)

    1,6,1), B3,5,2) and C3,2,1)" alt="A1,6,1), B3,5,2) and C3,2,1)" />

    n=AB*AC
    n=(OB-OA) * (OC-OA)
    n=((3i+5j+2k)-(i+6j+k)) * ((3i+2j+k)-(i+6j+k))
    n=(2i-j+k)*(2i-4j)

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


    n=i(0-4)-j(0-2)+k(-8+2)
    n=4i+2j+6k

    At plane (1,6,1)
    4(1)+2(6)-6(1)=d
    4+12-6=d
    d=10

    therefore equation is 4x+2y+6z=10<br />

    however book's answer is 2x+y-3z=5
    P.S
    There is a common factor of 2 on the LHS of your equation.

    4x + 2y + 6z = 10

    2(2x + y + 3z) = 10

    2x + y + 3z = 5.


    So you are correct.
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  3. #3
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    oh ok thanks, any ideas about question 1)
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  4. #4
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    The first question sounds strange to me. All equations of the form

    2x+y-3z=d

    are equations for planes with 2i+j-3k as a normal vector, regardless of what d is. Is there some extra information, like a point the plane must pass through?

    Where did the expression |d|=5 come from? Was P meant to be d?

    As for question 2, notice that you have a slight disagreement with the book's answer. While you have 2x+y+3z=5, the book has 2x+y-3z=5 (so notice the -3 on the z).
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  5. #5
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    Quote Originally Posted by HappyJoe View Post
    Is there some extra information, like a point the plane must pass through?

    Where did the expression |d|=5 come from? Was P meant to be d?
    well the full description of the question is:

    Find the equation of the plane normal to the vector 2i+j-3k and a distance 5 from the origin.

    This equation of the plane is given by P = \frac{|d|}{\sqrt{a^2+b^2+c^2}}


    Quote Originally Posted by HappyJoe View Post
    As for question 2, notice that you have a slight disagreement with the book's answer. While you have 2x+y+3z=5, the book has 2x+y-3z=5 (so notice the -3 on the z).
    my bad, typo my answer is correct.
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  6. #6
    Member HappyJoe's Avatar
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    Ah, I see.

    You write that the equation of the plane is given by P = \frac{|d|}{\sqrt{a^2+b^2+c^2}}. You probably didn't mean this. This expression is the distance from the plane to the origin.

    You are being told that this distance is 5, so that P=5. You already have a, b, and c:

    a = 2,
    b = 1,
    c = -3.

    So insert these pieces of information into the formula for the distance from the plane to the origin:

    5 = \frac{|d|}{\sqrt{2^2+1^2+(-3)^2}},

    or

    |d| = 5\sqrt{14},

    giving you the two required solutions.
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  7. #7
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    yes, but isn't it \frac{5}{\sqrt{14}}, how did you get 5\sqrt{14} that is what i don't understand
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  8. #8
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    Check that last bit of algebra carefully,

     5 = \frac{|d|}{\sqrt{14}}

     5\sqrt{14} = |d|

     |d| = 5\sqrt{14}
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  9. #9
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    oh =__+
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