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Thread: 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

    $\displaystyle 2x+y-3z = d$

    $\displaystyle |d|=5$
    $\displaystyle d= 5 $ or $\displaystyle -5$

    so

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

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

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

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

    $\displaystyle A1,6,1), B3,5,2) and C3,2,1)$

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

    $\displaystyle
    \begin{bmatrix}
    i & j & k \\
    2 & -1 & 1 \\
    2 & -4 & 0 \\
    \end{bmatrix}$


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

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

    therefore equation is $\displaystyle 4x+2y+6z=10
    $

    however book's answer is $\displaystyle 2x+y-3z=5$
    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

    $\displaystyle 2x+y-3z = d$

    $\displaystyle |d|=5$
    $\displaystyle d= 5 $ or $\displaystyle -5$

    so

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

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

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

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

    $\displaystyle A1,6,1), B3,5,2) and C3,2,1)$

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

    $\displaystyle
    \begin{bmatrix}
    i & j & k \\
    2 & -1 & 1 \\
    2 & -4 & 0 \\
    \end{bmatrix}$


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

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

    therefore equation is $\displaystyle 4x+2y+6z=10
    $

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

    $\displaystyle 4x + 2y + 6z = 10$

    $\displaystyle 2(2x + y + 3z) = 10$

    $\displaystyle 2x + y + 3z = 5$.


    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
    Member HappyJoe's Avatar
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    The first question sounds strange to me. All equations of the form

    $\displaystyle 2x+y-3z=d$

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

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

    As for question 2, notice that you have a slight disagreement with the book's answer. While you have $\displaystyle 2x+y+3z=5$, the book has $\displaystyle 2x+y-3z=5$ (so notice the $\displaystyle -3$ on the $\displaystyle z$).
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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 $\displaystyle |d|=5$ come from? Was $\displaystyle P$ meant to be $\displaystyle d$?
    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 $\displaystyle P = \frac{|d|}{\sqrt{a^2+b^2+c^2}}$


    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 $\displaystyle 2x+y+3z=5$, the book has $\displaystyle 2x+y-3z=5$ (so notice the $\displaystyle -3$ on the $\displaystyle z$).
    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 $\displaystyle P = \frac{|d|}{\sqrt{a^2+b^2+c^2}}$. 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 $\displaystyle P=5$. You already have $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$:

    $\displaystyle a = 2,$
    $\displaystyle b = 1,$
    $\displaystyle c = -3.$

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

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

    or

    $\displaystyle |d| = 5\sqrt{14},$

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

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

    $\displaystyle 5\sqrt{14} = |d| $

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