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Math Help - Scalar equation of a plane

  1. #1
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    Scalar equation of a plane

    Find the scalar equation of the plane that is perpendicular to the plane x+2y+4=0, contains the origin, and whose normal makes an angle of 30 with the z-axis.

    attempt:

    d1= (1,2,0)
    P(0,0,0)
    cos(theta)=z/|normal|
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  2. #2
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    Wow . . . what a stupid blunder!

    Sorry, everyone . . .

    Last edited by Soroban; April 13th 2008 at 08:07 AM.
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  3. #3
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    "Its normal vector, (a,b,c), is perpendicular to (1,2,4)"

    Wouldnt it be perpendicular to (1,2,0)? 4 is the D value. thnks.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, diet.pepsi!

    I thought I had a good approach . . . then I hit a wall.


    [size=3]The general equation of a plane is: . ax + by + cz + d \:=\:0

    Since it contains the origin: . ax + by + cz \:=\:0

    Its normal vector, \vec{n} \,=\,\langle a,b,c\rangle, is perpendicular to \langle1,2,4\rangle

    The normal vector is \langle1,2,{\color{red}0}\rangle


    ...
    I nearly finished this question ... but I couldn't find my mistake.

    I've attached the sketches I made so far.
    Attached Thumbnails Attached Thumbnails Scalar equation of a plane-scalarplane.gif  
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  5. #5
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    Hello earboth,

    With what software did you sketch this ?
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  6. #6
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    Quote Originally Posted by diet.pepsi View Post
    Find the scalar equation of the plane that is perpendicular to the plane x+2y+4=0, contains the origin, and whose normal makes an angle of 30 with the z-axis.

    attempt:

    d1= (1,2,0)
    P(0,0,0)
    cos(theta)=z/|normal|
    As I assume Soroban tried to attempt this problem algebraically. I've choosen a completely different way to do this question:

    1. From the equation of the plane p_1 you know that the normal vector is \overrightarrow{n_1} = (1,2,0) . That means p_1 is perpendicular to the x-y-plane.

    2. I looked for a line which lies simultaneously in p_1 and the x-y-plane and took the line which connects the intersection points of p_1 and the x-axis and p_1 and the y-axis. The intersection points are A(-4,0,0) and B(0,-2,0).

    3. I took the line \vec r= t \cdot (1,2,0) as a kind of axis. I then "tilted" the x-y-plane (which is perpendicular to p_1!) about this "axis" by 30 to get p_2. If I tilt the x-y-plane by 30 then the normal vector of p_2 and the z-axis include an angle of 30.

    4. I took the vector \overrightarrow{AB} =(-4,2,0) and rotated it by 30 to get the vector \vec b = \left(-4,2,\sqrt{\frac{20}3}\right) (By the way: You probably have noticed that there must exist a second vector \vec b ' if you rotate the x-y-plane in the opposite direction)

    5. Now I know 2 vectors which span the plane p_2. Since p_2 has to contain the origin the parametric equation of p_2 is:

    p_2: \vec r = s \cdot (1,2,0) + t \cdot \left(-4,2,\sqrt{\frac{20}3}\right) ....... or ....... p_2: \left\{\begin{array}{lcr}x&=&s-4t \\ y&=&2s+2t \\ z&=&\sqrt{\frac{20}3} \cdot t \end{array}\right.

    6. The cross product of the 2 direction vectors which span p_2 is the normal vector of p_2:

    (1,2,0) \times \left(-4,2,\sqrt{\frac{20}3}\right) = \left(\frac43 \cdot \sqrt{15}\ ,\ -\frac23 \cdot \sqrt{15}\ ,\ 10 \right)

    And therefore the scalar equation of p_2 is:

    \boxed{\frac43 \cdot \sqrt{15} \cdot x - \frac23 \cdot \sqrt{15} \cdot y + 10 \cdot z = 0}

    7. I finally used derive 6.0 to draw the 2 planes and the normal vector. This program is able to rotate the 3-D-graphs about an axis. I took 2 screenshots to show how the planes and the normal are placed.
    Last edited by earboth; April 13th 2008 at 10:50 PM.
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