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Math Help - Scalar Equation of a Plane

  1. #1
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    Scalar Equation of a Plane

    Please show me step by step solution!!!


    Question 1:
    Determine whether the following pairs of planes are coincident, parallel and distinct or neither.

    a) x + 3y - z - 2 = 0 and  2x + 6y - 2z - 8 = 0

    TEXTBOOK ANSWER: parallel and distinct

    I don't know how to put scalar equations into parametric equations...and how to find the direction vector...

    -----------------

    Question 2:
    The angle between 2 planes is defined as the angle between their normals. Determine the angle (0 ≤ θ ≤ 90), to the nearest degree, between the given planes.

    a) 2x + 3y - z + 9 = 0 and x + 2y + 4 = 0

    TEXTBOOK ANSWER: 17 degrees

    For #2, do you use the cross product?

    sinθ = (|u x v|) / (|u||v|)

    Again...for this question...I don't know how to put the scalar equation into parametric equation or find the direction vector..I know you use the algebraic cross product to find the normal vector, but I don't know what method to use to find the direction vector...I also know for a line, d = (d1, d2) and n = (-d2, d1)...


    Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Macleef View Post
    Please show me step by step solution!!!


    Question 1:
    Determine whether the following pairs of planes are coincident, parallel and distinct or neither.

    a) x + 3y - z - 2 = 0 and  2x + 6y - 2z - 8 = 0

    TEXTBOOK ANSWER: parallel and distinct

    I don't know how to put scalar equations into parametric equations...and how to find the direction vector...

    -----------------

    Question 2:
    The angle between 2 planes is defined as the angle between their normals. Determine the angle (0 ≤ θ ≤ 90), to the nearest degree, between the given planes.

    a) 2x + 3y - z + 9 = 0 and x + 2y + 4 = 0

    TEXTBOOK ANSWER: 17 degrees

    For #2, do you use the cross product?

    sinθ = (|u x v|) / (|u||v|)

    Again...for this question...I don't know how to put the scalar equation into parametric equation or find the direction vector..I know you use the algebraic cross product to find the normal vector, but I don't know what method to use to find the direction vector...I also know for a line, d = (d1, d2) and n = (-d2, d1)...


    Thanks.
    For the first one we need to find the normal vector to the plane.

    if the plane is in the form Ax+By+Cz+D=0 then the normal vector is Ai+Bj+Ck

    We also know that two planes are parallel iff their normal vectors are parallel.

    So the two normal vectors are i+3j-k and 2i+6j-2k these two vectors are parallel (the second is 2 times the first)

    so the planes are parallel since their normal vectors are parallel.

    The planes are distinct becuase they are not multiples of each other.

    For the second I would use the Dot Product

    the angle between two vectors is

    \theta=\cos^{-1}\left( \frac{u\cdot v}{|u||v|} \right)

    the normals are

    u=2i+3j-k \mbox{ and } v=i+2j+4k

    doing some computations

    u \cdot v = (2)(1)+(3)(2)+(-1)(4)=4

    u=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}

    v=\sqrt{1^2+2^2+4^2}=\sqrt{21}

    plugging into above we get

    \theta=\cos^{-1}\left( \frac{4}{\sqrt{14} \cdot \sqrt{21}}\right) \approx 76^{\circ}

    so the angle is 90^{\circ}-76^{\circ}=14^{\circ}

    P.s I think that 13.5 is a better answer
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