Planes

Key Questions

  • What can be said about two planes that are perpendicular to the same line?

    It can be said that those planes are parallel.

    You are here in a typical "sandwich" situation:

    enter image source here
    (Picture source: http://school.discoveryeducation.com/clipart/clip/sandwich.html)

    Your planes are the slices of bread. Both of them are perpendicular to the same line (the toothpick).

    As in the sandwich the first slice of bread must form an angle of 90° all around the toothpick; the second slice also must have the same property.
    As in the sandwich the two slices of bread will never "bend" to meet each other otherwise you'll have the "tortilla" effect!!!!
    enter image source here
    (Picture source: http://www.dunbarsystems.com/baked-unbaked-products/tortillas/default.html)

    Wataru · · Sep 28 2014
  • How many points are needed to establish a plane?

    Answer:

    Any three noncollinear points make up a plane.

    Explanation:

    Two distinct points, "A" and "B", determine a unique line in space.

    A third point "C" not on this line determines a unique plane that we can denote as "ABC".

    www.onemathematicalcat.org

    Ernest Z. · 2 · Aug 24 2015
  • How do I find the Cartesian equation of a plane containing two given lines?

    Answer:

    Given two lines, they define a plane only if they are:
    parallels non coincident or non coincident intersecting.

    Explanation:

    Given two lines, they define a plane only if they are:
    parallels non coincident or non coincident intersecting. If they are parallels, taking a point in one of them and the support of the other we can define a plane. If they intersect, with the normal to both directions and their intersection point, a plane can also be constructed.

    1) Parallel

    r_1->p = p_1 + lambda_1 vec v
    r_2->p = p_2 + lambda_2 vec v

    p_1^0 = p_1 + lambda_1^0 vec v
    p_2^0 = p_2 + lambda_2^0 vec v

    vec w = vec v xx (p_1^0-p_2^0) =vec v xx (p_1-p_2)

    the plane equation is given by

    Pi_1-><< vec w, (p-p_1)>> = << vec w, (p-p_2)>> = 0

    here p = {x,y,z}

    2) Intersecting

    Solve for the intersection point p_0

    p_0 = p_1 + lambda_1 vec v_1 = p_2 + lambda_2 vec v_2

    Once obtained p_0 the plane is built as follows

    vec w = vec v_1 xx vec v_2

    Pi_2 -> << vec w, (p-p_0) >> = 0

    Cesareo R. · 1 · Jun 14 2016
  • How do I find the Cartesian equation of a plane containing three given points?

    The Cartesian equation of a plane P is ax + by + cz + d = 0, where a, b, c are the coordinates of the normal vector vec n = ( (a), (b), (c) )

    Let A, B and C be three noncolinear points, A, B, C in P
    Note that A, B and C define two vectors vec (AB) and vec (AC) contained in the plane P. We know that the cross product of two vectors contained in a plane defines the normal vector of the plane.

    Now, we can use an example to illustrate the solution. Assume that the coordinates of the three points are the following:
    A(1,2,3), B(-2,1,0) and C(0,3,2)

    From the coordinates of the points A, B and C we can find the vectors vec (AB) and vec (AC):

    vec (AB) = (x_b - x_a)*hat i + (y_b - y_a)*hat j + (z_b - z_a)*hat k
    vec (AC) = (x_c - x_a)*hat i + (y_c - y_a)*hat j + (z_c - z_a)*hat k
    where hat i, hat j and hat k are the unit vectors on the cartesian axes of coordinates Ox, Oy and Oz.

    After plugging in the values of the coordinates, we have:
    vec (AB) = (-2 - 1)*hat i + (1-2)*hat j + (0-3)*hat k
    So, vec (AB) = -3 hat i - hat j -3 hat k

    vec (AC) = (0-1)*hat i + (3-2)* hat j + (2-3)*hat k.
    So, vec (AC) = -hat i + hat j - hat k

    Next, we'll find the normal vector from the cross product of vec (AB) and vec (AC)

    vec n = vec (AB) x vec (AC) = | (hat i,hat j,hat k),(-3,-1,-3),( -1,1,-1)| = 4 hat i - 4 hat k

    Therefore, the equation of the plane is 4x-4z+d=0. In order to find d, we can plug in the coordinates of point A:
    d= 4z - 4x => d = 4*3 - 4*1 = 8

    So, the cartesian equation of the plane is 4x-4z+8=0.

    Alex Boroda · 6 · Jun 5 2015
  • What is the general form of the equation of a plane?

    If a plane passes through the point (x_0,y_0,z_0) and has the normal vector (a,b,c), then an equation of the place can be written as:

    a(x-x_0)+b(y-y_0)+c(z-z_0)=0

    I hope that this was helpful.

    Wataru · 2 · Sep 26 2014

Questions

3-D Cartesian Coordinate System