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Prove that the intersection of two planes which are not parallel is a line

Prove that if M and M' are two planes which are not parallel then they intersect in a line.


Proof. Let the Cartesian equations of M and M' be given by

    \[ ax + by  + cz + d = 0, \qquad \text{and} \qquad a'x + b'y + c'z + d' = 0 \]

respectively. Then, the intersection is given by the common solutions (x,y,z) of these two equations. Since M and M' are not parallel, we know they do not have the same normal vector so that (a,b,c) \neq t (a', b', c') for all t. Further, since the normals are nonzero, we know each equation has at least one nonzero coefficient. Without loss of generality, let a \neq 0. Then,

    \[ x = \frac{-by - cz - d}{a}. \]

Substituting into the Cartesian equation for M' we have

    \[ (b'a - ba') y + (c'a - ca') z + (d'a - da') = 0 \]

is the set of solutions for the points on M \cap M'. But, we know at least one of (b'a - ba') or (c'a - ca') is nonzero, otherwise (a,b,c) = t(a',b',c'). Hence, we have the equation for a line. Therefore, M \cap M' is a line. \qquad \blacksquare

Prove a formula for the distance between a plane determined by three points and a point

  1. If a plane is determined by the points A,B,C prove that the distance from a point Q to this plane is given by

        \[ \frac{|(Q-A) \cdot (B-A) \times (C-A)|}{\lVert (B-A) \times (C-A) \rVert}. \]

  2. Using part (a) compute the distance in the case

        \[ Q = (1,0,0), \quad A = (0,1,1), \quad B = (1,-1,1), \quad C = (2,3,4). \]


  1. Proof. We know the distance from a plane containing a point P to a point Q not on the plane is given by the formula

        \[ \frac{(Q-P) \cdot N}{\lVert N \rVert}. \]

    Since the plane through A,B,C is the set of points

        \[ M = \{ A + s(B-A) + t(C-A) \} \]

    we have N = (B-A) \times (C-A). Thus, the distance from Q to M is

        \[ d = \frac{|(Q-A) \cdot (B-A) \times (C-A)|}{\lVert (B-A) \times (C-A) \rVert}. \]

  2. For the given points Q,A,B,C we have

        \[ d = \frac{|((1,0,0) - (0,1,1)) \cdot ((1,-1,1) - (0,1,1)) \times ((2,3,4) - (0,1,1))|}{\lVert ((1,-1,1) - (0,1,1)) \times ((2,3,4) - (0,1,1)) \rVert} = \frac{|-6+3-6|}{9} = 1. \]

Find a Cartesian equation for a plane parallel to a given plane and equidistant from a given point

Consider a plane given by the equation

    \[ 2x - y + 2z + 4 = 0. \]

Find the Cartesian equation for a plane parallel to this one and the same distance as this plane from the point (3,2,-1).


Since the requested plane is parallel to the given plane we know that they must have the same normal vector, N = (2,-1,2). Therefore, the Cartesian equation of the requested plane is of the form

    \[ 2x - y + 2z + d = 0. \]

From the previous exercise (Section 13.17, Exercise #19) we know the distance from (3,2,-1) to a plane ax + by + cz + d = 0 is given by the formula

    \[ \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2+b^2+c^2}}. \]

Therefore, the distance from the given plane to the point (3,2,-1) is

    \[ \frac{|6-2-2+4|}{\sqrt{9}} = 2. \]

Since the distance from the point to the requested plane must be the same we must have

    \[ \frac{|6-2-2+d|}{3} = 2 \quad \implies \quad d = -8. \]

(Since the d = 4 solution belongs to the other plane.)

Prove some equations about distances between points and planes

  1. Prove that the distance from the point (x_0, y_0, z_0) to the plane

        \[ ax + by + ca + d = 0 \]

    is given by the formula

        \[ \frac{| ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}. \]

  2. Find the point P on the plane 5x - 14y + 2z + 9 = 0 which is nearest to the point Q = (-2,15,-7).

  1. Proof. By Theorem 13.6 (page 476 of Apostol) we know that the distance from a point to a plane is given by

        \begin{align*}  d &= \frac{| N \cdot P |}{\lVert N \rVert} \\[9pt]  &= \frac{ | (a,b,c) \cdot (x_0, y_0, z_0) + d|}{\sqrt{a^2+b^2+c^2}} \\[9pt]  &= \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2+b^2+c^2}}. \qquad \blacksquare \end{align*}

  2. A normal to the plane is given by N = (5,-14,2). So, P \cdot N = 9 for any point P \in M. Further, the distance from M to a point Q not on M is minimal when X = tN where

        \[ t = \frac{P \cdot N}{N \cdot N} = \frac{9}{225} = \frac{1}{25}. \]

    Thus,

        \[ X = \frac{1}{25} (5,-14,2). \]

    Naming X to be the point P we have

        \[ P = \frac{1}{25} (5,-14,2). \]

Prove that the intersection of a line and plane which are not parallel contains exactly one point

Prove that the intersection of a line and a plane such that the line is not parallel to the plane contains one and only one point.


Proof. Denote the line by L and the plane by M. Let M be the set of points

    \[ M = \{ P + sB + tC \}. \]

Since L is not parallel to M w know that its direction vector A is not in the span of B and C. Further, by definition of a plane, we know the vectors B and C are linearly independent. Hence, A,B,C are linearly independent. Then, any point X = (x_1, x_2, x_3) in the intersection L \cap M must be a solution to the system of equations

    \begin{align*}  a_1 x_1 + a_2 x_2 + a_3 x_3 &= d_1 \\  b_1 x_1 + b_2 x_2 + b_3 x_3 &= d_2 \\  c_1 x_1 + c_2 x_3 + c_3 x_3 &= d_3. \end{align*}

By the linear independence of A,B,C we know this system has exactly one solution (x_1, x_2, x_3). Hence, L  \cap M contains exactly one point. \qquad \blacksquare

Find the parametric equation for a line through a point and parallel to two planes

We say that a line is parallel to a plane if the direction vector of the line is parallel to the plane. Let L be the line containing the point (1,2,3) and parallel to the planes

    \[ x + 2y + 3z = 4, \qquad 2x + 3y + 4z = 5. \]

Find a vector parametric equation for L.


The normal vectors of the planes are N_1 = (1,2,3) and N_2 = (2,3,4). So, the direction vector A = (a_1, a_2, a_3) of L will be perpendicular to both of these,

    \begin{align*}  N_1 \cdot A &= 0 & \implies && a_1 + 2a_2 + 3a_3 &= 0 \\  N_2 \cdot A &= 0 & \implies && 2a_1 + 3a_2 + 4a_3 &= 0. \end{align*}

From the first equation we have x = -2y - 3z. Plugging this into the second equation we obtain y = -2z, which then gives us x = z. Since z is arbitrary, we take z = 1 to obtain a direction vector A = (1,-2,1). Therefore, the vector parametric equation for the line is

    \[ X(t) = (1,2,3) +t(1,-2,1). \]

Compute the volume of a tetrahedron with given vertices

Consider the tetrahedron with vertices at the origin and at the points where the plane

    \[ x + 2y + 3z = 6 \]

intersects the coordinate axes. Compute the volume of this tetrahedron.


First, the intercepts of the plane are given by (6,0,0), (0,3,0), (0,0,2). Then from a previous exercise (Section 13.14, Exercise #13) we know that the volume of a tetrahedron with vertices A,B,C,D is

    \[ V = \frac{1}{6} | (B-A) \cdot (C -A) \times (D-A) |. \]

Letting A = (0,0,0), \ B = (6,0,0), \ C = (0,3,0), \ D = (0,0,2) we have

    \[ V = \frac{1}{6} | (6,0,0) \cdot (0,3,0) \times (0,0,2) | = \frac{1}{6} | (6,0,0) \cdot (6,0,0) | = 6. \]

Find a Cartesian equation for a plane through a point with normal vector making given angles

Let M be the plane whose normal vector N makes angles \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3} with the unit coordinate vectors and which contains the point (1,1,1). Find a Cartesian equation for the plane.


Since the normal vector to the plane makes angles \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3} with the unit coordinate vectors, we have

    \begin{align*}  N \cdot \mathbf{i} &= \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} & \implies && x_1 &= \frac{1}{2} \\  N \cdot \mathbf{j} &= \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} & \implies && x_2 &= \frac{\sqrt{2}}{2} \\  N \cdot \mathbf{k} &= \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} & \implies && x_3 &= \frac{1}{2}. \end{align*}

Hence, N = (1, \sqrt{2}, 1). So, the plane has a Cartesian equation of the form

    \[ x + \sqrt{2} y + z = d. \]

Since it contains (1,1,1) we have d = 1 + \sqrt{2} + 1 = 2 + \sqrt{2}. Therefore, a Cartesian equation of the plane is

    \[ x + \sqrt{2} y + z = 2 + \sqrt{2}. \]

Determine properties of a point whose movement in space is determined by a vector parametric equation

Consider a point moving in space with position at time t given by

    \[ X(t) = (1-t) \mathbf{i} + (2-3t) \mathbf{j} + (2t-1) \mathbf{k}. \]

  1. Prove that the motion of the point is along a line.
  2. Find a vector parallel to this line.
  3. Find the time t at which the point intersects the plane with Cartesian equation 2x+3y+2z+1 = 0.
  4. What is the Cartesian equation for the plane parallel to the plane in part (c) which contains the point X(3)?
  5. Let M be the plane perpendicular to L containing the point X(2). Find a Cartesian equation for M.

  1. Proof. We use the formula for the motion of the particle to compute

        \begin{align*}  X(t) &= (1-t) \mathbf{i} + (2-3t) \mathbf{j} + (2t-1) \mathbf{k} \\  &= \mathbf{i} + 2 \mathbf{j} - \mathbf{k} + t(-\mathbf{i} - 3\mathbf{j} + 2 \mathbf{k}) \\  &= (1,2,-1) + t(-1,-3,2).  \end{align*}

    This is the parametric equation for the line through (1,2,-1) parallel to the vector (-1,-3,2). \qquad \blacksquare

  2. From part (a) we have a vector N parallel to L given by N = (1,3,-2).
  3. First, the line on which the point moves is the set of points

        \[ L = \{ (1,2,-1) + t(-1,-3,2) \} = \{ (1-t, 2-3t, -1+2t) \}. \]

    So, to find the intersection with the plane 2x + 3y + 2z + 1 = 0 we compute

        \begin{align*}  && 2(1-t) + 3(2-3t) + 2(-1+2t) + 1 &= 0 \\  \implies && 2 - 2t + 6 - 9t -2 + 4t + 1 &= 0 \\  \implies && 7 -7t &= 0 \\  \implies && t &= 1. \end{align*}

  4. First, we have

        \[ X(3) = (1,2,-1) + 3(-1,-3,2) = (1,2,-1) + (-3,-9,6) = (-2,-7,5). \]

    Since we know the plane is parallel to the one in part (c) it has a Cartesian equation of the form

        \[ 2x + 3y + 2z = d. \]

    We compute d = 2(-2) + 3(-7) + 2(5) = -15. Hence, the plane has Cartesian equation

        \[ 2x + 3y + 2z + 15 = 0. \]

  5. Since the plane is perpendicular to the line L we know that it has a normal vector in the same direction as L, so N = (1,3,-2) (from part (b)). Thus, we have a Cartesian equation of the form

        \[ x + 3y - 2z = d. \]

    Since the point

        \[ X(2) = (1,2,-1) + 2(-1,-3,2) = (1,2,-1) + (-2, -6, 4) = (-1, -4, 3) \]

    is on the plane we have d = -1 + 3(-4) - 2(3) = -19. Therefore, the plane is given by

        \[ x + 3y - 2z + 19 = 0. \]