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Find the Cartesian equation for a plane parallel to a given vector and passing through the intersection of given planes

Let M be the plane which is parallel to \mathbf{j} and which passes through the intersection of the planes

    \[ x + 2y + 3z = 4 \qquad \text{and} \qquad 2x + y + z = 2. \]

Find a Cartesian equation for M.


First, we find the intersection of the two planes. This is the set of points which simultaneously satisfy

    \begin{align*}  x + 2y + 3z &= 4 \\  2x + y + z &= 2. \end{align*}

From the first equation we have

    \[ y = \frac{1}{2} (4 - x - 3z). \]

Plugging this into the second equation we have

    \[ 2x + 2 - \frac{1}{2} x - \frac{3}{2} z + z = 2 \quad \implies \quad z = 3x. \]

This then gives us y = 2-5x. So the line is the set of points (x, 2-5x, 3x) where x is arbitrary. Thus,

    \[ L = \{ (0,2,0) + t(1,-3,3) \}. \]

The plane must then contain this line and be parallel to the vector (0,1,0). Since it is parallel to (0,1,0) its normal must be perpendicular to (0,1,0), thus,

    \[ M = \{ (0,2,0) + s(1,-5,3) + t(b_1, b_2, b_3) \} \]

and we must have

    \[ (0,1,0) \cdot (1,-5,3) \times (b_1, b_2, b_3) = 0 \quad \implies \quad b_3 = 3b_1. \]

Therefore,

    \[ M = \{ (0,2,0) + s(1,-5,3) + t(1,0,3) \}. \]

This has the Cartesian equation

    \[ 3x - z = 0. \]

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

    \[ X(t) = (1,2,3) +t(1,-2,1). \]

Prove that three planes with independent normal vectors intersect in exactly one point

Prove that three planes with normal vectors which are linearly independent intersect in exactly one point.


Proof. Let the normals of the three planes be given by

    \begin{align*}  N_1 &= (a_1, b_1, c_1) \\   N_2 &= (a_2, b_2, c_2) \\  N_3 &= (a_3, b_3, c_3). \end{align*}

Then, the Cartesian equations of the three planes are given by

    \begin{align*}  a_1 x + b_1 y + c_1 z &= d_1 \\  a_2 x + b_2 y + c_2 z &= d_2 \\  a_3 x + b_3 y + c_3 z &= d_3. \end{align*}

Since the normals are independent we know that they span the zero vector uniquely (by definition of independence). BY Theorem 12.7 (page 463 of Apostol) this implies that they span every vector in \mathbb{R}^3. Hence, the vector equation x N_1  + y N_2 + z N_3 = (d_1, d_2, d_3) has a unique solution (x,y,z). Therefore, the system of equations has a unique solution. Hence, there is exactly one point (x,y,z) on the intersection of three planes with linearly independent normal vectors. \qquad \blacksquare

Find all points on the intersection of three given planes

Consider three planes given by the Cartesian equations

    \[ 3x + y + z = 5, \qquad 3x + y + 5z = 7, \qquad x - y + 3z = 3. \]

Find all of the points which are on the intersections of these three planes.


The points on the intersection must satisfy the system of equations

    \begin{align*}  3x + y + z &= 5, \\  3x + y + 5z &= 7, \\  x - y + 3z &= 3. \end{align*}

To find the points satisfying these equations we row-reduce the coefficient matrix of the system (if you don’t have any linear algebra, then you can use Gaussian elimination the long way),

    \begin{align*}  \begin{pmatrix*}[r] 3 & 1 & 1 & 5 \\ 3 & 1 & 5 & 7 \\ 1 & -1 & 3 & 3 \end{pmatrix*} &\rightarrow   \begin{pmatrix*}[r] 0 & 4 & -8 & -4 \\ 0 & 4 & -4 & -2 \\ 1 & -1 & 3 & 3 \end{pmatrix*} \\[9pt]  &\rightarrow \begin{pmatrix*}[r] 1 & -1 & 3 & 3 \\ 0 & 1 & -2 & -1 \\ 0 & 2 & -2 & -1 \end{pmatrix*} \\[9pt]  &\rightarrow \begin{pmatrix*}[r] 1 & 0 & 1 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & 2 & 1 \end{pmatrix*} \\[9pt]  & \rightarrow \begin{pmatrix*}[r] 1 & 0 & 0 & \frac{3}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} \end{pmatrix*}. \end{align*}

Therefore, we have x = \frac{3}{2}, y = 0, and z = \frac{1}{2}. So, the only point on the intersection is the point \left \{ \left( \frac{3}{2}, 0 \frac{1}{2} \right) \right\}.