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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 there is exactly one plane through a line and a point not on the line

Let L be a line and P a point not on L. Prove that there is exactly one plane through P containing every point of L.


Proof. Existence follows from a previous exercise (Section 13.8, Exercise #12) by letting Q,R be any two distinct points on L, then there is a plane M containing P,Q,R and every point on the line L through Q and R.
Now, to prove there is only one such plane we apply Theorem 13.10 to the points P,Q,R to conclude the plane that contains P,Q,R is unique. But, if we choose any other two points on L, the plane M must still contain Q and R (since it contains every point on L); hence, it is the same plane M. Thus, there is exactly one plane M containing P and L. \qquad \blacksquare

Find a Cartesian equation for a plane through a point containing a given line

Let L be the line through the point (1,2,3) and parallel to the vector (1,1,1) and let (2,3,5) be a point not on L. Find a Cartesian equation for the plane M which passes through (2,3,5) and entirely contains L.


The line L is the set of points

    \[ L = \{ (1,2,3) + t(1,1,1) \} \quad \implies \quad P = (1,2,3), \ Q = (2,3,4) \in L. \]

Then, the plane M is the set of points

    \begin{align*}  M &= \{ (2,3,5) + s((1,2,3) - (2,3,5)) + t((2,3,4) - (2,3,5)) \} \\  &= \{ (2,3,5) + s(-1,-1,-2) + t(0,0,-1) \}. \end{align*}

Then, to get the Cartesian equation, we have

    \begin{align*}  x &= 2-s \\  y &= 3-s \\  z &= 5 - 2s - t. \end{align*}

The first two equations give s = 2-x and so y = 3-2 + x. This implies x - y = -1, and z is arbitrary. So, the plane is described by the equation x-y = -1.

Determine whether a line is parallel to given planes

We say that a line L in the direction of a vector X is parallel to a plane M if X is parallel to M. Consider the line L through the point (1,1,1) and parallel to the vector (2,-1,3). Determine whether L is parallel to the following planes.

  1. The plane through the point (1,1,-2) and spanned by (2,1,3) and (\frac{3}{4}, 1,1).
  2. The plane through the points (1,1,-2), \ (3,5,2), \ (2,4,-1).
  3. The plane determined by the Cartesian equation x + 2y + 3z = -3.

  1. This asks if X = (2,-1,3) is in the span of \{ (2,1,3), (\frac{3}{4}, 1,1) \}, i.e., does there exist s,t such that

        \begin{align*}  2s + \frac{3}{4}t &= 2 \\  s + t &= -1 \\  3s + t &= 3. \end{align*}

    From the second equation we have s = -1-t. Then, from the first, 2 = 2 - 2t + \frac{3}{4}t which implies

        \[ t = -\frac{16}{5} \qquad \implies \qquad s = \frac{11}{5}. \]

    But then,

        \[ 3s + t = 3 \left( \frac{11}{5} \right) - \frac{16}{5} = \frac{17}{5} \neq 3. \]

    Thus, there is no solution, so the line is not parallel to the plane.

  2. The plane through the points (1,1,-2), \ (3,5,2), \ (2,4,-1) is the set of points

        \[ M = \{ (1,1,-2) + s((3,5,2) - (1,1,-2)) + t((2,4,-1) - (1,1,-2)) \} = \{ (1,1,-2) + s(2,4,4) + t(1,3,1) \}. \]

    For X to be in the span of \{ (2,4,4), (1,3,1) \} we must have s,t such that

        \begin{align*}  2s + t &= 2 \\  4s + 3t &= -1 \\  4s + t &= 3. \end{align*}

    From the first equation we have t = 2 - 2s. Then from the second we have 4s + 6 - 6s = -1 which implies

        \[ s = \frac{7}{2} \qquad \implies \qquad t = -5. \]

    But then,

        \[ 4s + t = 14 -5 = 9 \neq 3. \]

    Hence, L is not parallel to M.

  3. The plane with Cartesian equation x + 2y + 3z = -3 is the set of points

        \[ M = \{ (x,y,z) \mid x+2y + 3z = -3 \}. \]

    The points (-3,0,0), \ (0, -\frac{3}{2}, 0), \ (0,0,-1) are all in M. So,

        \begin{align*}  M &= \{ (-3,0,0) + s((0, -\frac{3}{2}, 0) - (-3,0,0)) + t((0,0,-1) - (-3,0,0)) \} \\  &= \{ (-3,0,0) + s(3, -\frac{3}{2}, 0) + t(3,0,-1) \}. \end{align*}

    Thus, we ask if X= (2,-1,3) is in the span of \{ (3,-\frac{3}{2},0), (3,0,-1) \}. This requires that there exist s,t such that

        \begin{align*}  3s + 3t &= 2 \\  -\frac{3}{2}s &= -1 \\  -t &=3. \end{align*}

    But, this fails since the second and third equations implies s = \frac{2}{3} and t = -3. But then

        \[ 3s+3t = 2 - 9 = -7 \neq 2. \]

    Hence, this line is not parallel to the plane.

Prove that there is exactly one point in the intersection of a given line and given plane

Consider the line L through the point (1,1,1) and parallel to the vector (2,-1,3). Then, let M be the plane through the point (1,1,-2) and spanned by the vectors (2,1,3) and (0,1,1). Prove that there is exactly one point in L \cap M, and determine the point.


Proof. First, we have that L and M are the sets of points

    \begin{align*}  L &= \{ (1,1,1) + r(2,-1,3) \} = \{ (1+2r, 1-r, 1+3r) \} \\  M &= \{ (1,1,-2) + s(2,1,3) + t(0,1,1) \} = \{ (1+2s, 1+s+t, -2+3s+t) \}. \end{align*}

Then, the points (x,y,z) in M are those which satisfy

    \[ x = 1+2s, \qquad y = 1+s+t, \qquad z = -2 + 3s + t. \]

This gives us the Cartesian equation

    \[ x+y-z = 4. \]

The points (x,y,z) on L are those such that

    \[ x = 1+2s, \qquad y = 1-s, \qquad z = 1+3s. \]

Thus, if (x,y,z) \in M \cap L then we have

    \begin{align*}  && (1+2s) + (1-s) - (1+3s) &= 4 \\  \implies && -2s &= 3 \\  \implies && s &= -\frac{3}{2}. \end{align*}

Hence,

    \[ x = -2, \qquad y = \frac{5}{2}, \qquad z = -\frac{7}{2} \]

is the unique point in L \cap M. \qquad \blacksquare

Prove that the intersection of two non-parallel lines is either empty or contains exactly one point

Let L(P;A) and L(Q;B) be two lines in \mathbb{R}^n which are not parallel. Prove that the intersection is either empty or contains exactly one point.


Proof. Since L(P;A) and L(Q;B) are not parallel we know A \neq B. Now, suppose there exist two distinct points X_1, X_2 \in L(P;A) \cap L(Q;B). This means there exist real numbers t_1, t_2, s_1, s_2 such that

    \begin{align*}  X_1 &= P + t_1 A & X_1 &= Q + s_1 B \\  X_2 &= P + t_2 A & X_2 &= Q + s_2 B. \end{align*}

Since X_1, X_2 are distinct points we also know t_1 \neq t_2 and s_1 \neq s_2. Then we have

    \begin{align*}  P + t_1 A &= Q + s_1 B & \implies && P-Q &= s_1 B - t_1 A \\  P + t_2 A &= Q + s_2 B & \implies && P-Q &= s_2 B - t_2 A. \end{align*}

But then

    \begin{align*}  s_1 B - t_1 A = s_2 B - t_2 A && \implies && (t_2 - t_1)A &= (s_2 - s_1)B \\  && \implies && A &= \frac{s_2 - s_1}{t_2 -t_1} B &(t_2 - t_1 \neq 0 \text{ since } t_2 \neq t_1) \\  && \implies && L(P;A) &= L(P';B) \end{align*}

for some P'. Thus, the lines are parallel, contradicting our assumption that the lines are not parallel. Hence, the intersection contains at most one point. \qquad \blacksquare

Prove some properties about a line and a point not on the line

Let Q be a point not on the line L(P;A) in \mathbb{R}^n.

  1. Consider the function

        \[ f(t) = \lVert Q - X(t) \rVert^2, \qquad \text{where} \qquad X(t) = P + tA. \]

    Prove f(t) is a quadratic polynomial in t and that there is a unique value of t, say t_0, at which this polynomial takes on its minimum.

  2. Prove that Q - X(t_0) is orthogonal to A.

  1. Proof. First, we compute

        \begin{align*}  \lVert Q - X(t) \rVert^2 &= (Q - X(t)) \cdot (Q - X(t)) \\  &= (Q - P - tA) \cdot (Q - P - tA) \\  &= (Q-P)\cdot (Q-P) - t \big( (Q-P)\cdot A + A \cdot (Q-P) \big) + t^2 A \cdot A. \end{align*}

    But, (Q-P)\cdot (Q-P), \ (Q-P) \cdot A, \ A \cdot (Q-P), A \cdot A are all just real scalars (by definition of the dot product); therefore,

        \[ f(t) = c_1 - c_2 t + c_3 t^2 \]

    for scalars c_1, c_2, c_3 given by

        \[ c_1 = (Q-P) \cdot (Q-P), \quad c_2 = 2A \cdot (Q-P), \quad c_3 = A \cdot A. \]

    Then,

        \[ f'(t) = -c_2 + 2c_3 t \]

    which implies f(t) has a unique minimum at t_0 = \frac{c_2}{2c_3}. \qquad \blacksquare

  2. Proof. We compute the dot product,

        \begin{align*}  (Q - X(t_0)) \cdot A &= (Q - P - t_0 A) \cdot A \\  &= (Q-P) \cdot A - t_0 A \cdot A. \end{align*}

    Then we have

        \[ t_0 = \frac{c_2}{2c_3} = \frac{2 A \cdot (Q-P)}{2A \cdot A}. \]

    Therefore,

        \[ (Q-X(t_0)) \cdot A = (Q-P) \cdot A - \frac{2A \cdot (Q-P)}{2A \cdot A} \cdot A \cdot A = 0. \qquad \blacksquare\]