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Prove an if and only if condition for two lines to intersect in Rn

by RoRi
  1. Let L(P;A) and L(Q;B) be two lines in \mathbb{R}^n. Prove that they intersect if and only if P-Q is in the linear span of A and B.
  2. Consider the two lines in \mathbb{R}^3,

        \[ L = \{ (1,1,-1) + t(-2,1,3) \}, \qquad L' = \{ (3,-4,1) + t(-1,5,2) \}. \]

    Determine whether they intersect.

  1. Proof. Assume L(P;A) and L(Q;B) intersect. Since L(P;A) = \{ P+tA \} and L(Q;B) = \{ Q + t'B\} we know there exists an X \in \mathbb{R}^n such that X = P+tA = Q + t'B. This implies

        \[ P-Q = t'B - tA = t'B + (-t)A. \]

    This implies P-Q is in the linear span of A,B.

    Conversely, assume P-Q is in the linear span of A,B. Then there exist t, t' \in \mathbb{R} such that

        \begin{align*} P-Q = tA + t'B && \implies && P-t'B &= Q+ tA \\ && \implies && P + (-t)A &= Q + t'B. \end{align*}

    Thus, there is some point in both L(P;A) and L(Q;B) so they intersect. \qquad \blacksquare

  2. The two given lines do not intersect. We know from part (a) that two lines L(P;A) and L(Q;B) intersect if and only if P-Q is in the linear span of A,B. In this case we have

        \[ P - Q = (1,1,-1) - (3,-4,1) = (-2,5,-2) \]

    and A = (-2,1,3), B = (-1,5,2). For P-Q to be in the linear span of A,B we must have r,s \in \mathbb{R} such that

        \begin{align*} -2r-s &= -2 \\ r + 5s &= 5 \\ 3r + 2s &= -2. \end{align*}

    But the second equation implies r = 5-5s. The third equation would then require 15 - 15s + 2s = -2 which gives s = \frac{17}{13}. Then, r = -\frac{20}{13}. But then from the first equation

        \[ -2 \left(-\frac{20}{13} \right) - \left( \frac{17}{13} \right) = \frac{23}{13} \neq -2. \]

    Hence, there are no such r,s so P-Q is not in the linear span of A,B. Hence, these two lines do not intersect.

Prove that two lines in R3 intersect and determine the points of intersection

by RoRi

Let

    \[ P = (1,1,1), \qquad Q = (2,1,0). \]

Consider the lines L_1 and L_2 where L_1 is the line through P parallel to A = (1,2,3) and L_2 is the line through Q parallel to B = (3,8,13). Prove that these two lines intersect and determine the point of intersection.

Proof. We have

    \begin{align*} L_1 = L(P;A) &= \{ (1+t, 1+2t, 1+3t) \mid t \in \mathbb{R} \} \\ L_2 = L(Q;B) &= \{ (2+3s, 1+8s, 13s) \mid s \in \mathbb{R} \}. \end{align*}

Then if L_1 and L_2 intersect we must have s,t \in \mathbb{R} such that

    \begin{align*} 1+t &= 2+3s \\ 1+2t &= 1+8s \\ 1+3t &= 13s. \end{align*}

From the second equation we have t - 4s =0 which implies t = 4s. Plugging this into the first equation we then have -3s + 4s - 1 = 0 which implies s = 1. Therefore, t = 4. Then we check these values of s,t satisfy the third equation,

    \[ 1+3t = 1+12 = 13 = 13(1) = 13s.\]

So, the point of intersection is (1+t, 1+2t, 1+3t) = (2+3s, 1+8s, 13s) = (5,9,13). \qquad \blacksquare

Determine all subsets of a given set of points which lie on the same line

by RoRi

Consider the set of points

    \begin{align*} A = (2,1,1), \quad B = (6,-1,1), \quad C = (-6,5,1), \quad D = (-2,3,1), \\ E = (1,1,1), \quad F = (-4,4,1), \quad G = (-13,9,1), \quad H = (14,-6,1). \end{align*}

The points A,B,C lie on a line. Determine all other subsets of three or more points which lie on the same line.

First, we notice that the third coordinate of all of the given points is 1. So, if we check that the x,y coordinates are on the same line, then the points will be on the same line. Next, we look at which other points are on the same line as A,B,C. This line is given by

    \[ L(A; A-B) = \{ (2,1,1) + t(-4,2,0) \mid t \in \mathbb{R} \} = \{ (2-4t,1+2t,1) \mid t \in \mathbb{R} \}. \]

Checking the remaining points we have
The point D = (-2,3,1) is on L since 2-4t = -2 implies t = 1 and 1+2t = 3. Thus, D = (2-4t,1+2t,1) for t = 1.
The point E = (1,1,1) is not on L since 2-4t = 1 implies t = \frac{1}{4}, but then 1 + 2t = \frac{3}{2} \neq 1.
The point F = (-4,4,1) is on L since 2-4t = -4 implies t = \frac{3}{2} and then 1+2t = 4. Thus, F = 2-4t,1+2t,1) for t = \frac{3}{2} so F is on L.
The point G = (-13,9,1) is not on L since 2-4t = -13 implies t = \frac{15}{4}, but then 1+2t = \frac{17}{2} \neq 9. SO G is not on L.
The point H = (14,-6,1) is not on L since 2-4t = 14 implies t = -3, but then 1+2t = -5 \neq -6. So H is not on L.

Therefore, we have that \{ A,B,C,D,F \} are all on the same line (hence, every subset of these with three or more elements is a subset with three or more points on the same line).

None of the other points E,G,H are on the same line since H is not on the unique line containing E,G.

Determine if a given set of points all lie on the same line

by RoRi

In each of the following cases determine if the given set of points all lie on the same line.

  1. P = (2,1,1), Q = (4,1,-1), R = (3,-1,1).
  2. P = (2,2,3), Q = (-2,3,1), R = (-6,4,1).
  3. P = (2,1,1), Q = (-2,3,1), R = (5,-1,1).
  1. The points P,Q,R are not on the same line since the unique line containing P,R is given by

        \[ L = \{ P + t(P-R) \mid t \in \mathbb{R} \} = \{ (2,1,1) + t(-1,2,0) \mid t \in \mathbb{R} \} = \{ (2-t,1+2t,1) \mid t \in \mathbb{R} \}. \]

    But, Q cannot be on this line since for any t, the third coordinate of every point on the line will be 1, while the third coordinate of Q is -1.

  2. The points P,Q,R are not on the same line since the unique line containing the points P,Q is given by

        \[ L = \{ P + t(P-Q) \mid t \in \mathbb{R} \} = \{ (2,2,3) + t (4,-1,2) \mid t \in \mathbb{R} \} = \{ (2+4t, 2-t, 3+2t) \mid t \in \mathbb{R} \}. \]

    But, R is not on this line since if -6 = 2+4t then we must have t = -2. However, with t = -2 we have 3+2t = -1 \neq 1.

  3. The points P,Q,R are not on the same line since the unique line containing the points P,Q is given by

        \[ L = \{ P + t(P-Q) \mid t \in \mathbb{R} \} = \{ (2,1,1) + t(4,-2,0) \mid t \in \mathbb{R} \} = \{ (2+4t,1-2t,1) \mid t \in \mathbb{R} \}. \]

    But then R is not on this line since if 2+4t = 5 then t = \frac{3}{4}, but then 1-2t = -\frac{1}{2} \neq -1. Hence, R is not on the line.

Determine which points are on a line containing (-3,1,1) and (1,2,7)

by RoRi

Let L be the line containing the points P = (-3,1,1) and Q = (1,2,7). Determine which of the following points are also on L.

  1. (-7,0,5);
  2. (-7,0,-5);
  3. (-11,1,11);
  4. (-11,-1,11);
  5. (-1, \frac{3}{2}, 4);
  6. (-\frac{5}{3}, \frac{4}{3}, 3);
  7. (-1, \frac{3}{2}, -4).
First, the line containing the points P = (-3,1,1) and Q = (1,2,7) is the set of points

    \[ L(P;P-Q) = \{ (-3,1,1); (-4,-1,-6) \} = \{ (-3-4t, 1-t, 1-6t) \mid t \in \mathbb{R} \}. \]

  1. The point (-7,0,5) is not on L since

        \[ -3-4t = -7 \quad \implies \quad t = 1. \]

    But then 1 - 6t = -5 \neq 5. Hence, there is no t \in \mathbb{R} such that (-7,0,5) = (-3-4t, 1-t, 1-6t), so (-7,0,5) is not on L.

  2. The point (-7,0,-5) is on L since

        \[ -3-4t = -7 \quad \implies \quad t = 1. \]

    And then 1-t = 0 and 1-6t = -5. Hence, (-7,0,-5) = (-3-4t,1-t,1-6t) where t = 1. Thus, (-7,0,-5) is on L.

  3. The point (-11,1,11) is not on L since

        \[ -3-4t = -11 \quad \implies \quad t = 2. \]

    But then 1-t = -1 \neq 1. Hence, there is no t \in \mathbb{R} such that (-11,1,11) = (-3-4t, 1-t, 1-6t), so (-11,1,11) is not on L.

  4. The point (-11,-1,11) is not on L since

        \[ -3-4t = -11 \quad \implies \quad t = 2. \]

    But then 1 - 6t = -11 \neq 11. Hence, there is no t \in \mathbb{R} such that (-11,-1,11) = (-3-4t, 1-t, 1-6t), so (-11,-1,11) is not on L.

  5. The point (-1,\frac{3}{2},4) is on L since

        \[ -3-4t = -1 \quad \implies \quad t = -\frac{1}{2}. \]

    And then 1-t = \frac{3}{2} and 1-6t = 4. Hence, (-1,\frac{3}{2},4) = (-3-4t,1-t,1-6t) where t = -\frac{1}{2}. Thus, (-1,\frac{3}{2},4) is on L.

  6. The point (-\frac{5}{3},\frac{4}{3},3) is on L since

        \[ -3-4t = -\frac{5}{3} \quad \implies \quad t = -\frac{1}{3}. \]

    And then 1-t = \frac{4}{3} and 1-6t = 3. Hence, (-\frac{5}{3},\frac{4}{3},3) = (-3-4t,1-t,1-6t) where t = -\frac{1}{3}. Thus, (-\frac{5}{3},\frac{4}{3},3) is on L.

  7. The point (-1,\frac{3}{2},-4) is not on L since

        \[ -3-4t = -1 \quad \implies \quad t = -\frac{1}{2}. \]

    But then 1 - 6t = 4 \neq -4. Hence, there is no t \in \mathbb{R} such that (-1,\frac{3}{2},-4) = (-3-4t, 1-t, 1-6t), so (-1,\frac{3}{2},-4) is not on L.

Determine which points are on a line containing the point (-3,1,1) and parallel to the vector (1,-2,3)

by RoRi

Let P = (-3,1,1) be a point in \mathbb{R}^3 and let L be a line containing P and parallel to the vector (1,-2,3). Determine which of the following points are also on L.

  1. (0,0,0);
  2. (2,-1,4);
  3. (-2,-1,4);
  4. (-4,3,-2);
  5. (2,-9,16).
Given a point P= (-3,1,1) and a vector A = (1,-2,3) the line containing P in the direction of A is given by

    \[ L(P;A) = L((-3,1,1);(1,-2,3)) = \{ (-3,1,1) + t(1,-2,3) \mid t \in \mathbb{R} \} = \{(-3+t, 1-2t, 1+3t) \mid t \in \mathbb{R} \}. \]

  1. If (0,0,0) were on L then we must have some t \in \mathbb{R} such that

        \[ -3+t = 0, \qquad 1-2t = 0, \qquad 1+3t = 0. \]

    From the first equation, this requires t = 3, but then neither of the other two equations can hold. Hence, there is no t \in \mathbb{R} such that (0,0,0) = (-3+t,1-2t,1+3t) so (0,0,0) is not on the line.

  2. If (2,-1,4) were on L then we must have some t \in \mathbb{R} such that

        \[ -3+t = 2, \qquad 1-2t = -1, \qquad 1+3t = 4. \]

    From the first equation, this requires t = 5, but then neither of the other two equations can hold. Hence, there is no t \in \mathbb{R} such that (2,-1,4) = (-3+t,1-2t,1+3t) so (2,-1,4) is not on the line.

  3. If (-2,-1,4) is on L then we must have some t \in \mathbb{R} such that

        \[ -3+t = -2, \qquad 1-2t = -1, \qquad 1+3t = 4. \]

    From the first equation we have t = 1. This value of t also satisfies the other two equations. Hence, (-2,-1,4) = (-3+t, 1-2t, 1+3t) for t = 1. Therefore, (-2,-1,4) is on the line.

  4. If (-4,3,-2) is on L then we must have some t \in \mathbb{R} such that

        \[ -3+t = -4, \qquad 1-2t = 3, \qquad 1+3t = -2. \]

    From the first equation we have t = -1. This value of t also satisfies the other two equations. Hence, (-4,3,-2) = (-3+t, 1-2t, 1+3t) for t = -1. Therefore, (-4,3,-2) is on the line.

  5. If (2,-9,16) is on L then we must have some t \in \mathbb{R} such that

        \[ -3+t = 2, \qquad 1-2t = -9, \qquad 1+3t = 16. \]

    From the first equation we have t = 5. This value of t also satisfies the other two equations. Hence, (2,-9,16) = (-3+t, 1-2t, 1+3t) for t = 5. Therefore, (2,-9,16) is on the line.

Determine which points are on a line containing (2,-1) and (-4,2)

by RoRi

Let L be a line containing the points P = (2,-1) and Q = (-4,2). Determine which of the following points are also on L.

  1. (0,0);
  2. (0,1);
  3. (1,2);
  4. (2,1);
  5. (-2,1);
First, we have P -Q = (6,-3), so the line containing the points P and Q is the set

    \[ L (P;P-Q) = \{ P + t(6,-3) \mid t \in \mathbb{R} \} = \{ (2+6t, -1-3t) \mid t \in \mathbb{R} \}. \]

  1. The point (0,0) is on L since

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

    Thus, (0,0) = (2+6t,-1-3t) for t = -\frac{1}{3}, so (0,0) is on L.

  2. The point (0,1) is not on L since

        \begin{align*} (0,1) &= (6t + 2, -1-3t) & \implies && 6t+2 &=0 & \implies && t = -\frac{1}{3} \\ && \implies && -1-3t &= 0 \neq 1. \end{align*}

    Thus, (0,1) \neq (2+6t,-1-3t) for any t \in \mathbb{R}, so (0,1) is not on L.

  3. The point (1,2) is not on L since

        \begin{align*} (1,2) &= (6t + 2, -1-3t) & \implies && 6t+2 &=1 & \implies && t = -\frac{1}{6} \\ && \implies && -1-3t &= -\frac{1}{2} \neq 2. \end{align*}

    Thus, (1,2) \neq (2+6t,-1-3t) for any t \in \mathbb{R}, so (1,2) is not on L.

  4. The point (2,1) is not on L since

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

    Thus, (2,1) \neq (2+6t,-1-3t) for any t \in \mathbb{R}, so (2,1) is not on L.

  5. The point (-2,1) is on L since

        \begin{align*} (-2,1) &= (6t + 2, -1-3t) & \implies && 6t+2 &=-2 & \implies && t = -\frac{2}{3} \\ && \implies && -1-3t &= 1. \end{align*}

    Thus, (-2,1) = (2+6t,-1-3t) for t = -\frac{2}{3}, so (-2,1) is on L.

Determine which points are on a given line

by RoRi

Let L be a line containing the points P = (-3,1) and Q = (1,1). Determine which of the following points are also on L.

  1. (0,0);
  2. (0,1);
  3. (1,2);
  4. (2,1);
  5. (-2,1);
By Theorem 13.4 (page 474 of Apostol)we know that given two points P, Q there is a unique line L between them which is the set of points

    \[ L = L(P; P-Q) = \{ P + t(Q-P) \} = \{ (-3,1) + t(4,0) \}. \]

So, the set of points on L are all points of the form (4t-3, 1) for t \in \mathbb{R}.

  1. The point (0,0) is not on L since there is no t \in \mathbb{R} such that (4t-3,1)= (0,0).
  2. The point (0,1) is on L since if we take t = -\frac{3}{4} then (4t-3,1) = (0,1).
  3. The point (1,2) is not on L since (1,2) \neq (4t-3,1) for any t.
  4. The point (2,1) is on L since if we take t = \frac{5}{4} then (4t-3,1) = (2,1).
  5. The point (-2,1) is on L since if we take t = \frac{1}{4} then (4t-3, 1) = (-2,1).