Tag: Lines

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

by RoRi

May 30, 2016

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

by RoRi

May 30, 2016

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

by RoRi

April 29, 2016

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

by RoRi

April 29, 2016

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 $x-y = -1$ .

Prove that the line through two distinct points on a plane is entirely contained in the plane

by RoRi

April 29, 2016

Let $M$ be a plane containing two distinct points $P$ and $Q$ . Prove that every point on the line through $P$ and $Q$ is in $M$ .

Proof. Fist, the line through $P$ and $Q$ is given by

$L = \{ P + r(Q-P) \}.$

Then, let $R$ be any point on $M$ not on $L(P;Q)$ . The plane $M$ is given by

$M = \{ P + s(Q-P) + t(R-P) \}.$

So, for any point $X \in L(P;Q)$ let $s = r$ and $t = 0$ . Then we have

$P + s(Q-P) + t(R-P) = P + r(Q-P) \in M \quad \implies \quad L(P;Q) \in M. \qquad \blacksquare$

Determine whether a line is parallel to given planes

by RoRi

April 29, 2016

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.

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

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.
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 $M$ .
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

by RoRi

April 29, 2016

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

by RoRi

April 29, 2016

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 that two parallel lines are either equal or have empty intersection

by RoRi

April 29, 2016

Let $L(P;A)$ and $L(Q;A)$ be two parallel lines in $\mathbb{R}^n$ . Prove that $L(P; A) \cap L(Q;A) = \varnothing$ or $L(P;A) = L(Q;A)$ .

Proof. Assume $L(P;A) \neq L(Q;A)$ . Suppose that the intersection is not empty, so that there exists a point $X \in L(P;A)$ and $X \in L(Q;A)$ . Then, we know there exist $s,t \in \mathbb{R}$ such that

$X = P + tA, \qquad \text{and} \qquad X = Q + sA.$

But then

$\begin{align*} P+tA = Q+sA && \implies && P = Q + (s-t)A \\ && \implies && P \in L(Q;A). \end{align*}$

But, by Theorem 13.2, this means $L(P;A) = L(Q;A)$ , contradicting our assumption. Hence, either $L(P;A) = L(Q;A)$ or the intersection is empty $. \qquad \blacksquare$

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

by RoRi

April 29, 2016

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

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

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 $t_0 = \frac{c_2}{2c_3}. \qquad \blacksquare$
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,

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