Tag: Planes

Find a vector parametric equation for a line containing a given point and perpendicular to a given plane

by RoRi

May 30, 2016

Let $L$ be the line which contains the point $(2,1,-3)$ and is perpendicular to the plane given by the equation $4x - 3y + z = 5$ . Find a vector parametric equation for $L$ .

From the Cartesian equation for the plane we have $N = (4,-3,1)$ is a normal vector. So, $L$ is the line through $(2,1,-3)$ which is parallel to $(4,-3,1)$ . Thus, the vector parametric equation for the line is

$X(t) = (2,1,-3) + t(4,-3,1).$

Find the Cartesian equation of plane through a given point and with a given perpendicular line

by RoRi

May 30, 2016

We say that a line parallel to a vector $N$ (non-zero) is perpendicular to a plane $M$ if $N$ is normal to $M$ . Given that a plane $M$ goes through the point $(2,3,-7)$ and that the line through the points $(1,2,3)$ and $(2,4,12)$ is perpendicular to $M$ find the Cartesian equation of $M$ .

First, $N = (2,4,12) - (1,2,3) = (1,2,9)$ . Therefore, the Cartesian equation of $M$ is of the form

$x + 2y + 9z = d.$

Since $(2,3,-7)$ is on the plane we have $d = 2 + 6 - 63 = -55$ . Thus, the Cartesian equation of $M$ is

$x + 2y + 9z + 55 = 0.$

Determine the angle between planes with given Cartesian equations

by RoRi

May 30, 2016

Consider two planes with Cartesian equations,

$x + y = 1, \qquad y + z = 2.$

Determine the angle between the planes.

For the two planes we have normals $N_1 = (1,1,0)$ and $N_2 = (0,1,1)$ . Therefore, the angle between the planes is

$\cos \theta = \frac{N_1 \cdot N_2}{\lVert N_1 \rVert \lVert N_2 \rVert} = \frac{1}{2} \qquad \implies \qquad \theta = \frac{\pi}{3} \ \text{or} \ \frac{2 \pi}{3}.$

Find the Cartesian equation of a plane given three points on the plane

by RoRi

May 30, 2016

Let $(1,2,3)$ , $(2,3,4)$ , and $(-1,7,-2)$ be three points on a plane. Find the Cartesian equation for the plane.

First, letting $P = (-1,7,-2)$ , $Q = (1,2,3)$ , and $R = (2,3,4)$ we compute a normal to the plane

$\begin{align*} N &= (P - Q) \times (P - R) \\ &= ((-1,7,-2) - (1,2,3)) \times ((-1,7,-2) - (2,3,4)) \\ &= (-2,5,-5) \times (-3,4,-6) \\ &= (-10,3,7). \end{align*}$

So the Cartesian equation of the plane is of the form

$-10x + 3y + 7z = d.$

Since $(1,2,3)$ is on the plane we have $d = -10 + 6 + 21 = 17$ . Hence, the Cartesian equation for the plane is

$-10x + 3y + 7z = 17 \quad \implies \quad 10x - 3y - 7z + 17 = 0.$

Find properties of a plane given three points that determine it

by RoRi

May 30, 2016

Let $M$ be the plane determined by the three points $(1,1,-1)$ , $(3,3,2)$ , and $(3,-1,-2)$ . Find the following:

A normal vector to the plane.
A Cartesian equation for the plane.
The distance between the plane and the origin.

Denote the points by $P = (3,-1,-2)$ , $Q = (1,1,-1)$ and $R = (3,3,2)$ , then we can compute a normal vector by
$\begin{align*} N &= (P-Q) \times (P-R) \\ &= ((3,-1,-2) - (1,1,-1)) \times ((3,-1,-2) - (3,3,2)) \\ &= (2,-2,-1) \times (0,-4,-4) \\ &= (4,8,-8). \end{align*}$

Therefore, $N = (1,2,-2)$ is a normal vector to the plane.
Since $(1,2,-2)$ is normal to the plane we have a Cartesian equation of the form
$x + 2y - 2z = d.$

Then, since $(1,1,-1)$ is on the plane we have $d = 5$ . Hence, the Cartesian equation is

$x + 2y - 2z = 5.$
The distance from the origin is
$d = \frac{5}{ \lVert N \rVert} = \frac{5}{3}.$

Establish properties of four planes with given Cartesian equations

by RoRi

May 30, 2016

Consider four planes with the Cartesian equations:

$\begin{align*} x + 2y - 2z & = 5 \\ 3x - 6y + 3z & = 2 \\ 2x + y + 2z &= -1 \\ x- 2y + z &= 7. \end{align*}$

Establish that two of them are parallel and the other two are perpendicular.
For the two parallel planes, find the distance between them.

The second and fourth planes are parallel since they have the same normal vector, $(1,-2,1)$ .
To see that the first and third are perpendicular, we denote the normal vectors by $N_1$ and $N_3$ , respectively, and compute

$N_1 \cdot N_3 = (1,2,-2) \cdot (2,1,2) = 2 + 2 - 4 = 0.$

Hence, they are perpendicular.
Denoting the second and fourth planes by $M_2$ and $M_4$ , respectively we have Cartesian equations
$M_2: \ x - 2y + z = \frac{2}{3}, \qquad M_4: \ x - 2y + z = 7.$

Therefore, the distance $d$ between them is

$d = \frac{7- \frac{2}{3}}{\lVert N \rVert} = \frac{19}{3 \sqrt{6}} = \frac{19 \sqrt{6}}{18}.$

Find the Cartesian equation of a plane passing through a point and parallel to a given plane

by RoRi

May 30, 2016

Let $M$ be a plane which passes through the point $(1,2,-3)$ and is parallel to the plane given by the equation $3x-y+2z = 4$ . Find the Cartesian equation of $M$ . Further, find the distance between the two planes.

Let $M'$ denote the plane given by the equation $3x-y+2z = 4$ . Since $M$ and $M'$ are parallel, we know they share a common normal vector. Therefore, the Cartesian equation of $M$ is of the form

$3x - y + 2z = d.$

Since $(1,2,-3)$ is on $M$ , we have

$d = 3 - 2 - 6 = -5.$

Hence, the Cartesian equation of $M$ is

$3x - y + 2z = -5.$

The distance between $M$ and $M'$ is then

$\left| \frac{d_1 - d_2}{\lVert N \rVert} \right| = \frac{9}{\sqrt{14}}.$

Give a Cartesian for planes through given points spanned by given vectors

by RoRi

May 13, 2016

Consider the vectors

$A = 2 \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k}, \qquad B = \mathbf{j} + \mathbf{k}.$

Find a nonzero vector $N$ perpendicular to both $A$ and $B$ .
Find a Cartesian equation for the plane through $(0,0,0)$ which is spanned by $A$ and $B$ .
Find a Cartesian equation for the plane through $(1,2,3)$ which is spanned by $A$ and $B$ .

Since $A$ and $B$ are independent, we can take
$N = A \times B = (3 - (-4), 0 - 2, 2 - 0) = (7,-2,2).$
From part (a) we have $N = (7,-2,2)$ is perpendicular to both $A$ and $B$ , so a Cartesian equation for the plane is given by
$7x - 2y + 2z = d$

Further, since the point $(0,0,0)$ is on the plane, we must have $d = 0$ . Hence, the Cartesian equation for the plane is

$7x - 2y + 2z = 0.$
Again, we have a Cartesian equation for the plane given by
$7x - 2y + 2z = d.$

Since $(1,2,3)$ is on the plane we must have

$d = 7(1) - 2(2) + 2(3) \quad \implies \quad d = 9.$

Hence, the Cartesian equation for the plane is given by

$7x - 2y + 2z = 9.$

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 $x-y = -1$ .