Tag: Analytic Geometry

Find a unit vector perpendicular to a given vector and parallel to a given plane

by RoRi

May 30, 2016

Find a unit length vector $A$ which is parallel to the plane with Cartesian equation $x - y + 5z = 1$ and perpendicular to the vector $\mathbf{i} + 2 \mathbf{j} - 3 \mathbf{k}$ .

The plane with Cartesian equation $x - y + 5z = 1$ has intercepts $(1,0,0), \ (0,-1,0), \ \left( 0,0, \frac{1}{5} \right)$ . Therefore, the plane is the set of points

$M = \left\{ (1,0,0) + s(1,1,0 + t\left( 1, 0, -\frac{1}{5} \right) \right\}.$

So, the vector $A$ being parallel to $M$ implies $A$ is in the span of $\left\{ (1,0,0), \ \left(1,0,-\frac{1}{5} \right) \right\}$ . Therefore,

$A = \left( s + t, s, -\frac{1}{5}t \right)$

for some $s,t$ . Then, $A$ perpendicular to $\mathbf{i} + 2 \mathbf{j} - 3 \mathbf{k}$ implies

$A \cdot (1,2,-3) = 0 \quad \implies \quad s+t + 2s + \frac{3}{5} t = 0 \quad \implies \quad s = -\frac{8}{15} t.$

Finally, since $A$ has unit length we have

$\begin{align*} \lVert A \rVert = 1 && \implies && \sqrt{ (s+t)^2 + s^2 + \frac{t^2}{25} } &= 1 \\ && \implies && s^2 + 2st + t^2 + s^2 + \frac{t^2}{25} &= 1 \\ && \implies && 2s^2 + \left( \frac{26}{25} \right)t^2 + 2st &= 1 \\ && \implies && t^2 \left( \frac{128}{225} + \frac{26}{25} - \frac{16}{15} \right) &= 1 \\ && \implies && t &= \frac{15}{\sqrt{122}}. \end{align*}$

This then gives us

$s = -\frac{8}{\sqrt{122}}.$

Hence,

$A = \frac{1}{\sqrt{122}} (7,-8,-3).$

Compute the volume of a tetrahedron with given vertices

by RoRi

May 30, 2016

Consider the tetrahedron with vertices at the origin and at the points where the plane

$x + 2y + 3z = 6$

intersects the coordinate axes. Compute the volume of this tetrahedron.

First, the intercepts of the plane are given by $(6,0,0), (0,3,0), (0,0,2)$ . Then from a previous exercise (Section 13.14, Exercise #13) we know that the volume of a tetrahedron with vertices $A,B,C,D$ is

$V = \frac{1}{6} | (B-A) \cdot (C -A) \times (D-A) |.$

Letting $A = (0,0,0), \ B = (6,0,0), \ C = (0,3,0), \ D = (0,0,2)$ we have

$V = \frac{1}{6} | (6,0,0) \cdot (0,3,0) \times (0,0,2) | = \frac{1}{6} | (6,0,0) \cdot (6,0,0) | = 6.$

Find a Cartesian equation for a plane through a point with normal vector making given angles

by RoRi

May 30, 2016

Let $M$ be the plane whose normal vector $N$ makes angles $\frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3}$ with the unit coordinate vectors and which contains the point $(1,1,1)$ . Find a Cartesian equation for the plane.

Since the normal vector to the plane makes angles $\frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3}$ with the unit coordinate vectors, we have

$\begin{align*} N \cdot \mathbf{i} &= \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} & \implies && x_1 &= \frac{1}{2} \\ N \cdot \mathbf{j} &= \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} & \implies && x_2 &= \frac{\sqrt{2}}{2} \\ N \cdot \mathbf{k} &= \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} & \implies && x_3 &= \frac{1}{2}. \end{align*}$

Hence, $N = (1, \sqrt{2}, 1)$ . So, the plane has a Cartesian equation of the form

$x + \sqrt{2} y + z = d.$

Since it contains $(1,1,1)$ we have $d = 1 + \sqrt{2} + 1 = 2 + \sqrt{2}$ . Therefore, a Cartesian equation of the plane is

$x + \sqrt{2} y + z = 2 + \sqrt{2}.$

Determine properties of a point whose movement in space is determined by a vector parametric equation

by RoRi

May 30, 2016

Consider a point moving in space with position at time $t$ given by

$X(t) = (1-t) \mathbf{i} + (2-3t) \mathbf{j} + (2t-1) \mathbf{k}.$

Prove that the motion of the point is along a line.
Find a vector parallel to this line.
Find the time $t$ at which the point intersects the plane with Cartesian equation $2x+3y+2z+1 = 0$ .
What is the Cartesian equation for the plane parallel to the plane in part (c) which contains the point $X(3)$ ?
Let $M$ be the plane perpendicular to $L$ containing the point $X(2)$ . Find a Cartesian equation for $M$ .

Proof. We use the formula for the motion of the particle to compute
$\begin{align*} X(t) &= (1-t) \mathbf{i} + (2-3t) \mathbf{j} + (2t-1) \mathbf{k} \\ &= \mathbf{i} + 2 \mathbf{j} - \mathbf{k} + t(-\mathbf{i} - 3\mathbf{j} + 2 \mathbf{k}) \\ &= (1,2,-1) + t(-1,-3,2). \end{align*}$

This is the parametric equation for the line through $(1,2,-1)$ parallel to the vector $(-1,-3,2). \qquad \blacksquare$
From part (a) we have a vector $N$ parallel to $L$ given by $N = (1,3,-2)$ .
First, the line on which the point moves is the set of points
$L = \{ (1,2,-1) + t(-1,-3,2) \} = \{ (1-t, 2-3t, -1+2t) \}.$

So, to find the intersection with the plane $2x + 3y + 2z + 1 = 0$ we compute

$\begin{align*} && 2(1-t) + 3(2-3t) + 2(-1+2t) + 1 &= 0 \\ \implies && 2 - 2t + 6 - 9t -2 + 4t + 1 &= 0 \\ \implies && 7 -7t &= 0 \\ \implies && t &= 1. \end{align*}$
First, we have
$X(3) = (1,2,-1) + 3(-1,-3,2) = (1,2,-1) + (-3,-9,6) = (-2,-7,5).$

Since we know the plane is parallel to the one in part (c) it has a Cartesian equation of the form

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

We compute $d = 2(-2) + 3(-7) + 2(5) = -15$ . Hence, the plane has Cartesian equation

$2x + 3y + 2z + 15 = 0.$
Since the plane is perpendicular to the line $L$ we know that it has a normal vector in the same direction as $L$ , so $N = (1,3,-2)$ (from part (b)). Thus, we have a Cartesian equation of the form
$x + 3y - 2z = d.$

Since the point

$X(2) = (1,2,-1) + 2(-1,-3,2) = (1,2,-1) + (-2, -6, 4) = (-1, -4, 3)$

is on the plane we have $d = -1 + 3(-4) - 2(3) = -19$ . Therefore, the plane is given by

$x + 3y - 2z + 19 = 0.$

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

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