Tag: Cartesian Equations

Find the Cartesian equation of the plane parallel to given vectors with given intercept

by RoRi

May 30, 2016

Consider the plane which is parallel to both of the vector $\mathbf{i} + \mathbf{j}$ and $\mathbf{j} + \mathbf{k}$ and intersects the $x$ -axis at the point $(2,0,0)$ . Find the Cartesian equation of this plane.

Since $M$ contains the point $(2,0,0)$ and is parallel to $(1,1,0)$ and $(0,1,1)$ we have

$M = \{ (2,0,0) + s(1,1,0) + t(0,1,1) \}.$

Thus, the Cartesian equations for $M$ are

$x = 2s, \qquad y = s + t, \qquad z = t.$

Hence,

$x - y + z = 2.$

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

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

Determine if two given planes are parallel and find points on their intersection

by RoRi

April 29, 2016

Let $P = (2,3,1), \ A = (1,2,3), \ B = (3,2,1)$ and let $M = \{ P + sA + tB \}$ . Let $M'$ be the plane determined by the Cartesian equation $x - 2y + z = 0$ .

Determine if the two planes are parallel.
If we define $M''$ to be the plane with Cartesian equation $x+2y+z = 0$ find two points on $M' \cap M''$ .

We know for any point $Q \notin M$ , there is a unique plane parallel to $M$ containing $Q$ . We pick a point $Q$ on $M'$ that is not on $M$ and show that the unique plane parallel to $M$ containing $A$ is, in fact, $M'$ .
The point $Q = (0,0,0)$ is on $M'$ since

$(0) - 2(0) + 0 = 0.$

It is not on $M$ since

$\begin{align*} 2+s+3t &= 0 \\ 3+2s + 2t &= 0 \\ 1+3s + t &= 0 \end{align*}$

has no solution $s,t$ . So, the unique plane parallel to $M$ containing $(0,0,0)$ is

$\{ (0,0,0) + sA + tB \} = \{ (s+3t, 2s+2t, 3s+t) \}.$

Then, we obtain the Cartesian equation of this plane. We have

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

This gives us $s = -3t + x$ . Then, $y = -6t + 2x + 2t$ which implies

$t = \frac{2x-y}{4}, \qquad \implies \qquad s = -3 \left( \frac{2x-y}{4} \right) + x.$

Hence,

$\begin{align*} z &= -9 \left( \frac{2x-y}{4} \right) + 3x + \left( \frac{2x-y}{4} \right) \\ &= 3x - 2(2x-y) \\ &= -x + 2y. \end{align*}$

Hence, this has the Cartesian equation $x - 2y + z = 0$ . But this is the plane $M'$ ; hence, $M'$ is parallel to $M$ .
The points $(0,0,0)$ and $(-1,0,1)$ are both in the intersection $M' \cap M''$ since they both satisfy the equations
$\begin{align*} x-2y+z &= 0 \\ x + 2y + z&=0. \end{align*}$

Determine which points are on the plane with Cartesian equation 3x – 5y + z = 9

by RoRi

April 29, 2016

Let $M$ be the plane with Cartesian equation $3x - 5y + z = 9$ .

Determine which of the points $(0,-2,-1), \ (-1,-2,2), \ (3,1,-5)$ are on the plane.
Find vectors $P,A,B$ such that $M = \{ P + sA + tB \}$ .

The point $(0,-2,-1) \in M$ since
$3(0) - 5(-2) + (-1) = 10 - 1 = 9.$

The point $(-1,2,2) \in M$ since

$3(-1) + (-5)(-2) + 2 = -3+10+2 = 9.$

The point $(3,1,-5) \notin M$ since

$3(3) + (-5)(1) + (-5) = 9 - 5 - 5 = -1 \neq 9.$
We know $(0,-2,-1), \ (-1,-2,2) \in M$ . So, let
$A = Q-P = (-1,-2,2) - (0,-2,-1) = (-1,0,3).$

Since $R = (3,1,5) \in M$ let

$B = R-P = (3,1,5) - (0,-2,-1) = (3,3,6).$

Then we have

$M = \{ (0,-2,-1) + s(-1,0,3) + t(3,3,6) \}.$

Determine a Cartesian equation for given planes

by RoRi

April 29, 2016

For each of the following planes, find a linear Cartesian equation of the form

$ax + by + cz = d$

that describes the plane.

The plane through $(2,3,1)$ spanned by $(3,2,1), \ (-1,-2,-3)$ .
The plane through the points $(2,3,1), \ (-2,-1,-3), \ (4,3,-1)$ .
The plane through the point $(2,3,1)$ parallel to the plan through $(0,0,0)$ spanned by $(2,0,-2)$ and $(1,1,1)$ .

The plane through $(2,3,1)$ spanned by $(3,2,1)$ and $(-1,-2,-3)$ is the set of points
$M = \{ (2,3,1) + s(3,2,1) + t(-1,-2,-3) \} = \{ (2+3s-t, 3 + 2s -2t, 1+s-3t) \}.$

Therefore, we have the parametric equations

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

Then we want to solve for $s,t$ in terms of $x,y,z$ . From the first equation we have

$x = 2+3s-t \quad \implies \quad t = 2+3s - x.$

From the second equation we then have

$y = 3+2s - 4 - 6s + 2x \quad \implies \quad s = \frac{-1 + 2x - y}{4}.$

Which gives us

$t = 2 + \frac{-3 + 6x - 3y}{4} - x.$

So, from the third equation we then have

$\begin{align*} z &= 1 + \frac{-1+2x-y}{4} - 6 + \frac{9-18x+9y}{4} + 3x \\ &= -5 + \frac{8-16x+8y}{4} + 3x \\ &= -3-x+2y. \end{align*}$

Thus,

$z = -3-x+2y \quad \implies \quad x - 2y + z = -3$

is the requested linear Cartesian equation.
The plane through the three points $(2,3,1), \ (-2,-1,-3), \ (4,3,-1)$ is the set of points
$M = \{ (2,3,1) + s(-4,-4,-4) + t(2,0,-2) \}.$

But, $(-4, -4, -4)$ and $(2,0,-2)$ are in the linear span of $(3,2,1), \ (-1,-2,-3)$ since

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

Thus, this plane is equal to the plane in part (a). Hence, we have the linear Cartesian equation,

$x - 2y + z = -3.$
Again, this is the same plane as in parts (a) and (b) since the span of $(2,0,-2)$ and $(1,1,1)$ is the same as the span of $(3,2,1)$ and $(-1,-2,-3)$ . Hence, the requested linear Cartesian equation is
$x - 2y + z = -3.$