Finding equation of a plane

**caelestis** · August 10th 2008, 04:30 AM

Hi

I'm sitting a quiz for maths and i'm having trouble with this question...

If anyone could help I would be very much appreciated!

So the question asks to...

Find the equation of the plane containing the line

x = (0, 0, 1) + t (0, -1, 0) and the point (1, 1, 0).

Here is my attempt:

This line passes through the pt P(0, 0, 1)

So we need a second vector which spans the plane together with the direction vector of the line.

If Q(1, 1, 0) then the second vector is:

PQ = (1, 1, 0) - (0, 0, 1) = (1, 1, -1)

So the equation of the plane is:
r = (0, 0, 1) + t (0,-1, 0) + s (1, 1, -1)

But I can't seem to get my head around which answer this comes to...
A. z - x = 1
B. z + x = 1
C. z - x = - 1
D. z + x = - 1

**o_O** · August 10th 2008, 08:00 AM

Remember that a plane can be defined by just a point and its normal: $\vec{n} \cdot \vec{p} = 0$

or $a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$ where (a,b,c) is your normal and $(x_{0}, y_{0}, z_{0})$ is a point.

We have the point (0, 0, 1) and we have the direction vector of the line (0, -1, 0). We have to find a vector perpendicular to this (i.e. the normal). This is simple as all we have to do is find a vector such that: $(0,-1,0) \cdot (a,b,c) = 0$ (remember, two vectors perpendicular to each other will give a dot product of 0).

So expand a bit: $0a - b + 0c = 0$ . Any solution will work. ex. (1, 0, 1). (In this case, b has to be equal to 0 since there's no other variable that can cancel it out).

So, plugging it all into $a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$ , we get ...

**Soroban** · August 10th 2008, 08:42 AM

Hello, caelestis!

Find the equation of the plane containing the line:
. . $x \:=\:(0, 0, 1) + t (0,\text{-}1, 0)$ and the point $(1, 1, 0).$

To write the equation of a plane, we need a point $(x_1,y_1,z_1)$
. . and its normal direction: $\vec{n}\:=\:\langle a,b,c\rangle$
Then: . $a(x-x_1) + b(y-y_1) + c(z-z_1) \;=\;0$

We have three points:
. . $t = 0:\;P_1 \:=\:(0,0,1)$
. . $t = 1:\;P_2\:=\:(0,\text{-}1,1)$
. . .and: . $P_3 \:=\:(1,1,0)$

Vector $\overrightarrow{v_1} \:=\:\overrightarrow{P_1P_2} \:=\:\langle 0,\text{-}1,0\rangle$ is in the plane.

Vector $\overrightarrow{v_2} \;=\;\overrightarrow{P_2P_3} \:=\: \langle 1,2,\text{-}1\rangle$ is in the plane.

The normal vector is their cross product:

. . $\vec{n} \;=\;\overrightarrow{v_1}\times \overrightarrow{v_2} \;=\;\left|\begin{array}{ccc} i & j & k \\ 0 & \text{-}1 & 0 \\ 1 & 2 & \text{-}1 \end{array}\right| \;=\;i(1) - j(0) + k(1) \;=\;i + k \;=\;\langle 1,0,1\rangle$

We have: . $P_1 = (0,0,1),\;\vec{n} = \langle 1,0,1\rangle$

The equation of the plane is: . $1(x-0) + 0(y-0) + 1(z-1) \:=\:0$

Therefore: . ${\color{blue}z + x \:=\:1\quad\text{answer B}}$

Thread: Finding equation of a plane

LinkBack

Thread Tools

Display

Finding equation of a plane

Similar Math Help Forum Discussions

Finding equation of plane.

Finding the equation of a plane

Finding an equation for the plane....

finding equation of plane

Finding an equation of the plane.

Search Tags