# 3 true and false questions

• Dec 12th 2006, 08:10 PM
Jenny20
3 true and false questions
Question 1
The plane given by the equation 3x-4y+z-1=0 is perpendicular to the vector
<9,-12,3> and contains the point (0,0,1).

Question 2
If A is a plane containing the point Psub1, Psub2, and Psub3 is any other point in 3-space, and Psub1Psub2 * Psub1Psub3 =0 , then Psub1Psub3 must be a normal vector for the plane.

Question 3
The four points (-2,1,1), (0,2,3), (1,0,-1), and (5,4,7) all lies in the same plane.

Could you please teach me how to solve them? Thank you very much.
• Dec 13th 2006, 04:25 AM
earboth
Quote:

Originally Posted by Jenny20
Question 1
The plane given by the equation 3x-4y+z-1=0 is perpendicular to the vector
<9,-12,3> and contains the point (0,0,1).

...

Hello, Jenny,

your equation is given in normal form. The normal vector which is perpendicular to the palne is: <3, -4, 1>.

The given vector is 3*<3, -4, 1> = <9, -12, 3> thus this vector is perpendicula to the plane too.

Plug in the coordinates of the point:
3*0-4*0+1-1=0 which is true. Thus the point belongs to the plane.

EB
• Dec 13th 2006, 04:40 AM
earboth
Quote:

Originally Posted by Jenny20
...
Question 3
The four points (-2,1,1), (0,2,3), (1,0,-1), and (5,4,7) all lies in the same plane.

...

Hello, Jenny,

three points, which don't form a straight line or are equal, form a plane:

Choose one point of the given 4 (I'll take the first one) and two other to set up the equation of this plane:

<x,y,z>=<-2,1,1>+ r*<<0,2,3>-<-2,1,1>> + s*<<1,0,-1>-<-2,1,1>>, r,s in IR.

<x,y,z>=<-2,1,1>+ r*<-2,1,2> + s*<3,-1,-2>

Now plug in the coordinates of the 4th point for <x,y,z>. You'll get 3 equations with 2 variables (r,s). You have to prove that there exist a unique(?) r and s for this system of equations:

EQU1 : 5 = -2 -2r + 3s
EQU2 : 4 = 1 + r - s
EQU3 : 7 = 1 + 2r - 2s

EQU1 + EQU3 : 12 = -1 +s <==> s = 13. Plug in in EQU3: 7 = 1 + 2r - 2*13
thus r = 16.

Now pug in these two values into the unuse EQU2: 4 =1 + 16 - 13
This is true, therefore the given point belongs to the plane.

EB
• Dec 13th 2006, 08:03 AM
Soroban
Hello, Jenny!

Here's another approach to #3 . . .

Quote:

3) The four points $\displaystyle P(-2,1,1),\:Q(0,2,3),\:R(1,0,-1),\:S(5,4,7)$ all lie in the same plane.

Find the plane that contains three of the points, say: $\displaystyle P,\:Q,\:R$
. . Then see if $\displaystyle S$ also lies on the plane.

We have vector $\displaystyle \vec{u}\:=\:\overrightarrow{PQ} \:=\:\langle 2,1,2\rangle$ and vector $\displaystyle \vec{v} \:=\:\overrightarrow{PR} \:=\:\langle3,\text{-}1,\text{-}2\rangle$

The normal vector of the plane is perpendicular to both $\displaystyle \vec{u}$ and $\displaystyle \vec{v}.$

. . Hence: .$\displaystyle \vec{n}\:=\:\begin{vmatrix}i & j & k \\ 2 & 1 & 2 \\ 3 & \text{-}1 & \text{-}2\end{vmatrix} \:=\:\langle 0,10,\text{-}5\rangle \:=\:\langle0,2,\text{-}1\rangle$

The plane through $\displaystyle P(-2,1,1)$ with $\displaystyle \vec{n} = \langle0,2,-1\rangle$ has the equation:

. . $\displaystyle 0(x + 2) + 2(y - 1) - (z - 1) \:=\:0\quad\Rightarrow\quad 2y - z \:=\:1$

Does $\displaystyle S(5,4,7)$ lie on this plane?

. . $\displaystyle 2(4) - 7 \:=\:1$ . . . Yes!

Therefore, the four points are coplanar.

• Dec 13th 2006, 08:29 AM
Soroban
Hello again, Jenny!

Quote:

2) If $\displaystyle A$ is a plane containing the points $\displaystyle P_1$ and $\displaystyle P_2$
and $\displaystyle P_3$ is any other point in 3-space, and: .$\displaystyle \overrightarrow{P_1P_2}\cdot\overrightarrow{P_1P_3 } \:=\:0$,
then $\displaystyle \vec{P_1P_3}$ must be a normal vector for the plane.

$\displaystyle P_1$ and $\displaystyle P_2$ are on the plane.
$\displaystyle P_3$ is any point not on the plane.
Code:

                    *P3                     |                 *---|-------------------*               /    |                  /               /    |                /             /      * - - - - *      /             /      P1        P2    /           / A                    /           *-----------------------*

Since $\displaystyle \overrightarrow{P_1P_2}\cdot\overrightarrow{P_1P_3 } \:=\:0$, then $\displaystyle \overrightarrow{P_1P_3} \perp \overrightarrow{P_1P_2}$

But $\displaystyle \overrightarrow{P_1P_3}$ is not necessarily perpendicular to the plane.

• Dec 13th 2006, 08:33 AM
earboth
Quote:

Originally Posted by Jenny20
...
Question 2
If A is a plane containing the point Psub1, Psub2, and Psub3 is any other point in 3-space, and Psub1Psub2 * Psub1Psub3 =0 , then Psub1Psub3 must be a normal vector for the plane.

The vector $\displaystyle \overrightarrow{Q_1, Q_2}$ (in black) is perpendicular to the vector $\displaystyle \overrightarrow{Q_1, Q_3}$ (in blue. Q3 doesn't belong to the plan). The normal vector which is perpendicular to the plane (in red) doesn't point to point Q3.