# Thread: 3 true and false questions

1. ## 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).

The answer is TRUE

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 answer is FALSE

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

The answer is TRUE

Could you please teach me how to solve them? Thank you very much.

2. 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).

The answer is TRUE
...
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

3. 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.

The answer is TRUE

...
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

4. Hello, Jenny!

Here's another approach to #3 . . .

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.

5. Hello again, Jenny!

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.

The answer is FALSE.

$\displaystyle P_1$ 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.

6. 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 answer is FALSE

...
Hello, Jenny,

I assume that the vector product of your problem should be the dot product.

I've attache a sketch to demonstrate why this statement is false.
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.

EB