# Important Vector question!!

• Apr 11th 2008, 10:17 AM
ashes
Important Vector question!!
Hi guys, I have this question which I really need help ...

The equation of a plane P1 in a cartesian coordinate system is given by 3x+y-z-6=0. THe cordinates of two points A and B are [1,1,3] and [2,1,2] respectively.

i)Find the unit vector normal to the plane P1.
ii)Find the vector AB.
iii)Find the equation of a plane P2 that passes thru the points A and B and is perpendicular to the plane P1.

so far i only got AB=OB-OA ..i think so which is equals to i-k.
• Apr 11th 2008, 10:21 AM
Moo
Hello,

If a plane has ax+by+cz=d as equation, a normal vector of it has coordinates (a,b,c)
To find the unit vector, divide each coordinate by the norm of the vector, given by $\sqrt{a^2+b^2+c^2}$

A vector $\vec{MN}$ joining 2 points $M(x_M, y_M, z_M)$ and $N(x_N, y_N, z_N)$ has its coordinates as follow :

$\vec{MN} \begin{pmatrix} x_N-x_M \\ y_N-y_M \\ z_N-z_M \end{pmatrix}$
• Apr 11th 2008, 10:35 AM
ashes
That means the vector normal to the plane is 3i+j-k andi divide it by its modulus so i will get its unit vector?
• Apr 11th 2008, 10:38 AM
Moo
Yes !

By "norm", i meant modulus yeah :)
• Apr 11th 2008, 10:41 AM
ashes
ok so i got 2 parts done..thanks a lot..What about the last part..?im clueless..
• Apr 11th 2008, 10:48 AM
Moo
Hm well you can start by stating that the equation of P2 is in the form ax+by+cz+d=0 then replacing the coordinates of A and B in it.

Then...i have to think about it >_<
• Apr 11th 2008, 10:56 AM
ashes
Moo i dont get you..you mean just put in the cordinates of A and B..that means i do a scalar product of the 2?and the result?
• Apr 11th 2008, 11:05 AM
Moo
No, you have the hypothetic equation of P2. You will have to determine a, b, c and d ;)
As A and B are in P2, the coordinates of A verify the equation of P2 and the coordinates of B verify the equation of P2.

This will give you two equations in the aim to solve for a, b, c and d. But you need 2 others to find the complete equation.

Oh, there can be a way of doing it... even 2. But i don't really like it because it doesn't use the previous questions...

I don't remember if there are other properties we can get from the equation of a plane... With orthogonality.

Perhaps you can try a thing :

The vector (a,b,c) is orthogonal (perpendicular) to P2 (same rule as for the 1st question).
As $\vec{AB}$ is in P2, you can say that the scalar product of $\vec{AB}$ with the vector (a,b,c) is 0...

I'm not sure... can you try ?
• Apr 11th 2008, 11:24 AM
ashes
I just learned about orthogonal matrices.but cant seem to figure out equations of perpendicular planes.would the plane by any chancebe parallel to to the vector normal to the plane?
• Apr 11th 2008, 12:11 PM
earboth
Quote:

Originally Posted by ashes
Hi guys, I have this question which I really need help ...

The equation of a plane P1 in a cartesian coordinate system is given by 3x+y-z-6=0. THe cordinates of two points A and B are [1,1,3] and [2,1,2] respectively.

i)Find the unit vector normal to the plane P1.
ii)Find the vector AB.
iii)Find the equation of a plane P2 that passes thru the points A and B and is perpendicular to the plane P1.

so far i only got AB=OB-OA ..i think so which is equals to i-k.

1. The normal vector of P1 is $\vec n = (3, 1, -1)$ as Moo has posted.

2. The vector $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1, 0, -1)$

3. That means the plane P2 contains the point A and the 2 vectors which span the plane:

$\vec r = (1, 1, 3) + s \cdot (1, 0, -1) + t \cdot (3, 1, -1)$ which yield the equation of the plane in parametric form:

$\left|\begin{array}{lcr}x&=&1+s+3t \\ y&=&1 + t \\ z&=& 3-s-t\end{array}\right.$

4. To get this equation in coordinate form(?) calculate the normal vector of the 2 direction vectors which span the plane:

$(1,0,-1) \times (3,1,-1) = (1,-2,1)$

The plane contains point A. Calculate the dot-product to get the constant summand. Therefore the equation of the plane P2 is:

$x-2y+z-2=0$
• Apr 11th 2008, 12:42 PM
TheEmptySet
A different, but longer way
Quote:

Originally Posted by ashes
need the maths experts' help...

let $ax+by+cz=d$ be the perpendicular plane $3x+y-z=6$

then its normal vector is

$$

so since the planes are perpendicular the dot product of their normals must be zero. so

$ \cdot <3,1,-1> =0 \iff 3a+b-c=0$

Now if we evaluate the plane at the two given points we get
(1,1,3) and (2,1,2)

$a+b+3c=d \mbox{ and } 2a+b+2c=d$

This is an underdetermined system

solving the system
$\begin{bmatrix}
1 && 1 && 3 && -1 && 0 \\
2 && 1 && 2 && -1 && 0 \\
3 && 1 && -1 && 0 && 0 \\
\end{bmatrix}$

$\begin{bmatrix}
1 && 1 && 3 && -1 && 0 \\
0 && -1 && -4 && 1 && 0 \\
0 && 0 && 2 && -1 && 0 \\
\end{bmatrix}$

so d=2c c=c b=-2c a=c

plugging into our equation we get

$ax+by+cz=d \iff cx-2cy+cz=2c$ divide by c and we get

$x-2y+z=2$
• Apr 11th 2008, 06:48 PM
ashes
just check with you, so the unit vector normal to the plane, $\hat{n}= [\frac{3}{\sqrt{11}},\frac{1}{\sqrt{11}},\frac{-1}{\sqrt{11}}]$ where $\sqrt{11}$ is the modulus of the normal vector
just check with you, so the unit vector normal to the plane, $\hat{n}= [\frac{3}{\sqrt{11}},\frac{1}{\sqrt{11}},\frac{-1}{\sqrt{11}}]$ where $\sqrt{11}$ is the modulus of the normal vector