# Math Help - Important Vector question!!

1. ## 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.

2. 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 :

$\vec{MN} \begin{pmatrix} x_N-x_M \\ y_N-y_M \\ z_N-z_M \end{pmatrix}$

3. ## reply

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?

4. Yes !

By "norm", i meant modulus yeah

5. ## reply

ok so i got 2 parts done..thanks a lot..What about the last part..?im clueless..

6. 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.
It's already 2 equations ^^

Then...i have to think about it >_<

7. ## reply

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?

8. 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 ?

9. ## reply

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?

10. 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:

$x-2y+z-2=0$

11. ## A different, but longer way

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

$x-2y+z=2$

12. ## reply

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

13. Originally Posted by 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
Yes.