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.

A different, but longer way

Quote:

Originally Posted by

**ashes** need the maths experts' help...

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

then its normal vector is

$\displaystyle <a,b,c>$

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

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

Now if we evaluate the plane at the two given points we get

(1,1,3) and (2,1,2)

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

This is an underdetermined system

solving the system

$\displaystyle \begin{bmatrix}

1 && 1 && 3 && -1 && 0 \\

2 && 1 && 2 && -1 && 0 \\

3 && 1 && -1 && 0 && 0 \\

\end{bmatrix}$

$\displaystyle \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

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

$\displaystyle x-2y+z=2$