1. ## Perpendicular vectors

1. Given the points O (0, 0, 0), P (2, 6, −1), Q (1, 1, 1) and R (4, 6, 2), find a vector that is perpendicular to both

(a) OP and OQ (b) PQ and QR (c) RP and QR

2. Find an equation of the plane passing through point P and having vector n as a normal.
P (−1, 3, −2); n = (-2,1,-1)

3. Find an equation of the plane that passes through the given points.
P (5, 4, 3), Q (4, 3, 1), R (1, 5, 4)

2. Originally Posted by dorwei92
1. Given the points O (0, 0, 0), P (2, 6, −1), Q (1, 1, 1) and R (4, 6, 2), find a vector that is perpendicular to both

(a) OP and OQ (b) PQ and QR (c) RP and QR

2. Find an equation of the plane passing through point P and having vector n as a normal.
P (−1, 3, −2); n = (-2,1,-1)

3. Find an equation of the plane that passes through the given points.
P (5, 4, 3), Q (4, 3, 1), R (1, 5, 4)

1. Use the cross product:

$\vec n = \overrightarrow{OP} \times \overrightarrow{OQ}$

$\vec n = (\overrightarrow{OQ}-\overrightarrow{OP}) \times (\overrightarrow{OR} - \overrightarrow{OQ})$

and so on

2. If $\vec x = (x,y,z)$ and $\vec p$ is the staionary vector of P then the plane in question has the equation:

$\vec n(\vec x - \vec p)=0$

Plug in the given values.

3. The plane has the equation:

$\vec x = \vec p + s \cdot (\vec q- \vec p) + t \cdot (\vec r - \vec p)$

3. I'm thankful for the reply but i still dont understand part 2 and 3 solution, could u explain it in a simpler manner?
as in what does vector p is stationary vector of P and
why is there a vector n beside it?

for part 3;
i totally got no idea what it means
with the equation..i know nothing about it.
so sorry, could u explain more specifically?
thanks and sorry for the troubles.

4. Originally Posted by earboth
...
2. If $\vec x = (x,y,z)$ and $\vec p$ is the staionary vector of P then the plane in question has the equation:

$\vec n(\vec x - \vec p)=0$

Plug in the given values.

Originally Posted by dorwei92
I'm thankful for the reply but i still dont understand part 2 and 3 solution, could u explain it in a simpler manner?
as in what does vector p is stationary vector of P and
why is there a vector n beside it?
$\vec n$ is the given normal vector of the plane; $(\vec x - \vec p)$ describes any vector which is placed in the plane and therefore the dotproduct of these 2 vectors must be zero. Thus you get:

$(-2,1,-1) \cdot ((x,y,z)-(-1,2,-3))=0$ Multiply the brackets:

$-2(x+1)+1(y-2)-1(z-3)=0~\implies~-2x+y-z=1$

for part 3;
i totally got no idea what it means
with the equation..i know nothing about it. <<<<<<<< what exactly have you done in vector geometry so far?
so sorry, could u explain more specifically?
thanks and sorry for the troubles.
3. The plane has the equation:

$\vec x = \vec p + s \cdot (\vec q- \vec p) + t \cdot (\vec r - \vec p)$
Two points define a vector (and of course a straight line), three points define a plane. To determine a plane you need one fixed point P and two different directions (exactly: non-collinear). Then any point of the plane (which has the coordinates (x, y, z)) is described by:

$(x,y,z)= \underbrace{(5,4,3)}_{\vec p} + s \cdot (\underbrace{(4,3,1)}_{\vec q}-\underbrace{(5,4,3)}_{\vec p}) + t \cdot (\underbrace{(1,5,4)}_{\vec r}-\underbrace{(5,4,3)}_{\vec p})$

$(x,y,z)= (5,4,3) + s \cdot (-1,-1,-2) + t \cdot (-4,1,1)$

5. thanks..now i really do understand quite a fair bit.
but can u explain what does the s and t means
for the equation of the plane in part 3?
thanks^^

6. Originally Posted by dorwei92
thanks..now i really do understand quite a fair bit.
but can u explain what does the s and t means
for the equation of the plane in part 3?
thanks^^
s, t are real numbers which you can choose to calculate the coordinates (or the staionary vector) of a point located in the plane.

For instance s = 2 and t = -1 produce the point P(7, 1, -2)