Question

Points P=(1,0,0), Q=(0,2,0), R=(0,0,3). Is the vector PQ orthagonal to the vector QR why or why not?

Answer 1

create the 2 vectors and dot them

Answer 2

Find an equation of the plane containing P, Q, and R.

Answer 3

cross the 2 vectors from the first question you asked to determine a "normal" vector

then create the equation using the normal vector and any of the given points

Answer 4

\[\rightarrow PR X rightarrowPQ \]

Answer 5

I'm trying to show you here what I have on my paper..

Answer 6

that looks like its going in the right direction; i tend to use a different setup tho

first lets check that your vectors are good

P=(1,0,0), Q=(0,2,0)
-1 -1
-----------------
0 0 0 -1 2 0


Q=(0,2,0), R=(0,0,3)
-2 -2
-----------------
0 0 0 0 -2 3

i see you used PR instead of PQ ... which is fine, any 2 vectors from the set will work

Answer 7

i tend to do a column setup instead of a row setup; and i think that QR will be easier for me to play with

x -1 0 ; x = 6
y 2 -2 ; -y = -3
z 0 3 ; z = 2

Answer 8

of course your vectors produce the same results

x -1 -1 ; x = 6
y 2 0 ; -y = -3
z 0 3 ; z = 2

Answer 9

normal (6,3,2,), attach this to a given point (a,b,c) to produce

6(x-a) + 3(y-b) + 2(z-c) = 0

Answer 10

I have attached a document showing my work and final answer please let me know if it is correct or of any errors I made. Thank you.

Answer 11

your cross is alittle off, and once you find the crossing vector, you have to apply to develop a dot product to all vectors of a given point.

Answer 12

spose we have a point: (xo,yo,zo)

all vectors from this point to any given (x,y,z) point is

(x-xo, y-yo, z-zo)

to define all the vectors of this set that are perpendicular to a given vector (a,b,c), we do a dot product that results in

a(x-xo) + b(y-yo) + c(z-zo) = 0

Answer 13

the crossing of 2 vectors will produce the (a,b,c) vector we need to define the plane, and any given point can be used as the (xo,yo,zo) to anchor it to

Answer 14

a cross vector is only as good as the method employed to produce it. your setup has a few flaws in it

Answer 15

\[\begin{vmatrix}2&0\\0&3\end{vmatrix}i-\begin{vmatrix}-1&0\\-1&3\end{vmatrix}j+\begin{vmatrix}-1&2\\-1&0\end{vmatrix}k\]
\[6i-(-3)j+2k\]
\[6i+3j+2k\to~<6,3,2>\]

Answer 16

so 6i + 3j+ 2k would be my final answer?

Answer 17

of course not, that is only the normal vector of the plane equation.

Answer 18

you need to determine which point you want to use to fill in the actual plane equation with: P Q or R ?? this will be the values of (xo,yo,zo) in the setup

6(x-xo) + 3(y-yo) + 2(z-zo) = 0

Answer 19

=( I'm confused I thought I was on the right track...

Answer 20

on the right track, yes

at the end of the track, no

Answer 21

the "final" result is the construction of the plane equation itself; we need the normal vector as part of it

Answer 22

Find an equation of the tangent plane to the surface z+ y/z = 1 at the point (4, 3, -1).