Question

Where does the line through (1,0,1) and (4,-2,2) intersect the place x+y+z=6 ?

Answer 1

create the vector from point to point; and calibrate it to either of the given points to use in the plane equation

Answer 2

(1, 0,1)
-(4,-2,2)
--------
<-3, 2, -1>

x = 1-3t
y = 0+2t
z = 1- t

plug these in for x, y and z in the plane equation and soplve for t to find the point

Answer 3

so t is -2 right ?

Answer 4

dunno, lets see :)

Answer 5

1-3t+ 0+2t+1- t = 6

2-3t+2t- t = 6

2-2t = 6

-2t = 4; t = -2, i agree

Answer 6

x = 1-3t
y = 0+2t
z = 1- t


x = 1-3(-2) = 7
y = 0+2(-2) = -4
z = 1- (-2) = 3

(7,-4,3) should be the point of intersection if we did it right

Answer 7

so if there is an equation of a plane with three points, for example if a question is like this
A(1,-2,4) is parallel to 3i + 5j + 2k, how do i find the vector equation, parametric equation and Cartesian equation?

Answer 8

is A a point of a vector?

Answer 9

*of = or

Answer 10

its a point on the plane

Answer 11

ok, it a point on the plane; what does "find the vector equation" mean in this context? or are there other points ?

Answer 12

<3,5,2> is the vector, i see it now

Answer 13

x = Ax + 3t
y = Ay + 5t
z = Az + 2t

is the vector equation

Answer 14

i kept seeing that as an equation of the plane .... doh!!

Answer 15

how do i find the vector, parametric and cartesian equation with three points ? for example
three points on a plane have cordinates A(1,-1,0) , B(0,1,1) and C(2,1,-2) referred to an origin. find a vector equation of the plane

Answer 16

create 2 vectors from the points; cross them to find the normal; and use the normal and a given point to establish the plane

Answer 17

A(1,-1,0) B(0,1, 1)
-C(2,1,-2) -C(2,1,-2)
---------------------
<-1,-2, 2> <-2,0,3>

these are the vectors produced from Ato C and B to C

Answer 18

\begin{array}c
x&y&z\\
-1&-2&2\\
-2&0&3
\end{array}

the cross produces a normal vector that is just the determinante of this matrix

Answer 19

x(-2(3)-0(2)) -y(-1(3)--2(2)) + z(-1(0)--2(-2))

x(-6) -y(-3+4) + z(-4)

<-6,1,-4> i sour normal; might wanna dbl chk it tho

Answer 20

forgot a negative; <-6,-1,-4>

Answer 21

we can drop the negatives all the same ....

<6,1,4> is just the same vector in the opposite direction

6(x-Px)+1(y-Py)+4(z-Pz) = 0

where (Px,Py,Pz) is any of the given points

Answer 22

we have to give it in this form
\[r= a + \lambda b\]

Answer 23

ah, that would be parametric i believe. and i aint to confident in those yet

Answer 24

class is about to start; taking a final in discrete today; good luck :)

Answer 25

alright, thanks so much for your help ! :)