Question

multivariable calculus equation of line help:
a) Find parametric equations for the line through (2,4,6) that is perpendicular to the plane x-y+3z=7

b) In what points does this line intersect the coordinate planes?

Answer 1

the gradient of the equation of the plane will give you a vector \(\vec n\) perpendicular to the plane
what do you have for the gradient?

Answer 2

the parametric equation of a line requires a point, and a vector parallel to the line
in this case we want a vector parallel to \(\vec n=\nabla f(x,y,z)\)

Answer 3

we never learned gradiant yet...just basic vectors (dot product, cross product, etc)

Answer 4

then choose three random points in the plane
(1,0,2) (7,0,0) (0,-7,0)
and get two vectors from those points
what do you get for that?

Answer 5

what vector can you get between (1,0,2) and (7,0,0) ?

Answer 6

oh okay! so we just choose random points? (0,0,0) would work right?

Answer 7

no, because (0,0,0) does not satisfy x-y+3z=7

Answer 8

oh!!!

Answer 9

\[0-3(0)+3(0)\neq7\]

Answer 10

typo, but you get the point
so choose some logical points by inspection

Answer 11

so we get a point that satisfies that equation, then subtract it from the other point (2,4,6) to get a vector

Answer 12

then use dot product?

Answer 13

no, we are to going to use the given point just yet; that's for later

Answer 14

aren't*

Answer 15

first we need 3 points in the plane
call them P1 P2 and P3
they must all be (x,y,z) that satisfy x-y+3z=7

Answer 16

so, for instance, it looks like \((3,-1,1)\) is in the plane, because\[3-(-1)+3(1)=7\]

Answer 17

okay

Answer 18

so let\[P_1=(3,-1,1)\]\[P_2=(4,-2,0)\]\[P_3=(1,0,2)\]what vector do we get between\(P_1\) and \(P_2\) ?

Answer 19

<1,-3,-1>?

Answer 20

<1,-1,-1>

Answer 21

btw my the points i chose were...
(7,0,0)
(0,-7,0)
(4,0,1)
but yeah i wna know the process.. my points are right?

Answer 22

yeah, those work too

Answer 23

oh yeah my bad lol -2-(-1) = -1 lol

Answer 24

okay so after you get a vector...

Answer 25

we get the vector P2P3?

Answer 26

yes^
(thinking if sign matters; I think it does not)

Answer 27

then you get the cross product of the two? since it's asking for a perpendicular plane?

Answer 28

you got it :)

Answer 29

oh and thats the answer for part A?! Thanks! and for part B we just set them equal to each other and solve for x,y,z right?

Answer 30

whoa now, we're not done with A just yet
we got the vector, but they want the \(line\) that passes through the given point, which we haven't used yet
to do that you need to know that the parametric equation of a line parallel to a vector \(\vec v\) is given by\[\vec r(t)=P_0+t\vec v\]

Answer 31

in our case \(\vec v=\vec n\) which is the cross product of the two vectors we just found, and \(P_0\) is the given point

Answer 32

oh so we use P0 (which was given) +t(the vector we found from the 3 points)

Answer 33

exactly

Answer 34

I was wondering where the equation of a line concept came in, since it is equation of line section. Thanks!

Answer 35

my I ask, how did you know right away to pick any points that satisfy x-y+3z=7? Like why would it make sense to use points that agree to that equation? Nothing in the problem gives it off for me.

Answer 36

well I am familiar with the behavior of vectors in planes, and I know that if we pick 3 points in the same plane like so...

Answer 37

oh it lies in the same plane...

Answer 38

we can draw vectors between them