Question

Hey. This is my first time using OpenStudy. Here's my two questions (from Calc III):

1. Let P = (1, 2, -1). Find the point of intersection of the plane
3x-4y+z=2,
with the line through P, perpendicular to that plane.
2. Let Q = <1, -1, 2>, P = (1, 3, -2), and N = <1, 2, 2>. Find the point of the intersection of the line through P in the direction of N, and the plane through Q perpendicular to N.

Can you show me how to work the problems out? Thanks!

Answer 1

Does anyone have an idea of how to approach this question?

Answer 2

yep

Answer 3

pull the coeffs off your plane eq for me ... what are they?

Answer 4

No plane coeffs are given, what you see is what the question asks. N = the Normal. P = a Point, Q = a vector.

Answer 5

maybe this might be more productive; set up a vector using the coeffs and variables from the plane eq as components

Answer 6

3x-4y+z=2

<3x,-4y,1z>

now take away the variabels

<3,-4,1> this is the vector that is perp to the plane; this is the normal to the plane

Answer 7

use this vector and anchor it to the point given so that we can define all x y and zs on the line

Answer 8

So would I set it up like this:
x = 1 + 3t
y = 2 -4y
z = -1 +1z

Answer 9

***x = 1 + 3t
y = 2 -4t
z = -1 +1t

Answer 10

yes, very good :)

now we know what x y and z has to equal for any given point; and we want to know which point is in the plane itslef

Answer 11

substitute you xyz eqs into the variables of the plane and solve for t

Answer 12

What variables am I substituting for t?

Answer 13

or x, y, z rather

Answer 14

knowing:

x = 1 + 3t
y = 2 -4t
z = -1 +1t

then the point in the line in the plane is satisfies:

3x -4y +z = 2

3(1+3t) - 4(2 -4t) + (-1 +1t) = 2


solve for t to know how to find the xyz point

Answer 15

oh awesome, thanks! - simple enough. how about the next question?

Answer 16

lets see

Q = <1, -1, 2>
P = (1, 3, -2)
N = <1, 2, 2>

Find the point of the intersection of the line
through P in the direction of N,

and the plane through Q perpendicular to N.

I take it P is a point and QN are vectors

Answer 17

set up the line then using P as the anchor and N as the direction vector

Answer 18

x= Px +Nx t
y= Py +Ny t
z= Pz +Nz t

Answer 19

the plane thru Q?

Answer 20

perp to N means N is our normal vector; do we include Q in the plane? hard to see what its asking for clearly

Answer 21

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

Answer 22

yeah I understand the question is worded a bit weird. it's from a textbook. if it helps, this is the answer: (1, 3, -2) - I just want to know how that answer was gotten

Answer 23

is what im thinking but there is no point to anchor any of this to

Answer 24

in fact we can slide this plane up and down the N vector; Im missing a way to read how to get a point to anchor the plane to

Answer 25

Well, the question wants us to find a point, and the answer is, according to the book, (1, 3, -2). Interesting to note that that is the same as the point given

Answer 26

(1,3,-2) is on one of an infinite number of planes

Answer 27

yeah, but there is no way to determine that from my point of view other than saying P is in the plane

Answer 28

and it doesnt specify ANY point to anchor this thing down with

Answer 29

yeah what I wrote initially is all it says. i see why it's an odd question though

Answer 30

read the question again and make sure you havent left anything out

Answer 31

2. Let Q = (1, -1, 12), P = (1, 3, -2), and N = (1, 2, 2). Find the point of the intersection of the line through P in the direction of N, and the plane through Q perpendicular to N.

Answer 32

maybe use of the cross product is required?

Answer 33

ok, those are 3 points in space in that notation; which means we find the line running thru N and P