Question

Find the eq, of the plane through the point P(-3,3,3), and the line of intersection of the two planes: R x+y-z=5 and Q 3x-y+5z=4

Answer 1

Im doing something wrong with my vectors and getting the wrong answer, last attempt yielded: 25x+9y+43z=81

Answer 2

A bit confused. Is the intersection of P and Q perpendicular to the plane you are trying to find?

Answer 3

I took the cross product of the normal vectors of the two planes to get direction vector (A), set z as a parameter and found a point on the line (U), computed the direction vector from P to U, then took the cross product of PU and A to get the normal vector of the plane (N), Then, using the components of N as coefficients plugged in P to the generic plane equation to get the specific plane.

Answer 4

first of all, you must find out the equation of the intersection line

Answer 5

*the plane contains the line of intersection

Answer 6

Excuse me for any confusion, the unknown plane contains the intersection of R and Q

Answer 7

It might be a better idea to pick two points on the intersection line. Then form two vectors from P to those two points. Cross these two new vectors and you have the normal vectors to the plane.
Since the plane contains two points on the line it must contain all

Answer 8

let n1 is normal vector of R and n2 is normal vector of Q, n1 X n2 = d is vector direction of the intersection line, is it right? and what you have from that?

Answer 9

Yes that sounds right to me, from that i have d=<4,8,-4>, which i am tempted to scale as <1,2,-1> but im not sure if this would effect the final equation of the unknown plane

Answer 10

no no friend, how can you get d like that, the cross product from those vectors is not that, please, check

Answer 11

yes, you are right, sorry

Answer 12

Isn't a plane just uniquely defined by three points? You have P and then you have any 2 points on the line of intersection between R and Q

Answer 13

oops, i got a sign error on my j term!

Answer 14

no, it's right

Answer 15

I think i can use three points too, the "watch it" (watch how someone does it) on the homework site shows it done the way i did it, but as we have already seen it lends alot to computation error!

Answer 16

ok, like what Xavier talk, just pick any point on the line. to get the plane

Answer 17

you have direction d of the line, so can you have the equation of the line?

Answer 18

so ide have L(9/4,11/4,0) and H(4,9/4,1)

Answer 19

Im not entirely sure how to find the eq of the line in such a way that would be use able, the notion of vectors having no starting point is really messing me up here.

Answer 20

why do you choose a fraction to manipulate?

Answer 21

i choose z equal to 1 and zero and thats just what came out

Answer 22

the parametric equation of the line is x =3t, y = -4t, z = -2t. let t =1 you have a point (3,-4, -2). let t =2 you have another point (6,-8,-4), now 3 points easy to go

Answer 23

How did you get that parametric equation with out a point on the line?

Answer 24

just base on the direction vector.

Answer 25

see thats whats confusing me most, is the direction vector somehow "on top" of the line? or is it at the orgin?

Answer 26

wait few minutes, let me re check from my notebook,

Answer 27

ok, thanks

Answer 28

oh my god. i go another way too far from what i study to get the points on intersection line, I dkn . it's weird but it's true. i have 2 points on the line are (29/8,0,-11/8) and (25/8,1.-7/8). I'm sorry, I forgot how to use the direction vector d to figure out the points. mine in previous is wrong. since I assume the plane through origin. . but those points I figure out by algebraic way is correct for sure. from those points you can write out the plane (of course accompany with the point on the problem) hope this helps

Answer 29

I check, and you can check again, those points satisfy both equation plane of R and Q, it means they are on the intersection line

Answer 30

sorry, I check it and know how to get the parametric equation from 2 planes. no way to avoid solve for x, y , z from those equations:
x +y-z=5
3x-y+5z=4
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4x +4z=9 ---> x =9/4 -z
replace to x +y -z =5 ---> y = 11/4 +2z .
Let z =t you have parametric equation is x = 9/4 -t; y = 11/4 +2t; z = t.
From that parametric equation, you let t =0, t =1 or any number else to get as many points as you want on the intersection line.
Bingo. At last, I got it.