Question

The plane that contains the line x : 3t, f : 1 + t, z, :2t
and is parallel to the intersection of the planes y + z : - 1
and2x -.r'- *=0.

Answer 1

might need some clarification on what you are actually posting

Answer 2

find the equation of the plane passing through the given points
The plane that contains the line x : 3t, f : 1 + t, z, :2t
and is parallel to the intersection of the planes y + z : - 1
and2x -.r'- *=0.

Answer 3

sorry this was complete question i want to ask

Answer 4

The plane that contains the line

x = 3t
f = 1 + tz,
2t

im having trouble with your notation

im wanting to see it as:

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

but i just can be sure that is what its saying

Answer 5

is it possible to post a picture of the question?

Answer 6

wait im sending you

Answer 7

find the equation of the plane that passing through the points
... further question is written in atachment

Answer 8

beforehand, some thoughts

we most likely have 2 lines that are skew, the direction vectors of these lines will need to be crossed, to find a normal vector that is perpendicular to them both. this is then the normal vector for the plane, and we can use the anchor point of the first line to establish the equation with

Answer 9

the picture is much clearer to read thnx

Answer 10

thanks should be mine

Answer 11

we determine the line of crossing planes by crossing their own normals, and anchoring it to a point. but we only need the direction vector itself and the line is immaterial to us

this make sense?

Answer 12

sir we have parametric equations of line and two eq.s of plane

Answer 13

plane1: 0x + 1y + 1z = -1
plane2: 2x - 1y + 1z = 0

what are our normal vectors for these planes?

Answer 14

(0,1,1)
(2,-1,1)

Answer 15

good, and we need to cross them ... what is our cross?

Answer 16

cross product ???

Answer 17

yes, the cross product of the normal vectors

Answer 18

2i +2j-2k

Answer 19

very good, since length is immaterial to us: (1,1,-1) is fine. or (2,2,-2) any scaled version of it

Answer 20

sir can i have your personal contact type thing ? to contact to you for study ?

Answer 21

since length is immaterial to us: (1,1,-1) is fine. or (2,2,-2) any scaled version of it
i could not understand sir ?

Answer 22

now we have 2 lines to play with

L1 L2
x = 0 + 3t x = a + 1t
y = 1 + 1t y = b + 1t
z = 0 + 2t z = c - 1t

direction (3,1,2) direction (1,1,-1)

we need the cross product of the direction vectors now

Answer 23

we want direction, not magnitude

Answer 24

the vector (2,2,-2) is equal to 2(1,1,-1)

Answer 25

use either one ... both are the same direction

Answer 26

-3i-5j+2k
cross product

Answer 27

good, (-3,5,2) is the normal to our desired plane

using the anchor point from line 1 (L1) we can create the equation of the plane

what was our anchor point?

Answer 28

P(0,1,0)

Answer 29

good, L1 is passing thru the point (0,1,0)



we have a point and a normal for our plane now