Question

find the vector and the cartesian equation of the line passing through (1,2,3) and parallel to the planes
x-y+2z=5 and 3x+y+z=6

Answer 1

@Sauravshakya

Answer 2

let l,m,n be the direction cosines of your line, the line is parallel to the planes
which means the line is perpendicular to the dc of the normal of the line.

Answer 3

would be plse explain more or do the sum fro me..because i have some problem in these sums.

Answer 4

when two vectors are perpendicular, their dot product is zero. the set of direction cosines of lines <l,m,n> and the set of direction cosines of normal of plane <a,b,c> are perpendicular.

Answer 5

find the direction cosines of the normal of the plane/

Answer 6

Plane parallel to x-y+2z=5 and that passes through (1,2,3) is
x-y+2z=5 ....i
Plane parallel to3x+y+z=6 and that passes through (1,2,3) is
3x+y+z=8 ...ii

Answer 7

Now, u need to find equation of line i.e intersection of plane i and ii

Answer 8

works perfectly that way too ... since a line is a intersection of 2 planes in 3d

Answer 9

@sauravshakya , plse continue

Answer 10

x-y+2z=5.....i
3x+y+z=8 ....ii
Let x=0 then
-y+2z=5 ..iii
y+z=8...iv
This gives z=13/3 and y=11/3
So, one of the coordinate of point that lies in the line is (0,11/3 ,13/3)

Answer 11

Similarly, Let z=0
and u will get x=13/4 and y=-7/4
So, another coordinate of the point is (13/4 , -7/4,0)

Answer 12

Now, can u do the rest?

Answer 13

srry
can u plse do further also..

Answer 14

use the definition of the equation of line
http://answers.yahoo.com/question/index?qid=20080830195656AA3aEBr

Answer 15

i m not getting it..

Answer 16

@Sauravshakya

Answer 17

can anyone help me

Answer 18

@AravindG

Answer 19

this is the equation of your line. x0,y0,z0 and x1,y1,z1 are your two points.