Question

Hi can someone help me with this ?
Write a vector equation and parametric equations of lines:
The line is the intersection of the planes
x + y + z = 1 and
3x-7y + z = 11

Answer 1

you need to find the equation of the intersection of given planes

Answer 2

for that you need below :
1) direction cosines
2) any point on the line

Answer 3

I started off by minusing the 2 equations

Answer 4

you can find direction cosines by the fact that the "line" is perpendicular to both the "planes normals"

Answer 5

start by finding direction cosines

Answer 6

I ended up getting x = 5+4y and z = -4 -5y ..

Answer 7

the answer is r=5i-4k + t(4i+j-5k).. I got the \[r _{0}\] but i don't know how to get the t(4i+j-5k)

Answer 8

start by finding direction cosines

Answer 9

I'm confused ..

Answer 10

its easy, lets start from beginning :)

Answer 11

first step :
observe that you're given two "planes"

Answer 12

and you're asked to find the equation for intersection of those planes.

Answer 13

thats the problem at hand, ok ?

Answer 14

Alright

Answer 15

wat do u knw about equation of plane ?

Answer 16

in the plane equation :
\(ax + by + cz = d\)

the direction consines of its normal are : \(\left< a, b, c\right >\)

Answer 17

right ?

Answer 18

Agreed

Answer 19

and also we knw that, EVERY line on the plane is "perpendicular" to the plane's normal

Answer 20

Right

Answer 21

good, next see that the required line lies on both the planes

Answer 22

so, it is "perpendicular" to both the plane's normals

Answer 23

so, if u take "cross product" of given planes' normals, you wud get the direction cosines of the required line

Answer 24

planes :
x + y + z = 1 and
3x-7y + z = 11

Answer 25

But how do you do that when you have a d ?

Answer 26

direction cosines of any line perpendicular to "normals of these planes"
= <1, 1, 1> x <3, -7, 1>

Answer 27

d will not affect the direction.
d oly changes the distance from origin.

Answer 28

so we get 8 2 -10?

Answer 29

Then when do we integrate d in our calculations ?

Answer 30

yes, thats the direction vector of the required line.

Answer 31

you just need some "point on the line", then u can write the equation of line as :

r = <point> + t<8, 2, -10>

Answer 32

Which i already found at the start .. awesome!

Answer 33

I substracted both equations, got rid of z, replaced it in the first equation and got z = -4-5y .. and x = 5 + 4 y ... I think I answered to fast lol I just confused myslef i think

Answer 34

another easy way for finding the "some point" is to simply plugin z = 0
u wud get two equations in x and y
solve them

Answer 35

then, the "some point" wud be : <x1, y1, 0>

Answer 36

u may use which ever method u feel comfortable... :)

Answer 37

i got x = 5 and y = -4

Answer 38

ButWhich gives d(5 , -4 , 0) but my answer says its's ( 5, 0, -4)

Answer 39

plugin ur point in given plane equations,
does it satisfy both equations ?

Answer 40

Did i miss something ? lol

Answer 41

Ohh i got it cool thanks again :D

Answer 42

u wlc :)