Question

the plane through (-1,4,-3) that is perpendicular to the line x-2=t , y+3=2t , z=-t.

Answer 1

the direction vector of the line is:
<1,2,-1>
since the plane is perpendicular to line, then the normal vector to the plane is same as line vector
normal = <1, 2, -1>
equation of plane
\[(x-x_o) + 2(y-y_o) -(z-z_o) = 0\]
where (x0,y0,z0) is point on plane

Answer 2

is it correct ?

Answer 3

i believe so

Answer 4

ye iss ka solution ni :(

Answer 5

how you find 1,2,-1 point ?

Answer 6

oh sorry they are the "t" coefficients of line
x = t + 2
y = 2t -3
z = -t

Answer 7

Let's step back for a minute and determine what we're looking for here, as well as how to find it. @duaali, what is your goal in this problem?

Answer 8

actually it is my calculus assignment and my goal is to submit it tomorrow :p
i have no idea that what to solve and how to solve :(

Answer 9

what I was hoping to hear from you was, "My goal is to find the equation of the plane through (-1,4,-3) that is perpendicular to the line (given above)."

As Dumbcow has pointed out, there's a standard equation of a plane that passes through a given point and has a given normal vector.

Look at the given line. Can you obtain the direction vector from that, as Dumbcow has correctly done?

Answer 10

i think he is right i was confuse with coefficients

Answer 11

Would you mind actually doing the work? What is the direction vector of that given line? Please let's think in terms of doing the necessary work.

Answer 12

wait i have another qestion if u dont mind?

Answer 13

i solved it

Answer 14

I am uncomfortable here because you're not answering the questions I've asked you.
Please type out your solution, here, and explain how you got it.

Answer 15

ok wait m typing ...

Answer 16

(x+1)+2(y-4)-(z-2)=0
x+1+2y-8_z+2

Answer 17

x+2y-z=5

Answer 18

(x+1) + 2(y-4) - is correct, but not the last term in your equation. Could you fix that, please?
Note that x+1+2y-8_z+2 is not an equation, as it must be. Please fix this also.

Answer 19

actually i put z=2 which is wrong
x+1+2y-8-z-3

Answer 20

that's better, but your expression is not an equation, as it should be. sorry to be picky, but we do have to be precise in writing out math solutions.

Answer 21

the equation of a plane in space is just that: an equation.

Answer 22

yeah... thnxx :)

Answer 23

now can i ask another question?

Answer 24

Unfortunately, I need to get off the Internet now. But if y ou'll post your question, either someone else or I will respond. In my case, that'd have to be later on today.
Where are you and what time is it where you are?

Answer 25

m from pakistan and 10:10 pm

Answer 26

Well, go ahead then and post your 2nd question; I'll see what I can do.

Answer 27

the plane through (1 , 2, -1) that is perpendicular to the line of intersection of the planes 2x+y+z=2 and x+2y+z=3

Answer 28

I wish we had more time to discuss this. But briefly: if you have two planes that intersect, you can obtain their respective normal vectors from their equations. If you then take these two normal vectors and cross them (find their cross product), the result will be a new vector which will be the normal vector of your new plane. Then you use the same formula for the equation of a plane in space as last time to find the equation of the plane.

Answer 29

The first given plane has the equation 2x+1y+1z=2, so its normal vector is <2,1,1>.
The second plane has equation 1x+2y+1z=3. Find its normal vector, please.

Answer 30

ok doing

Answer 31

1 , 2 , 1

Answer 32

Fine. Now find the cross product of <2,1,1> and <1,2,1>.

Answer 33

ok

Answer 34

-i -j +3k

Answer 35

that's great. This is the vector that is normal to your new plane.
Take info from this normal vector, along with info from the given point, to write the equation of this new plane.

Answer 36

ok thanks can i ask another question?

Answer 37

As before, I would like very much to get off the Internet now. I know you're under pressure of time to finish this assignment, and am sorry. You have a choice: Post another question (separately) and wait for another person to respond and help you, or wait until I'm online again. At this point I don't know when that will be.

Please post questions a bit earlier next time, so that you won't feel as much pressure to finish them.

You learn fast and I've enjoyed working with you.

Answer 38

thanks a lot sir u r the best teacher :)

Answer 39

thank you ... I truly hope to be able to work with you again. So long for now!

Answer 40

same here sir ... :)