Equation of lines and planes

[eternal wagon]

given are the planes: V1:4x-2y+z=8 and V2: 2x-3y+z=5, the point p(0,1,10) and the line R: x=2y-3=2z-15 give a system of parametric equations for the line L through p that lies in V1 and is perpendicular to R

[left acorn]

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[eternal wagon]

so I was thinking about this exercise and I thought if we find the cross product between the two vectors of alpha and the line R, we find the vector that is perpendicular to R which means we can find the line L

However it didn’t seem to be right, and I have difficulties on how to start with those exercises when they give you like 3 requirements

[drifting patrol]

what is alpha?

[eternal wagon]

ups that was V1

Sorry

[drifting patrol]

start with a sketch of the problem

that usually simplifies things

also what is this?

oh i see

name a direction vector of R

[eternal wagon]

yeah so the direction of R would be $$\vec{v} = <1,2,2>$$

[eternal wagon]

Alright let me try and do that

[drifting patrol]

are you sure about that?

[eternal wagon]

let me think

so for x it is surely 1

Ups

one sec let me write ti down

oh wait

$$\vec{v_{R}}=<1,1/2,1/2>$$

[ripe wigeon]

Reyess

[eternal wagon]

And then multiply everything by 2 to get rid of the fractions

$$\vec_{v_R}= <2,1,1>$$

[ripe wigeon]

Reyess
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[drifting patrol]

good

[eternal wagon]

Ups wrote it wrong in latex, but you get my point

[drifting patrol]

you have to write the line equation as x-a / .. = y-b / .. and so on

[eternal wagon]

yeah

do now

I have to find out the red vector basically

[drifting patrol]

well that is easy

[eternal wagon]

Yes

[drifting patrol]

any nonzero vector will do

[eternal wagon]

Cross product

[drifting patrol]

why cross product?

cross product of what?

[eternal wagon]

if I take the vector of R and take a cross product with the normal vector of the plane

Wouldn’t that give a vector perpendicular to R

[drifting patrol]

it does

[eternal wagon]

hmm so what did you mean then?

[drifting patrol]

you want the line to be on V1 and perpendicular to R

take two different points on V1 such that their direction is perpendicular to R

[eternal wagon]

yeah so basically the line is the red one in the figure I drew

And we clearly see that the red line or I look at it as a vector is both perpendicular to the line R and the normal vector of the plane

so if I take the cross product of the normal vector and the direction vector of R

That would give me a perpendicular vector to both of these vectors, and thus finding the line is easy

[drifting patrol]

alright, find the cross product

[eternal wagon]

yup got the correct answer

If they tell me something that the sphere touches the plane

is that the same as basically the intersection?

[drifting patrol]

plane is tangent to the sphere

it means the radius is normal to the plane

[eternal wagon]

Oh yeah that’s true my bad

Thanks!

[drifting patrol]

[eternal wagon]

Sorry to disturb you again, but how would you approach that to find the radius? I was thinking that the scalar product must be equal to 0 but you have 3 unkowns

To find the radius

[drifting patrol]

how is the sphere defined

[eternal wagon]

(x-1)^2+(y-2)^2+(z-8)^2=r^2

[drifting patrol]

is r something you need to determine?

[eternal wagon]

Yeah

[drifting patrol]

ok, what is the center of the sphere?

[eternal wagon]

(1,2,8)

[drifting patrol]

and its distance from plane

something about cross products here

[eternal wagon]

Where it touches?

Well that distance I need to find

[drifting patrol]

yeah but you don't know the tangent point

but you do know what the plane is, right?

[eternal wagon]

yeah

yes

[drifting patrol]

"distance of a point from the plane" formula

[eternal wagon]

oh I totally forgot about that one

[drifting patrol]

since the radius is normal to the plane, it must be collinear to a normal vector of the plane

once you know what the radius is, you should be able to solve this problem

[eternal wagon]

oh I only need to find the equation of the sphere with center s and it touches the plane V2

[drifting patrol]

did you calculate the radius yet?

[eternal wagon]

well

I have slightly different answer not sure why

They get radius 1/14^2

but if the distance from the center to the point is 1/sqrt(14)

Then r= 1/sqrt(14)

[drifting patrol]

right now you are assuming the point 0 is on the plane

but it clearly isn't

[eternal wagon]

hmm what do you mean?

[drifting patrol]

name a point on the plane

any point will do

[eternal wagon]

yes

so if the sphere touches the plane, call it point P, we know the center of the sphere and the distance

[drifting patrol]

no

any point

name any point on the plane

[eternal wagon]

okay, not where it touches right?

[drifting patrol]

any pointttt

[eternal wagon]

yes I did that

Where it touches

[drifting patrol]

something that satisfies this equation

[drifting patrol]

dont worry about that now

just name a point on the plane

[eternal wagon]

(1,1,6)

[drifting patrol]

that works

now calculate vector from (1,1,6) to the center of the sphere

[eternal wagon]

okay

<0,-1,-2>

[drifting patrol]

now |(0,-1,-2) . (2,-3,1)|

[eternal wagon]

gives 1

[drifting patrol]

1/sqrt{14}

[eternal wagon]

this is the dot product right?

[drifting patrol]

yes

[eternal wagon]

Yes so it should give 1

[drifting patrol]

and it does

[eternal wagon]

yeah

what does this tell me?

[drifting patrol]

means r = 1/sqrt{14}

or r^2 = 1/14

[eternal wagon]

Yeah

[drifting patrol]

now you already know a normal vector of the plane

(2,-3,1)

[eternal wagon]

yup

ah and then crossproduct with that position vector

[drifting patrol]

its length squared is 14, which means that the radius^2 is scaled by a factor for 14^2

[eternal wagon]

Sorry, now I am quite lost. Give me a second

[drifting patrol]

call the tangent point T(a,b,c) on the plane and C(1,2,8)

do you agree that the vector TC is normal to the plane?

[eternal wagon]

Yes

[drifting patrol]

(2,-3,1) is also normal

[eternal wagon]

yes

[drifting patrol]

they must be collinear

[eternal wagon]

yes I agree

[drifting patrol]

collinear means one is a scaled version of the other

[eternal wagon]

Ah we defined it as 3 or more points that lie on the same line

[drifting patrol]

but we also know the length of the vector TC

[eternal wagon]

Yes

[drifting patrol]

$$ |TC|^2 = (1-a)^2 + (2-b)^2 + (8-c)^2 = \frac{1}{14} $$

[ripe wigeon]

aL

[eternal wagon]

Yup I agree with this so far

[drifting patrol]

but the coordinates of TC are all scaled by the same factor

relative to the normal (2,-3,1)

put these two together and tell me what the coordinates of the tangent point are

[eternal wagon]

my excuses that it takes a little bit long

im thinking

[drifting patrol]

take your time

[eternal wagon]

Hmm, do you have some resources on this stuff to actually understand this well? I am quite struggling a bit on how to think

I looked into some yt videos and a book, but it still very hard to visualize those things

[drifting patrol]

i don't really know, I just know some basic definitions

[eternal wagon]

hmm okay

Because I tried a simpler exercise, but I still struggle visualizing and finding the relationships

[drifting patrol]

keep practicing, but don't just look at it like "what formula do I use here" kind of way

draw a sketch of the problem as I asked you to do earlier so you'd have a better idea of what's going on

[eternal wagon]

yeah I was doing it just now with a picture

[drifting patrol]

some knowledge comes in handy, like knowing that the tangent plane is always perpendicular to the radius

[eternal wagon]

yeah that was p useful

[drifting patrol]

but if you don't believe that, you should prove it

[eternal wagon]

If you could help me with visualizing the following: I have to find the equation of a line L that goes through p(1,2,3) and is parallel with alpha: 4x-5y+4z-3=0 and intersects the line A which has the following system {11x-2y+9=0
{9x-2z-7=0}

so I did this

If the plane is parallel to the line, does that mean that the normal vector of alpha is perpendicular to the vector of L

[drifting patrol]

this is where cross product comes into play again

the line A is the intersection of two planes and its direction is collinear to the cross product of the normals of the planes

[eternal wagon]

Hmm

[drifting patrol]

the normals are (11,-2,0) and (9,0,-2), yes?

but be careful of your sketch

right now your sketch seems to suggest that p lies in A

this is not given in the text

[eternal wagon]

Yes

[drifting patrol]

find the direction of A

[eternal wagon]

okay one second

So A is a system of two planes, and the intersection of those two gives a line?

[drifting patrol]

yes

[eternal wagon]

That’s quite interesting, how that works

[drifting patrol]

two planes either coincide, are parallel or they intersect at a line

[eternal wagon]

Okay and in this case the two planes intersect at a line L, and the line L is parallel with the plane alpha

[drifting patrol]

they intersect at A

[eternal wagon]

Yeah

[drifting patrol]

" line A which has the following system {11x-2y+9=0
{9x-2z-7=0}"

[eternal wagon]

yeah line A

The two planes intersect at line A

And line A intersects with the line L

[drifting patrol]

yes

1. find direction of A
2. is p on the plane alpha or is it on the line A?

[eternal wagon]

doesn’t p go through line L?

[drifting patrol]

it does

but check just in case whether it's on line A

to justify your sketch

or maybe it's on the plane alpha, who knows

[eternal wagon]

see thats the hard part

the drawing to see everything

[drifting patrol]

just check whether p is on the line A

or on the plane alpha

[eternal wagon]

Oh

I see what you’re trying tod au

[drifting patrol]

your drawing doesn't have to be 100% accurate

[eternal wagon]

sorry

yeah

It’s neither

I plugged in the point in both of them, and it didn’t satisfy

[drifting patrol]

alright good

[eternal wagon]

something like this?

now it’s clearly that the point P isn’t on A

[drifting patrol]

ok, find direction of A

[eternal wagon]

Okay

It’s basically <1,-1,1>

[drifting patrol]

what's gonna happen now is this

since L must be parallel to the plane alpha, it means L is on a plane with the same normal vector as alpha, but the plane also contains the point p

but secondly, the line L must intersect with A

[eternal wagon]

So if a line is parallel to a plane, it means their normal vectors are the same?

[drifting patrol]

plane is parallel to plane

and yes, two different planes are parallel if and only if they have the same normal vectors

[eternal wagon]

Ah because it said the line is parallel to the plane, so we just say that the line is on a plane we don’t know

and thus if they’re parallel, they have the same normal vectors

okay makes sense now

so we know the direction of A, we know the normal vector of alpha

and we know it should go through the point (1,2,3)

[drifting patrol]

L goes through p

[eternal wagon]

Yeaht

so we have to find our vector now, but we also have to keep in mind that the line A intersects L

[drifting patrol]

think about it like this

without the line A data we just know that we have a parallel plane to alpha with point p on it

the line L goes through p, but there are infinitely many lines like this

and they are all parallel to alpha

[eternal wagon]

Yeah

[drifting patrol]

somehow line A must determine the direction of L uniquely now

can you find a point on line A as well

so you can write out the line equations for A

[eternal wagon]

(1,1,1) would be a point for A

Ups

No

(-1,-1,-1)

[drifting patrol]

now remember what I said about the parallel plane?

if A intersects with L and L is entirely on the plane, then at the intersection point, A must also be on the plane

[eternal wagon]

Yes

[drifting patrol]

1. find the parallel plane to alpha with point p on it
2. find line equations for A
3. find a point on A that is also on the parallel plane

and that then determines the direction of L

[eternal wagon]

Okay

One sec

Also what do you mean by line equations

Just that satisfy the condition?

I got the point A

[drifting patrol]

x-a/.. = y-b/..

that stuff

or parametrised line equation if you want

[eternal wagon]

yea

So the direction of A is <1,-1,1>

And the point that satisfy A, is (-1,-1,-8)

so I got t-1= x, 1-t = y and t-8=z

so by finding the point, the line L will also go through that point

Thanks btw

[drifting patrol]

A:
{11x-2y+9=0
{9x-2z-7=0}

[eternal wagon]

yeah

[drifting patrol]

-9 +16 - 7

yeah ok

checks out

[eternal wagon]

so I plugged it in the equation of my plane

and t = 3.61

[drifting patrol]

which plane

[eternal wagon]

the plane where p is in and the line L

[drifting patrol]

ok

[eternal wagon]

which was 4(x-1)-5(y-2)+4(z-3) = 0

[drifting patrol]

$$ (1-x,2-y,3-z) \cdot (4,-5,4) = 0 $$

[ripe wigeon]

aL

[drifting patrol]

looks correct yeah

[eternal wagon]

so then I plugged in my x, the line equation

and y and z, and found t to be 3.61

[drifting patrol]

lemme calculate one moment

[eternal wagon]

alright

[drifting patrol]

i got a different result

[eternal wagon]

oh wait one sec

[drifting patrol]

(11,-2,0) x (9,0,-2)

[eternal wagon]

I thought they were points so I just did the position vector thingy

my bad

[drifting patrol]

cross product of the normals of the planes is in the direction of A

[eternal wagon]

<4,22,18>

[eternal wagon]

yeah I was confused on how you would get the direction of the line A

but seems like you just take the cross product, not sure why exactly that is

[eternal wagon]

this right?

t = -37/22

[drifting patrol]

yes, that's also what i got

[drifting patrol]

4x - 5y +4z -6 = 0

[eternal wagon]

yes

[drifting patrol]

(-1,-1,-8) is on A so

$$ L : \frac{x+1}{2} = \frac{y+1}{11} = \frac{z+8}{9} $$

[ripe wigeon]

aL

[drifting patrol]

$$ (x,y,z) = (2t-1, 11t-1, 9t-8) $$

[ripe wigeon]

aL

[eternal wagon]

yep got the same

,w solve 4(-1+4t)-5(-1+22t)+4(-8+18t)-6=0

[ripe wigeon]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by reyess77

[drifting patrol]

,w 4(2t-1) -5(11t-1) + 4(9t-8) - 6 = 0

[ripe wigeon]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by al8628

[drifting patrol]

ok yeah, it's correct

now you found the point on A that is also on the same plane as L

and you also know another point on L, hence you have direction of L

[eternal wagon]

ohhh

im sorry for a lot of questions, I hope you didn't mind but

could you maybe explain visually why the cross product of the normals of A gives the direction of the line?

[drifting patrol]

by definition of cross product, A x B is perpendicular to both A and B

yes?

[eternal wagon]

yes

[drifting patrol]

and only the direction of the intersection line can be perpendicular to both normals at once

[eternal wagon]

hmm a little bit confusing but I'll get there

[drifting patrol]

fold a paper in half

[eternal wagon]

yes

[drifting patrol]

now the folding line is the intersection line

[eternal wagon]

but don't you have two planes?

[drifting patrol]

one half is one plane, second half is another

leave it at an angle

this picture should illustrate my point

[eternal wagon]

ah I see yeah

also, just to make sure do you also get x = 1+96t, y = 2 +440t, z = 3+454t

[drifting patrol]

i didn't calculate, looks too tedious, but it should be obvious from here

p(1,2,3) and the other point give the direction

[eternal wagon]

yeah I did that

however how do I go from a line to a system of planes

[drifting patrol]

then it should be fine

[eternal wagon]

because thats what they wrote in the solution

like A how they gave the line, they made a system with two planes and the intersection gives the line

[drifting patrol]

"find the equation of a line L that goes through p(1,2,3) and is parallel with alpha: 4x-5y+4z-3=0 and intersects the line A which has the following system {11x-2y+9=0
{9x-2z-7=0}"

[eternal wagon]

yeah

[drifting patrol]

this asks to find line equation for L

you did that

that's all

[eternal wagon]

ah alr, because this was the answer and I just thought I was wrong but maybe they went a little over it

[drifting patrol]

there are infinitely many planes with the same intersection line L

you know the direction of L

pick any two planes such that the cross product of their normals is collinear to the direction of L

and of course p(1,2,3) is on both planes

[eternal wagon]

ah I see, alrighty that helps out

ty for ur help and ur time!

[drifting patrol]

[drifting patrol]

@eternal wagonif your questions have been answered close the topic please

+close

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[eternal wagon]

+close

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