What are the Coordinates of the School at the Armageddon Point?

[t_n_p]

Homework Statement

In this question, Earth is a plane described by the equation
x + y + z = 18.

Earth will be destroyed by an explosion that occurs at the point A = (1, 1, 1), also known as the “armageddon point”. It so happens that the school (considered as a point) will be the first place on Earth that is destroyed by this explosion.

(a) Calculate the coordinates of the school.


The attempt at a solution
I don't get what the question is asking, if the school is first place to be destroyed, isn't it simply located at point A (1,1,1) ?

 

[kydong_kevin]

(1,1,16) Is it right?

 

[pardesi]

@t_n_p
if u read the lat line then u will note that the question is asking u the point on [y]earth[/u] that will be destroyed first.and note that (1,1,1) is not on earth.so the poin on Earth taht will be destroyed firsyt should be nearest to (1,1,1) ,hence should be on the perpendicular from that point to the plane of Earth and must lie on the plane of earth

 

[t_n_p]

(1,1,16) Is it right?

Not sure what the answer is yet..

@t_n_p
if u read the lat line then u will note that the question is asking u the point on [y]earth[/u] that will be destroyed first.and note that (1,1,1) is not on earth.so the poin on Earth taht will be destroyed firsyt should be nearest to (1,1,1) ,hence should be on the perpendicular from that point to the plane of Earth and must lie on the plane of earth

Ok, that's cleared up now.
I'm still unclear on how to find the coordinates of the school though..

 

[learningphysics]

Not sure what the answer is yet..



Ok, that's cleared up now.
I'm still unclear on how to find the coordinates of the school though..

I recommend drawing a sketch...

You need to find the intersection of the line through (1,1,1) perpendicular to the given plane, and the plane... itself.

What is the equation of the line?

 

[t_n_p]

Not sure how to find the line that is perpendicular through (1,1,1) and the plane x + y + z = 18.

 

[learningphysics]

Not sure how to find the line that is perpendicular through (1,1,1) and the plane x + y + z = 18.

What is the normal to the plane? You can get this directly from the equation of the plane.

Remember that the normal is a vector perpendicular to the plane.

 

[kydong_kevin]

I know the answer :the coordinates of the school is (6,6,6);
The direction vector of Earth n= (1,1,1) ,and A=(1,1,1) ;Suppose the school P=(x,y,z); the line perpendicular to the plane and through points A and P is :
x-1=y-1=z-1, denoted as equation (1). In addition, we know x+y+z=18;
so we can get the coordinates (6,6,6).
Because of the particularity of the number , in this case, we only need two equations. In general, we can get another equation reflecting the shortest distance from A to the plane.

 

[t_n_p]

What is the normal to the plane? You can get this directly from the equation of the plane.

Remember that the normal is a vector perpendicular to the plane.

Well if ax + by + cz = d is the equation of a plane, then the vector (a, b, c) is a normal vector of the plane.

In this case, the vector is (1,1,1) which is point A

What now to find the coordinates of the school?

 

[HallsofIvy]

Find the equations of the line through (1,1,1) with direction vector the same as the normal vector to the plane x+ y+ z= 18. That should be easy and that line will, of course, be perpendicular to the plane. All you have to now is find where that line hits the plane x+ y+ z= 18

 

[t_n_p]

I'm bloody confused at the moment. Can you go through it real slow?

 

[kydong_kevin]

the point A = (1, 1, 1), also known as the “armageddon point”.

 

[t_n_p]

the point A = (1, 1, 1), also known as the “armageddon point”.

?

learningphysics said "You need to find the intersection of the line through (1,1,1) perpendicular to the given plane, and the plane... itself"

But I don't know what to do!

 

[kydong_kevin]

When you know the direction of a line and one point on this line, you can write
an equation for that line . Then the intersection both belong to the line and the
plane. you can get that point's coordinates

 

[t_n_p]

Ok, so step by step.
I know the one point on this line (1,1,1) how do I find the direction of the line?

 

[kydong_kevin]

Suppose the coordinate of one point ,on a line ,is (a,b,c), and the direction
vector of the line is n=(A,B,C). Then that line's expression is ((x-a)/A)=((y-b)/B)=((z-c)/C);

 

[kydong_kevin]

When you know the expression for a plane ,Ax+By+Cz=D. Then the direction vector of this plane is (A,B,C), which also is the direction vector of the line
perpendicularing to that plane .

 

[kydong_kevin]

Just see from the expression for the plane

 

[learningphysics]

Ok, so step by step.
I know the one point on this line (1,1,1) how do I find the direction of the line?

You know that the normal to the plane is (1,1,1)... so that is a vector perpendicular to the plane. So a line that has the direction vector of the normal will be perpendicular to the plane.

So what is the equation of the line that has the point (1,1,1) and has the direction vector (1,1,1) It is : (x,y,z) = (1,1,1) + (1,1,1)t .

Find the intersection of this line with the given plane... You'll be solving for t. Then you can get x,y,z.

 

[t_n_p]

You know that the normal to the plane is (1,1,1)... so that is a vector perpendicular to the plane. So a line that has the direction vector of the normal will be perpendicular to the plane.

So what is the equation of the line that has the point (1,1,1) and has the direction vector (1,1,1) It is : (x,y,z) = (1,1,1) + (1,1,1)t .

Find the intersection of this line with the given plane... You'll be solving for t. Then you can get x,y,z.

So equation of the line is (1+1t)x + (1+1t)y+ (1+1t)z?
How do I find intersection?

So unbeleivably confused by this topic at the moment...

 

[learningphysics]

So equation of the line is (1+1t)x + (1+1t)y+ (1+1t)z?
How do I find intersection?

So unbeleivably confused by this topic at the moment...

No, the parametric equations for the line are
x = 1 + t
y = 1 + t
z = 1 + t

substitute these values for into your plane equation:

x + y + z = 18

then solve for t.

 

[t_n_p]

No, the parametric equations for the line are
x = 1 + t
y = 1 + t
z = 1 + t

substitute these values for into your plane equation:

x + y + z = 18

then solve for t.

I solved for t (t=3). How do I now find the coordinates of the school given I now know t?

 

[learningphysics]

I solved for t (t=3). How do I now find the coordinates of the school given I now know t?

Check your algebra. It's not t=3.

You're trying to find the intersection of the line and the plane. In other words you're trying to find x,y,z that satisfy the equations for the line:

x = 1 + t
y = 1 + t
z = 1 + t

AND the equation for the plane

x + y + z = 18

This is a system of 4 equations and 4 unknowns. So you're trying to find x, y and z.

Once you've found t, just calculate x, y and z. Just use the equations.

 

[t_n_p]

Check your algebra. It's not t=3.

You're trying to find the intersection of the line and the plane. In other words you're trying to find x,y,z that satisfy the equations for the line:

x = 1 + t
y = 1 + t
z = 1 + t

AND the equation for the plane

x + y + z = 18

This is a system of 4 equations and 4 unknowns. So you're trying to find x, y and z.

Once you've found t, just calculate x, y and z. Just use the equations.

my bad, that was a typo and meant to be 5! Hence x=6, y=6 and z=6 and coordinates of school are (6,6,6).

I know it may seem a stupid question, but I need to refresh my memory. How do I find the distance between 2 parallel lines (specifically, point (1,1,1) and (6,6,6))?

 

[learningphysics]

my bad, that was a typo and meant to be 5! Hence x=6, y=6 and z=6 and coordinates of school are (6,6,6).

I know it may seem a stupid question, but I need to refresh my memory. How do I find the distance between 2 parallel lines (specifically, point (1,1,1) and (6,6,6))?

Two lines or two points? The distance between two points (x1,y1,z1) and (x2,y2,z2) is \sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2}

The distance between two parallel lines is a more complex formula... your text might have it...

Here's a link:
http://www.owlnet.rice.edu/~comp360/lectures/ApplicationsofVG.pdf

There's a section that talks about distance between two parallel lines...

 

[HallsofIvy]

I know it may seem a stupid question, but I need to refresh my memory. How do I find the distance between 2 parallel lines (specifically, point (1,1,1) and (6,6,6))?

How to find the distance between two parallel lines is certainly not a stupid question but what do you mean by "specifically, point (1,1,1) and (6,6,6)"? Two points do not determine two parallel lines.

To find the distance between any two lines, choose any point on one of the lines. Write the equation of the normal plane to the line through that point. Find the point at which the second line goes through that plane. Finally, find the distance between it and the original point.

For example to find the distance between x= 2t+ 1, y= 3t+1, z= t+ 1 (which goes through (1,1,1)) and x= 2t+ 6, y= 3y+ 6, z= t+ 6 (which goes through (6,6,6)) (they are parallel since the coefficients of t are the same), first choose any point on the first line: I'm going to take (1,1,1) itself. The equation of the normal plane to that line at (1,1,1) is
2(x-1)+ 3(y-1)+ (z- 1)= 0 or 2x+ 3y+ z= 6.
Now, where does the second line, x= 2t+ 6, y= 3y+ 6, z= t+ 6, intersect that plane? Putting those values in for x, y, z, the equation of the plane becomes 2(2t+ 6)+ 3(3y+6)+ (t+6)= 6 or 4t+ 9y+ t= 14t= -30. t= -30/14= -15/7. Putting that into the equation of the second line, the point where the second line intersects the plane is x= -30/7 + 6= -12/7, y= -45/7+ 6= -3/7, z= -15/7+ 6= 27/7. (The numbers aren't very "nice" because I didn't try to choose a line that gave "nice" answers.) The distance between the two parallel lines is the distance between that point and (1, 1, 1).

 

[t_n_p]

my bad, I totally confused myself on that one. Basically I wanted to find distance betwen 2 points and a point and a plane. I managed to figure that out myself.

Thanks for the help to all,
might have some more questions in the near future!

 

[t_n_p]

Got another question. How do I find the parametric equation of a line in which Earth intersects the plane 4x+3y-z=-3?

Basically I let n be the normal vector of earth, hence (1,1,1)
and n' be the normal vector of the second plane, hence (-4,5,-1)

I cross product those two and get the direction vector of the line, i.e. (-4,5,-1) but I'm not sure where to go from there...

 

[HallsofIvy]

First, the normal vector of the plane you give is (4, 3, -1), not (-4, 5, -1)! Second, the cross product of nether of those with (1, 1, 1) is (-4, 5, -1). How are you calculating the cross product?

In any case you don't need to use the normal vectors at all. The "earth" is represented, you said, by the plane x+ y+ z= 18. Now you want to find the intersection of that with 4x+ 3y- z= -3. If you were finding the intersection of two lines, you would solve their equations simultaneously. Do the same with the two equations of the planes. Since they involve 3 variables, x, y, and z, and there are only you can only solve for two of them in terms of the third. You can use that third variable as the parameter.

By the way, people have written the equation for the straight line from (1, 1, 1) normal to x+ y+ z= 18 in three different ways:
Symmetric form: (x-1)/1= (y-1)/1= (z-1)/1
Vector form: (t-1)i+ (t-1)j+ (t-1)k
parametric form: x= t-1, y= t-1, z= t-1

Those are all valid but which form does your textbook prefer?

 

[t_n_p]

doh, my bad again with the typo..
I got that idea from notes, under "Let’s determine a vector equation
for the line of intersection of two planes".

I think I just got confused with the wording and thought that was what I was after. Anyhow I understand what you mean (sort of)..
Basically, Gaussian elimination?

They want this answer is parametric form. I've only got 2 equations, 3 unknowns, me thinks this is where I introduce the free variable and get the parametric form?

 

[learningphysics]

doh, my bad again with the typo..
I got that idea from notes, under "Let’s determine a vector equation
for the line of intersection of two planes".

I think I just got confused with the wording and thought that was what I was after. Anyhow I understand what you mean (sort of)..
Basically, Gaussian elimination?

They want this answer is parametric form. I've only got 2 equations, 3 unknowns, me thinks this is where I introduce the free variable and get the parametric form?

Yes...get the matrix of equations in row echelon form using gaussian elimination... then let z = t and then get x and y in terms of t...

 

[t_n_p]

Ok, I did that (I think the correct way )
and got z=t, y=21-2t and x=(-66+7t)/4

 

[learningphysics]

Ok, I did that (I think the correct way )
and got z=t, y=21-2t and x=(-66+7t)/4

I think there's some error. The two planes are:

x + y + z = 18 and
4x + 3y - z = -3 right?

One way to check your answer is to see if your direction vector is perpendicular to the normals... your direction vector of your line is (7/4, -2, 1)... the two normals are (1,1,1) and (4,3,-1)... the direction vector is perpendicular to the second normal but not the first. It should be perpendicular to both.

 

[t_n_p]

Yeah silly mistake again..

x=-57 + 4t, y=75-5t, z=t

Should be better now :|

 

[learningphysics]

Yeah silly mistake again..

x=-57 + 4t, y=75-5t, z=t

Should be better now :|

Looks good.

 

[t_n_p]

cool, now i want to find the distance from the line to the point (-5,10,13)

Basically I subtracted the point from the equation of the line giving..(-52+4t, 65-5t, -13+t), now I know the shortest distance is going to be perpendicular, but I don't know exactly what to do next..

 

[learningphysics]

cool, now i want to find the distance from the line to the point (-5,10,13)

Basically I subtracted the point from the equation of the line giving..(-52+4t, 65-5t, -13+t), now I know the shortest distance is going to be perpendicular, but I don't know exactly what to do next..

Yeah, you've calculated the vector from the point to any point on the line which you need... you want to find the point on the line, when this vector is perpendicular to the line... ie perpendicular to the direction vector of your line.

So you want the direction vector of the line to be perpendicular to (-52+4t,65-5t,-13+t)... so find t when this happens. ie: solve for t to make the dot product zero.

Then calculate the x,y,z values of the point using your t.

Then finally calculate the distance between the two points.

 

[t_n_p]

By direction vector, you mean of the line we found out before (the one in parametric form), i.e. (-57 + 4t, 75-5t, t)?

 

[learningphysics]

By direction vector, you mean of the line we found out before (the one in parametric form), i.e. (-57 + 4t, 75-5t, t)?

The direction vector comes from the coefficients of t... so in this case

x = -57 + 4t
y = 75 - 5t
z = t

The direction vector is (4,-5,1)... You can also see this when you write the equation of the line in this form:

(x,y,z) = (-57, 75, 0) + (4,-5,1)t

So this is a line that goes through (-57,75,0) with direction vector (4,5,-1). All lines with the same directon vector are parallel... the direction vector gives the direction in which the line is moving.

Might be helpful to sketch to see this.

(x,y,z) = (0,0,0) + (4,-5,1)t
(x,y,z) = (1,2,7) + (4,-5,1)t
(x,y,z) = (7,9,23) + (4,-5,1)t

These are all parallel lines. When two lines are perpendicular, that means that their direction vectors are perpendicular.

Also important:

To get the direction vector of a line... just take two points of the lines, and subtract them to get the direction vector (that's what you did before with the point you're trying to find the distance from and the line)... so you can get many direction vectors, but they are all parallel

 

[t_n_p]

so (4)(-52+4t)+(-5)(65-5t)+(1)(-13+t)=0?

and solve for t?

 

[learningphysics]

so (4)(-52+4t)+(-5)(65-5t)+(1)(-13+t)=0?

and solve for t?

Yeah, solve for t, get the coordinates of the point on the line with that t... then get the distance between that point and (-5,10,13)

 

[t_n_p]

After subbing t back in, I get (-5,10,13).

How do I find distance for 3d?

 

[learningphysics]

After subbing t back in, I get (-5,10,13).

How do I find distance for 3d?

Hmmm... that doesn't make sense... what did you get for t? Remember you should sub it into the equation for the line ie: sub in here:

x = -57 + 4t
y = 75 - 5t
z = t

 

[t_n_p]

I got t=13

 

[learningphysics]

Hmmm... ok that point (-5,10,13) is actually on the line. Are you sure that's the question? Because then the distance is just 0.

 

[t_n_p]

How far away is the point (−5, 10, 13), from this line (the one with the parametric...)?

the weird thing is we found the equation of the line to be (-5,10,13)..

 

[learningphysics]

How far away is the point (−5, 10, 13), from this line (the one with the parametric...)?

the weird thing is we found the equation of the line to be (-5,10,13)..

The distance is 0.

We were trying to get the point on the line so that the line from (-5,10,13) to the line would be perpendicular to the line... that point turned out to be (-5,10,13). The distance between (-5,10,13) and itself, is 0.

The point (-5,10,13) is on the line:

x = -57 + 4t
y = 75 - 5t
z = t

Maybe that's what they wanted you to find.

We could have checked at the beginning to see if the point was on the line... but I didn't think it would be, since it's kind of a strange question.

 

[t_n_p]

Interesting...

 

[learningphysics]

Interesting...

Are there other parts afterward?

 

[t_n_p]

There are more point, line and plane questions, but none directly related to that question. Interesting isn't it...