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

[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...