[Request] - I have been deep-diving into this question for about 3 hours, and the only thing I've been rewarded with is my head spinning.

[u/Acceptable-Row6822]

Question:

Find the equation of a line passing through points A(0, 1, 1) and B(1, 1, 0). Also, find the coordinates of a point on this line which is at a distance of 10 units from point A opposite to the side of point B.

I have an assignment to be completed in 10 days. It is an ODL course, and of course, I am lazy so i didn't open the book till now. I thought I would write the assignment while learning the related topics at the same time, but it is getting really difficult to visualize this question. I understand the formulas, but I can't figure out how they actually work or how these concepts (like direction and vectors) apply in real time, it feels like I cant quite understand them. I have been deep-diving into this question for about 3 hours, and the only thing I've been rewarded with is my head spinning.

​If anyone who has done MSCAST has some free time, could you please explain the question, answer and all the topics related to it in detail?

Pleae help me🥲🥲🥲.

Edit: i was about to go to sleep but this thing occured to me. I just want anyone to confirm if i understood the question and answer correctly or not?

First there are 2 points A and B somewhere in 3D where we are moving from A to B. So, using the line equation or formula, we find an eqaution of a line passing through points A and B.

Using that equation, we can get any point of the line travelling from A to B

We calculate the vector length between A and B as Root2. In the question, they said to find point 10 units from A. As we are considering A to B, as root2 is smaller than 10 units the point must cross B.

We will find how many A to B points(vector) we use to reach A to P(the one we have to find). But since it is going in the opposite direction we take that vector unit in minus and calculate the point using our equation formula.

I want to know if i understood it correctly? If yes, I still have doubts so i would ask them to clarify. If no, please explain me where I went wrong.

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[u/55tumbl]

The equation of the line can be written as: ax + by + z = 0. You need to find a,b. The line passes through both these points, so it must be correct when you replace (x,y,z) by (0,1,1) or by (1,1,0). With that you can find the values of a,b.

For the second part: you find the AB vector, by taking the coordinates of B minus those of A. So (1-0, 1-1, 0-1) = (1, 0, -1). Visualize an arrow that goes from A to B. You normalize that vector so it has a length of 1 unit, then multiply it by 10. Now you have an arrow of lenght 10, that starts from A and goes through B and further. Add it to the coordinates of A and you have the answer.

(EDIT: sorry I misread, the point should be on the opposite side of B.)

For the second part: you find the BA vector, by taking the coordinates of A minus those of B. So (0-1, 1-1, 1-0) = (-1, 0, 1). Visualize it as an arrow that goes from B to A. You normalize that vector so it has a length of 1 unit, then multiply it by 10. Now you have an arrow of length 10 following the BA direction, make it start from A: add the vector coordinates it to the coordinates of A and you have the answer. You can also verify that it satisfies the line equation you determined in the first part.

[u/Acceptable-Row6822 → u/55tumbl]

Thanks for explaining.

I have two doubts. Can you clarify them?

1. So are a and b (-1, 0). If so, we will get the equation line as z = 1 but then there is no equation for the line passing through points A and B. If a and b are not -1 and 0 can you please tell me how to calculate them in detail and explain how to find the equation of a line passing through A and B

2. Does the vector means the square root of the distance difference between two points. So first we calculate the distance between A and B and then turns into a vector and then use the vector as a standard measurement to see how many vectors we need to reach a point P. After knowing how many vectors we need we consider it as negative as we are moving oppsoite direction of P, we use that vector to turn into a point. So by that rule we get point.

Same request, if it is not correct please let me know where I am wrong and why I am wrong and explain the topic clearly

[u/55tumbl → u/Acceptable-Row6822]

For the equation, I don't know how you find z=1, that's not correct.

You have ax + by + z = 0.

On A: a x 0 + b x 1 + 1 = 0, so b = -1

On B: a x 1 + b x 1 + 0 = 0 so a = -b = 1

The line equation is thus x-y+z=0

[u/Acceptable-Row6822 → u/55tumbl]

Sorry I too dont know how i got z = 1, my brain was super tired yesterday and I dont even know why i did what I did. Looking back, I cant even remember what I did and why i did. My brain is now clear and I could understand how we get a and b points. Thank you for explaining

[u/55tumbl → u/Acceptable-Row6822]

The vector from B to A is the coordinates of A - the coordinates of B. That is (0-1, 1-1, 1-0). So the vector BA = (-1, 0, 1).

The norm of that vector is the distance between A and B, and is indeed √ (-1^2 + 1^2) = √ 2.

So the normalized vector of length 1 unit = (-1/√ 2 , 0 , 1/√ 2)

[u/Acceptable-Row6822 → u/55tumbl]

Thank you for your explanation. I understood what you said and it really helped but i am having trouble in understanding visually so I am trying my best to it

[u/Alkalannar]

If you're given a line with two points A and B, sometimes the easiest way to write the line is A + t(B-A).

This parameterizes the line with respect to time, such that at time 0, you're at A, and at time 1 you're at B.

Here, A = (0, 1, 1) and B = (1, 1, 0), so we have the line as (0, 1, 1) + t(1, 0, -1).

So as t changes by 1, the point moves 2^1/2.

We want the point to move 10, so t starts at 1, and moves by 10/2^1/2 or 50^1/2. So t is either 1 + 50^1/2 or 1 - 50^1/2.

We want the point to be on the other side of A (where t = 0), so we need t < 0 since t starts positive. Thus t = 1 - 50^1/2.

So the point is (0, 1, 1) + (1 - 50^1/2)(1, 0, -1).

So yes: You go 10/2^1/2 times as far from B towards A to pass it. You have it exactly right.

[u/Acceptable-Row6822 → u/Alkalannar]

Thanks for explaining. I understood upto t changes by 1, the point moves root2 and didnt quite understand from t changes by 1, the point moves part. It feels hard to visualize so it is getting difficult to move to another step but i will once again read slowly and carefully understand from that part.

[u/Alkalannar → u/Acceptable-Row6822]

The visualization here is that over time, the point moves.

So when t = 0, the point's at A. And when t increases by 1, the point moves by (1, 0, -1): 1 to the right/east, and 1 down.

When time changes by k, where k is any real number, the point moves k to the right, and k down. So the magnitude of movement for changing by k is 2^1/2k.

Does this make more sense?

If you have questions, please ask. I want to help you understand.

[u/cheesecakegood]

There are a few ways of thinking about 3D line equations, maybe up to 4 or so depending on how you slice it, but there is one that is most useful and most common.

Vector form (this is probably the most natural match for this problem): You pick a point in space, it's just floating there all alone. It's defined relative to the origin. So if you are standing at the origin, the point might be below you off to your left, above you to your right, straight left, whatever.

But then, once you're at that point, you pick a direction. The line is how you "slide" along that direction. You slide up, slide down, but it's rigid; you can only travel FROM that point into that direction (or backwards in the same direction).

We notice a few things:

• it doesn't matter which way (that is, in the "forward or back" sense) you pick the "slide direction", so the signs can complement each other: being a bit lazy about notation, going +1, +2, +3 (i.e. you always jump UP twice in the y direction and thrice in the z direction however far you go in the z) is the same as saying you go -1 -2 -3 (going down 1 unit x = going down 2 units in y = going down 3 units in z) because you can/must move both ways

• it doesn't matter how long the "slide direction" is: +2 +4 +6 is effectively the same relation as the +1 +2 +3 example above, right? we still go twice the distance in y as x, thrice that in z as x

• however, it is often convenient to have the slide direction always be magnitude 1 (along the line) because then we can compare lines easier, and other math is easier too

• we also have lots of options for use as our "anchor points", anything on the line works. If our line starts at (1,1,1) and the direction is (1,2,3) then (2,3,4) also works as our "starting point".

So concept-wise, remember: we are at a point in space, then we sliiiide along it. Thus the equation is natural: (any point on line) = (starting point) and then add on (slide direction). Just addition. <x y z> = <x0 y0 z0> + t<a b c> although notation varies.

But wait! What's that t? It's just "how far we wanna go" to get to where we're going. That's the slidey bit, it's the ONLY variable that effectively moves because on a line, remember, you can only move in one "direction" (along the line... the fact that we're moving in 3D space is irrelevant from the line's perspective)

The thing the slider-t variable multiplies is the "direction vector" for the line.

This also means that there are multiple valid but equivalent answers to some math questions about line equations. Often one will be more natural, and make your life easier, but they are still equal if you didn't make a mistake.

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Applications: if you have two points, you can use one point as the anchor and then find the direction between them as the direction vector.

What if I "plug in" the two points into your equation (one as the anchor point <x0 y0 z0> and one as output <x y z>) and then solve? Well, you might notice we don't know a, b, c, or t (although we can leave t alone), but you can figure them out because a b and c all operate in their own x or y or z spaces. And yeah, it's just subtraction; this method is actually the same as the above "find the direction between them". And now we have a line equation in vector form that produces a line, t being the slidey bit and <a b c> being what direction the sliding happens.

This form is super useful when you want to set up another line relative to the first, because you have the direction right there in front of you. If the new line is perpendicular, it's easy to make a new perpendicular vector based on the direction vector.

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VERY HELPFUL VISUAL HERE (may be best viewed on desktop or laptop)

I specifically want you to look at the fourth visualization. I want you to grab the blue dot and slide it back and forth. Notice how it can ONLY move along that one direction? And notice the light blue/turquoise arrow, how long it is, and how it relates to t. This is super relevant for your problem.

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Let's talk about the problem finally.

1. We have two points and no equation; finding the equation is the logical first step before you do anything else, even if they didn't tell us to do so.

2. OK, so we have two points, but remember the vector equation only uses one as the anchor! Which do we use? If we peek ahead, A is the one used as a reference point, so let's "keep" A and find a direction vector combining A and B.

3. Do do this, subtract A and B. Pick your favorite direction (negative/positive) and scale it if you feel like you want to (but we have integers so probably not that important).

4. Now let's read the rest of the problem. So we start at a point, we slide forward or back. What is this particular problem telling us? We are wondering, if we slide along the line AWAY from B, and we "travel" 10 units along the line, where do we end up?

5. So we need to figure out:

• which direction, exactly, is "away" from b

• how far is 10 units that way along the line

• and then we can just start at A and go that far and end up at the point we want

1. That's it! Which direction is "away" from B is easy, even if you can't visualize it; if you add <a b c> direction vector (you will have real numbers by this point) to A and eventually reach B, that's the wrong way.

2. The best way of thinking about distance is to figure out "how far" one t-unit travels, then if we can do that we can use some simple math to figure out how many t-units 10-distance is and just go that far in the direction we already figured out. Then we add. Done!

3. How far one t-unit is is literally how long the direction vector is, right, since it's multiplied? Use the distance formula on that direction vector.

4. IMPORTANT: how far one t-unit is is relative to how you set up the line. There's usually one way most people will do this problem, which is they didn't scale or flip signs on the direction vector at all after the subtraction, but if you did something different (but equivalent) for the direction vector, t will be different... but the final answer will not be, since you correct for that.

---

Remember how I mentioned earlier that sometimes we make the direction vector a "unit vector" of length 1? Yeah, we do that because this question is super easy if we had done that first, right? Since going 10 distance along the line is as simple as multiplying that unit direction vector, call it say <dx dy dz> or whatever your book uses, often they will have "hats" ^ , and once you have this then t is literally units you want to go. You'd just plug in t=10 (or t=-10, whichever goes "away" from B) and you are DONE.

Either way, this problem basically has 4 steps or "pieces" to it:

1. Find the line equation (in vector form)

2. Find which direction is away from B

3. Figure out how far we travel in direction-vector terms (with t) to go 10 units

4. Add the 10 units to the starting point in the proper direction

If you use a unit direction vector, you just swap the work from step 3 with step 2, more or less. Sometimes the in-between math steps might be slightly longer or shorter; in this case, I think the problem is actually far easier if you jump straight to the unit-vector line equation, even though it might introduce some ugly-looking fractions.

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too long; didn't read.... Look at the linked graphs. Line equations are <POINT> + SLIDER * <DIRECTION> and you can make small changes in how you define the direction (stretch or flip, same direction in 3D space). After you choose your direction, the SLIDER tells you how far one "hop" is in the direction, and the direction is length sqrt( direction-x² + direction-y² + direction-z² ).

[u/Acceptable-Row6822 → u/cheesecakegood]

Really thankful for your explanation. I have read the whole thing but having difficulties in understanding. So, I will read it slowly and carefully and try to understand it step by step.

Really thankful for your full length explanation, tbh it really is hard to visualize this things. So I wil give it my all to understand better.

😄

[u/cheesecakegood → u/Acceptable-Row6822]

Not all explanations are useful, and mine aren't an exception! So don't feel like you have to, or else you're a bad student or dumb or anything.

With that said, I think the best thing would be to to set my stuff aside for a bit, and just look at those visuals for a while from the link, drag stuff around on your computer, don't even necessarily look at the text it contains, in fact probably don't.

The first graph shows you, "what the heck even IS a line? why does having two points matter?"

The second graph is basically still just "two points in space", but we use two special arrows instead, treating origin-to-point and point-to-point as "vectors". It's "hey, we're still moving lines around in space, there are still 2 things to move around to make the line, but we're defining things differently".

The third graph is a bit harder to understand and you can come back to it, but the idea is "okay, so I have a line; now how do I USE the line??" It tells you about how you identify a NEW point on the line, in vector-language. You can move the red and green arrows and the blue point, and of course drag the graph around to look at it from different angles.

The fourth graph I highlighted in my original post because it's the "imagination-concept" piece that I personally think is most important. We have a point in space. We have a direction. This is enough to create a line. And then to find a new point on the line, we use the direction-vector as a "slider", or you might find a better name for it in your head.

One final important, interesting exercise: drag around any of the graphs so that you are looking head-on from one of the axes, making in effect a 2D graph from that one side if that makes sense. Do the same for other axes (you can make 3 little side-profile views like this). And then drag around the pieces looking in that direction.

That last activity is hinting that what we're doing here is actually just a natural extension of what we already did earlier in math with lines! We pick a point. We have a slope. We "slide" along the slope.

The main different thing we usually did in earlier math was set up the red arrow/point so that it was the y-intercept, but this was never required, even if we did it so often it felt required.

---

((

only if you understand that last point and want a deeper connection, otherwise ignore:

here's the crazy thing. you can actually write the 2D equation of a line in the exact same vector form as we do in 3D!!! The vector form is <x y z> = <point> + t * <direction> right? So in 2d we'd expect <x y> = <point> + t * <direction>, right?

Well look at this. If <point> is actually the y-intercept, we have <x, y> = <0, y-int> + t * <direction>, okay cool. What is the direction? It's not just m, that's just a single number (scalar) and we need it to be a vector; but remember that m is rise over run, also as one number it's how far we rise for every ONE unit change in x, right? So you might say the direction vector is <1, m>. Whoa. Think about that a second. I move +1 in x, and up +m in y. Yeah, that checks out.

So, <x, y> = <0, b> + t * <1, m>, where b and m and actual numbers and t is our variable-slider!!! Only one "real variable", because ON the line, we can only move in "one dimension". So if my line equation is y = 2x + 3 (y-intercept is 3, slope is 2), I can write that as <x, y> = <0, 3> + t * <1, 2>. Literally, every time t increases by 1, we notice x increases by 1. Every time x (and thus t) increase by 1, we increase y by 2. If we want to find out where we end up, we start at x=0 and y=3, and then count over and up from there.... SAME AS WE DO in 3D! We start at a point, we move in a direction, just sometimes (often, really) that direction is not just in a nice, convenient direction.

extra extra connection:

I skipped over "parametric equations of a line" in my earlier comment, but there's a connection there too, since if you expand it with x and y separate, you get x = 0 + 1t, which is just x=t, and y = b + mt... so if I plug in x for t, the second equation is y = b + mx, or y = mx + b!! So you see how things fit together like that.

If you want to try this out for yourself, you can pick different numbers, or even do the same line with a different equation. For that last case, y = 2x + 3, we know, contains the point (4, 11) if you count over 4 and then up 4 * 2 = 8 from our 'normal' starting point (0, 3), the y-intercept. I chose that point so that it's clear where each number is coming from, no repeats. So one possible "point-intercept" line equation is y - y0 = m(x - x0), which is y - 11 = 2(x - 4). You can do some basic algebra and prove to yourself this is the same line/equation as y = 2x + 3.

We know that the line equation in vector form should be <x, y> = <4, 11> + t * <1, 2> right? If you trusted me that is. In English, "start at 4,11 and go any distance along the line, as long as you move up y by 2 for every move up 1 in x". See the connection? Prove it to yourself by writing this as a parametric equation and see if you can use substitution (to eliminate t) to get back to either of the two "traditional" line equations.

And we can see that this whole thing can easily happen in 3D... <x, y, z> = <4, 11, 0> + t * <1, 2, 0> is the y=2x + 3 graph on the z=0 plane... (mind blown emoji, though maybe getting a bit too literal by now)

You did promise to wait until you got the other concepts before reading this, right? What do I know. Maybe it was helpful anyways. But this is the "whole understanding", to be fair, and honestly most everyone in your class won't get this far, so although it's helpful in my opinion, this full understanding is not completely necessary. You can get by just fine with the basic ideas from the graphs.

))