JEE ADVANCED 2026 PAPER 2 || QUESTION 6

Añadido: 2026-05-27

[00:00:00]All right. Hell yeah. So, this is J Advanced. This is 2026, paper 1, question six. And hell yeah, I went on a little 3-day vacation. And for this problem, you know what I mean? I just um I didn't know how to do it. And um and so we just went went for it. Now, I'm not going to show you any of these. I'm just kind of just showing you the work and effort we put in. And then my first question is like did I learn anything?

[00:00:30]Uh my first question is did I learn something intuitive? So if I do see this, you know what I mean? Something like this, do I know what it means? Um but sometimes I don't like to learn things that way. One is like I just like to kind of know I want to you know know the fundamentals. And so we have we're going to have a plane P. And so I'm just going to draw that. And sometimes, you know, your drawing isn't going to be you're you're going to have to change your drawing up um you know, accordingly.

[00:01:05]Um especially if other other things are going to happen. We're going to have we're going to have like uh two different planes, three different planes. Hell yeah. So, we're just going to say this is plane P and it contains a line. Cool. We're just going to put a line in it. Um hell yeah. And uh and it's and this plane is going to be perpendicular to another plane. And so I'm just going to come in here and do that. So that's another plane. And then we have perpendicular.

[00:01:32]And so this plane P is going to be perpendicular to this plane. Now this plane has an equation. Cool. It has this equation. And um and then we're going to let P1 um which isn't, you know, so this wording is very tricky, too, because it's like, hey, this is plane P. They didn't label what this plane was. They just said it's perpendicular. And then all of a sudden they're saying P1. And it seems like P1 might be that it's not.

[00:02:01]But it says let P1 be the plane which passes through a point and is parallel to P. But then even from intuition and not getting flustered. Well, they just said that this plane was perpendicular to P. And then they just said P1 is parallel. So like obviously this perpendicular plane can't also be parallel in some way. And then so I'm just going to go straight up like this.

[00:02:26]And um and then hell yeah. So now this is P1.

[00:02:32]And so and then we have a point we have a point on here. And then so the questions it says hey maybe what's the equation of this plane? Um cool. and then um and then and then the distance between the planes. So I guess I want to start here on the equation of a straight line because this is the most difficult especially equation of a straight line in in three um in three directions in three dimensions because we just have a proportion you know we have three equations that are just proportions and and so from an intuition why do I know that they're a proportion or from the proof and let's go over the proof that's probably the most important one of the day is that we're going to have and it just actually just first starts out with a line. So, we're just going to have three dimensions and we're going to have a line and then maybe up here I'm going to have a vector. So, this so from the proof you would have some sort of vector um and then this would be L. You'd have line L and then you would have a point on line L and then you would have any point you know what I mean on line L. So the any point is going to be X Y Z in three dimensions uh in more dimensions just a continuation of that. And this point is going to be a chosen point. So, it's going to be x knot y z knot and um you know they're going to put this is a vector r and then this is a vector um r and then so anytime that you're doing proofs of lines and planes like draw your plane first draw your line first and then draw the coordinate system like put your origin because you know what I mean you're going to have vectors to there and then you're going to have vectors to there. And if you had vectors to here and here, then you would have a vector here, you know, and either you could put the arrow either way. But then you can say that um you can say that this vector plus and I would just do vector addition because you know we don't know what this is. So I use a little box. We know that this vector is r and this vector is r. And so we're just going to say r plus this box is equal to r. So let's actually do that.

[00:05:05]That's that's super fundamental. Um because this is a subtraction and I can't even see it as a subtraction. It's like we need to get to R. So this vector um do is that right? I mean this vector uh starts here ends at r. So this vector should have r um but then it's um um minus r. See I have a hard time doing vector subtraction from an intuition perspective. And you know would that mean that um yeah and this is a transmissible vector.

[00:05:48]I could put this vector right here. And so that's another cool thing like always do that if you have a triangle and you're working with vectors like make it a parallelogram because this is the same thing right there. Hell yeah. Okay. So but vector addition we're going to do R plus this unknown vector is going to equal R. And so now we can do vector subtraction. Let's just solve for this.

[00:06:12]And it's just R - R. Hell yeah.

[00:06:18]Um, cool. And it should it it seems Yeah. Um, anyway, I'm just going to leave it at that. Hell yeah. Uh, and now this vector right here, so I'm just going to say that it's r - r. Now, from the proof is this these are not equal. So this vector right here is not equal to a. So I can say that a is not equal to r - r but it's um it's proportional. I can say, you know what I mean? And this would this was getting in the um the equation of a line, but I can say that a this vector a that I do know is equal to r - r multiplied by some scalar. So t and this is kind of you call it a parameter, but just any any scaler. Let's say you let's say you lucked out and like this was just twice what a was then a is going to be this times a half you know what I mean this vector times a half if this you know and so forth maybe a is actually going in the other direction you know what I mean um so then this this uh would uh you would have this vector and then this parameter would be negative and then some value to scale it to be the same length length or dimension is that vector. Okay. Hell yeah. So, and then because it's symmetric, then we have all the components of a. So, we have ax is symmetrical to um and then we're looking at this one. Um these two are they're symmetrical. And so we have ax is proportional to rx minus rx.

[00:08:14]Um and and this is what this is how you would pro set up the proportion like that ratio of the x's is the same ratio of the y's u is the same ratio as the z's.

[00:08:34]And so um you know this is just any point and so uh let's write it again.

[00:08:41]This is going to be um uh and they have this proportion flipped. You know what I mean? You could have you you don't have to put up the proportion this way. You could flip these. And so I just did that there. And so um uh let's flip it back. And so what we have is that um rx is x you know I mean that is the x value r y is the y value and then rz is the zvalue. Um the x component of r is just that and that's just x knot. The the y component of r is y knot and z knot. And then so if I said the x component of this vector, well we're labeling it x, the y component and the z component. Cool. So we have this proportion that we have the x component minus the x knot component divided by um the ax component and I will leave it like that. Then we have y - y kn / a y and then zus z kn / a z.

[00:09:56]All right. The other fundamental of a line and this vector A is is and we we showed that it we can proportion it. Um but it's just kind of interesting that it's it's like a map, you know, in two dimensions. If I said um you know, you can only move one block this way and then two blocks that way. Cool. One and then two. And then that's going to be your vector A.

[00:10:29]All right. Hell yeah. And then we have a a city that maybe has, you know, city blocks um just in a grided uh a grid fashion.

[00:10:41]And so I can say wherever you are, you're going to be right there. You can only move, you know, one this way and two that way. Now, you can you don't have to move um you know that far every time. You can move twice as far. You can move half as far. You could move in the negative direction, but you can only move along that uh line. And so, um and so I guess it's, you know, if you did say you're going to move that way and that way, you're technically only moving, you know, from there to there.

[00:11:16]Um but you could look at it like you move this way and then you move that way and those are perpendicular.

[00:11:22]And so the same thing is in three dimensions. Um is that you can have any point in space and if uh and then here's the x direction, the y direction and the z direction. And if I have this vector A that has an X or yeah, I'm just going to say A B C. And these are X Y and Z directions, then I'm going to move this way in the X, this way in the Y, and then this way in the Z. So there's my vector. And I'm going to do that again.

[00:11:52]This way X, this way Y, this way Z. And there's my vector. I could go X, negative Y, negative Z. And then there's my new location.

[00:12:04]umx, - y, negative z. And then so these are all just going to be points on a line. And so this vector a in three dimensions or higher dimensions is the is the magnitude that you'll move in the x, the y, you know, the x, the y, the z.

[00:12:22]Uh, and you can move in the negative direction. Negative x negativex y z. And then you can scale that um as much as you want. But this vector is, you know, kind of like that direction.

[00:12:38]So that's what that is. And so we see that vector right here. And so vector A typically gets um this A B C. So now here is kind of the formal proof of a line is that you have this proportion between um between a point on a line and the actual equation um you know any point on the line and then just a vector that we know on the line. And this is very similar to being like, hey, if we the equation of a line requires that we know um we So here here's how Yeah. If you just had an equation of a line with a slope of two, cool. Do we know that that slope like there's there's a line with a slope of two. There's a line with a slope of two. And so right here is we've shown that we have one piece of information that's the slope. And so that's that's that's what this vector is. This is basically the slope or the trajectory.

[00:13:52]And this is cool because you can use this for higher dimensions. The slope in two dimension is just a change in x, change in y where the slope in higher dimension is um um and even though you have higher dimensions, a line in multi-dimensions is still a line and it can still have a you know if you move in a in a shadow direction where if you shine a shadow downward um on it, you know, you're still you're still your rise is still that distance and your run is still that distance. So even in higher dimensions you can take a line and still have um kind of two a rise and a run to show the slope. So slope in three dimensions um is still just going to be two aspects but it's much simpler to do the slope based on um trajectory. So like right here you would have this would be your A trajectory in the X and this would be your B trajectory in your Y.

[00:15:00]Cool. We know the slope of this line, but we don't know anything about it until we've chosen a point. And so that's where we're going to choose a point like r not on this line. If we say that this point is on that line, then this line becomes our equation. And maybe this is, you know, down here at three. So that would be y is = 2x - 3.

[00:15:24]And now we have the full equation of the line.

[00:15:28]And so um and if we just knew the point on the line that doesn't give us anything you I mean you know that's just one piece of information. Just like here I know the point if we know the point on this line is xnot y znot which is um these values right here. uh it would be 1 3 -2 um based on the way the equation is written this is x - x knot and so um the x knot is just there there and then we need to change this to a -2 uh and that makes the positive2 and so cool so now we we would have um what was I saying so those are the points on the line these are the points on the line. Uh, this setup is is showing you that, hey, that's one point on the line. That's one piece of information. This is just any arbitrary XYZ. That's not a piece of information.

[00:16:31]It's just kind of the structure of how a line is is um is made, but we need that extra piece of information, and that's the slope. And so, yeah. Hell yeah. And I think I'm doing good on this because I did not go over that, you know, on my little vacation. Um, and so that was useful to be like, hey, a line has kind of two things. It has a trajectory, which is the slope, and then it has a point that describes um, you know, where at least one point where that line goes through. And those are the two two things you need.

[00:17:05]So now up here, I can see that um, you know what I mean? My rn vector x knot y knot z knot has a value of of 1 3 -2 and then I know my vector a has a value of a b c which is 2 3 1 hell yeah and back here here's my vector a here is my you know maybe origin here is my r here is my r this vector is r - r kn um hell yeah and so now we can kind of move on to the next thing.

[00:17:50]All right the next thing that I'm going to show is that hey if we have the equation of a plane this is 1 x + 2 y + 3 z those values 1 2 3 are going to be the normal directions. So if here's our plane, um here's a vector that's perpendicular to that plane and it's going to have uh and this can have any magnitude but a vector that describes this and it's transmissible but be can be put anywhere on this plane. Uh but one one um one magnitude or is just going to be these values. It's going to be one two three. And I think it's it's important to know where that proof comes from because it's just a simple dot product, but the literal the dot product might be the Achilles heel of most students. So we're going to take a look at that. So we're done with the equation of a line.

[00:18:44]This is the proof for the equation of a line in in whatever dimensions you know we got three dimensions.

[00:18:51]So, this NASA aspect is we're going to Why is this the equation of a um Oh, yeah. And and that's crazy because even if they just give you the equation of a plane, you have to backtrack and be like, well, um you know what I mean? So, we're going to choose a plane like this, and then we're still going to have a line on the plane.

[00:19:18]Uh is that right?

[00:19:21]Um, oh yeah, there's there's a few ways to do this. Um, but yes, you here's one line on the plane. And what's important for this, if I just walk myself through it, is that you need a perpendicular, and there could be multiple ways to do this, but I need one line on the plane, and then I need another um line on the plane. Um, and I guess I don't need those to be um perpendicular.

[00:19:53]Oh, I have to take a pause because there's a way to do this using the dot product. And um, yeah, let me take a pause.

[00:20:05]All right. Hely, I just took like a 10-second pause. But what's going to happen is if you just have line one and then you're going to try to find a perpendicular direction. I mean, you might luck out and it's going to be perpendicular to this actual plane. Um, but perpendicular could be this direction. It could be in line. It could be still in the plane, you know. So, perpendicular is um uh you can wheel this vector in a full circle. And so one one way to get this vector, this normal vector, and we're just going to say um you know, let's say this is plane P and we're going to say this vector is the normal of P. Then you're going to take an equation for this line um or a vector. Let's say this is a vector on this line A and this is a vector on this line B. Then you can take a cross b and that is going to be a vector um of some magnitude in the normal direction. And so that will equal a normal vector. It's not going to be a unit vector, but then you could make it a unit vector. So lowercase n p uh with a little hat unit vector. That's going to be a cross b divided by the magnitude of a cross b.

[00:21:27]And that would be um a unit vector in that direction.

[00:21:34]All right. Hell yeah. So that was my Achilles heel. And we're actually going to get a new sheet of paper. All right.

[00:21:42]Hell yeah. And so yeah, this is very difficult for me to see is that if we have a plane, a plane can be described by a point in the plane. Um this is going to be and then um and so pick an origin. So we have our vectors.

[00:22:03]This is going to be a point in the plane r which is going to be represented as x knot y z knot. And then this is going to be every point in the plane um x y z and this is sorry about that. This is vector r and um and a plane is not just represented as a line you know it's every every location. So even just this u put our origin in and so again this is r this is r using vector addition like what is this vector so we have r plus this unknown vector is equal to r.

[00:22:48]So we can use now subtraction. So this is going to be r minus r kn. I have to do this almost every time. Use vector addition to figure out vector subtraction. I think there might be a point where it clicks. I just I I don't care to make it click using vector subtraction. I just use vector addition and then rearrange to show the vector subtraction.

[00:23:12]So hell yeah. So this vector is r - r.

[00:23:15]The one the one thing though is like hey you are ending this vector at r so it has to be that and then it's minus r because um you know you're starting over here um instead of the origin or something um you're going r u minus r um oh yeah cuz if you went r and then here's r and if you subtracted that then you would be going in the opposite direction. So using the parallelogram method, you're going to go R and then subtract R. You're going to go this direction. That's that's subtraction. So the subtraction of R. And guess where you are from the origin. You're that which is this vector. Um and I didn't draw it perpendicular. Sorry about that.

[00:24:06]There you go. So there's your parallelogram. Uh and this is R - R.

[00:24:12]Cool. But yeah, it was hard for me to see that just r this point in a plane r and then any point in the plane. I guess this vector can be just any direction.

[00:24:23]And so if here's our plane and here's our r then you know we could go out to there to there to any point in the plane and all of these can be represented as r this xy just any point that is going to be a point in the plane and because of that then you can have a perpendicular vector this n uh and we're going to give it a bc and it's going to be the same they're going to use in my book they're going to use the same ABC is they used on the equation of a line when we just had a line and we had a point R kn x knot y z knot and then any arbitrary point which was xyz and that's represented r and then a vector a that's just in line with that line and then we're going to give that abc. They're going to give the same abc there as they're going to give the same abc here for this normal vector.

[00:25:21]But this sets up a dotproduct. If we take this vector and that was confusing.

[00:25:26]This is not just the ve this is not a line. This is an equation representing every point in the plane.

[00:25:35]Um and it seems very similar because this is this is r - r.

[00:25:42]Um you know if we have an origin here this is r this is r. This vector is r minus r.

[00:25:52]um parallelogram method. This this vector right here is r minus r and that's going to be that vector. So this vector right here is r - r. Uh this would be representing um and I guess this vector uh oh yeah yeah so I probably take another pause because this vector could be anywhere in space. I didn't realize that. Um is that this Oh yeah. And then cuz in both of these were in 3D space.

[00:26:28]You know here's a plane in 3D space.

[00:26:30]Here's a line in 3D space. this vector r - r kn this x this yeah sure you picked a point in 3D space in both of these and then you have a vector an arbitrary xyz I guess I just didn't realize that's totally arbitrary that's that point could be any anywhere in 3D space if you're trying to find the equation of a line then it's hey let's we have to figure out the slope you chose this point cool but then the slope says it can only have one trajectory Tory from that, you know, and in 3D space, if I choose this point, it can have a trajectory in any 3D location. And so that's that's why you have this trajectory. And what this sets up is um you know, and I guess I get it. It sets up equations that are like this. They're symmetric equations, proportional equations. And in three dimensions, you're going to have three of them. In two dimensions, you're just going to have a y equals, you know, mx plus b.

[00:27:35]Um, but that's going to be the um kind of the same same uh format.

[00:27:43]Um, hell yeah.

[00:27:46]All right. So, that that was actually Yeah, I'm just actually learning that for the first time. So that was there was Achilles heel on this and the Achilles heel on this R minus R for sure was that I couldn't see that this R minus R vector is any vector in 3D space.

[00:28:05]Um and same thing here if it's and and then you're and then you're you know to actually get the line you're saying hey we have a trajectory to get the plane you know if you wanted a line in this plane there's a line in this plane and you could have this trajectory you would need a vector that's along that trajectory. Um you could also find just another um and then you could do that by uh a bunch of different ways but just finding another point on this plane and then finding this vector and then trying to find another point on the plane um that has another vector let's say B and then take the crossroduct. So you could easily take the crossroduct of A and B and get a perpendicular vector. Um I think it's actually more difficult. This was my Achilles heel for this one was to see that this was a point on the plane and this was just any arbitrary XYZ.

[00:29:04]Now the thing about this arbitrary XYZ is it has to be a point in the plane. I mean I guess it it doesn't have to be when you first set up R minus R. This can be anywhere in 3D space but then we have to like you know kind of squeeze it or constrain it down that that point perfectly lands in the plane. This is the same thing as having this r being a point on a line and then this um this xyz the r can be any point in 3D space and it's only until we have this trajectory that if you did choose this point it has to be the points now have to be along this this line and so and we do that by similarity equations here we can do it by the dotproduct or the crossroduct but that's really difficult Because the trap for me was just to see, hey, I just drew this vector on the page. I can't understand that this vector could be even up here. You know, I should probably draw this vector XYZ up here just for kicks just to show that when we um when we do take the dotproduct um well, you wouldn't want to do that because we do have to say that this point lies in the plane and it could be here, it could be over there. Um, and sure what the I I Googled this. If you did, if you just chose one point, if you chose a a point here, R, and then another XYZ, like an XYZ input, you're going to have that vector. And if you did the dotproduct of one, you know, and you could you could um you could get result that vector.

[00:30:55]But then if you actually have this vector right here and you're going to set that up as your XYZ um then perpendicular to this would have to be another perpendicular direction. You could say it's this, but then that's going to be perpendicular.

[00:31:17]So the perpendicular to this that's in the plane um you know and then hey this is another point in the plane and the dotproduct is going to be not zero it's um um yeah wait does that make sense if uh this is the line then I could say that the the dotproduct between this vector and that vector these dot products are zero.

[00:31:49]And so I'd be like, well, this is perpendicular to this vector. Cool. I know that. Um, but then let's choose this is our line and this vector isn't perpendicular because to this vector. So I I think that makes sense to me kind of is that hey, this is an XYZ. I'm going to I need a perpendicular vector. I could choose that way. I could choose straight down. I mean I could choose any direction that's perpendicular to that.

[00:32:18]And let's show that. So I have this vector and then I just need it perpendicular. That's perpendicular.

[00:32:24]That's perpendicular. So it's all these vectors are perpendicular.

[00:32:30]Um it's just the fact that this XYZ has to represent every point on the plane. And so this point is on the plane. Oh, let's let's take like let's take take a step back. When I have this vector, I'm trying to find a perpendicular vector to it and I want to call it this. I want to have it be perpendicular.

[00:32:56]And so I can take the dotproduct of this and it could be this one. It could be this one, but it could be that one.

[00:33:06]Um, but then if I actually move this point over here and take a perpendicular to this vector, it couldn't have been that one. It could have been this one.

[00:33:16]You know what I mean? Uh, but it for sure now couldn't have been that vector.

[00:33:21]And I think we could play that game um all the way around this thing.

[00:33:28]Okay. And yeah, let's play that game at a 45 because let's say here is a 45 degree angle to this plane and it's still perpendicular to this vector.

[00:33:41]Cool. Now let's take and then here's the shadow of that 45° vector. So I just took this r - r kn and dotted it with this vector um call it r star and I got zero. So I dotted those two vectors together and I get zero. So I think hey I'm pretty pretty chill. But then I come over here and I take this vector um which is still another r minus r and then I'm going to dot it with um n star.

[00:34:21]But guess what? These aren't perpendicular. We have a 45° angle. So they're going to result in some sort of value as a multiplication between these.

[00:34:31]So these are not zero. And so even though if I started out with one vector and I can have um I can I can take a dotproduct and find a vector that's perpendicular. All right. Hell yeah.

[00:34:48]Sorry about that. But hell yeah. All right. So final conclusion is that if I have this as my r minus r kn I can find a perpendicular vector to that one you know line and the dot product is going to be zero but if I change this r minus r kn to be over here you know what I mean I still need hey that's not 90 so let's do this so hey this is 90 but then if I move this vector over here hey this angle is not 90 no more and so I need to move it to 90. I could do over here, but then when I move this vector over here, hey, this isn't 90. And so what's going to result is this being completely perpendicular to all the r R R minus R knots in the plane. And so I'm glad I spent some time on that because that's I mean that has to be fundamental. Like that just has to be I mean for me to to to go a month without studying this and then coming back to it. I'm so glad that I just spent the 10 minutes kind of playing around with that to be like, "Oh yeah."

[00:35:56]Um yeah. So there was a few things there. I don't want to reiterate. I just want to move on actually. And so hell yeah.

[00:36:03]We're going to um you know, this video is kind of getting long enough, but we're going to actually try to use now those um and we really didn't actually shoot.

[00:36:14]We actually need to get the equation of that. And so let's uh yeah, let's let's do this.

[00:36:25]So now we can kind of whoop this out.

[00:36:27]I'm going to get a new pen. All right.

[00:36:29]Hell yes, we're back. So, we have a plane.

[00:36:34]Um, we're going to have a point in the plane. And then we're going to have uh an xyz that represents any point in the plane.

[00:36:46]That's going to be r and this is going to be r.

[00:36:49]Any chosen point in the plane.

[00:36:53]Here's the vector between those.

[00:36:56]R minus r. Um, here's our origin.

[00:37:03]R are kn you could do the parallelogram method. I think that's useful. Um, cool.

[00:37:12]And now we're going to try to find this normal NP normal to the plane normal to plane P.

[00:37:21]This is going to have values ABC.

[00:37:23]And so we're going to set up the dotproduct. The dot productduct says that this vector r - r dotted with np is going to be zero.

[00:37:37]And then so let's write this out again.

[00:37:39]This is going to be x y z minus x knot y z knot dotted with a b c.

[00:37:52]And you could um show this would be a great time to actually the dotproduct showing that this is going to be and let's do it up here. This is the same thing as XY Z dotted with ABC minus um X knot Y knot Z knot dotted with ABC.

[00:38:19]This is the assoc is it assoc distributive distributive property.

[00:38:25]You're kind of doing the distribution of that. It' be the same thing as 1 + x * 2 which would be 1 * 2 + x * 2 or 1 - x in our case * 2 which is 1 * 2 - x * 2. So that's that's all we're doing there. If you did it in one go, you could say that this these two vectors just do this subtraction and just subtraction. You know, also subtraction is just the same thing as vector addition. Vector addition with this changing signs.

[00:39:07]Hell yeah. Negative x knot negative y z knot. Cool. So you do that, but at the end of the day, we're going to get a new vector that's going to be x - x knot. Ah, see I'm So it's going to be x - x knot as your x value. Uh y - y kn as your y value and z minus z knot as your zvalue dotted with abc.

[00:39:35]If you're not quite used to this vector notation, you know, go back to, hey, this is x ihat plus y jhat plus z khat.

[00:39:47]Um, minus x knot ihat plus y jhat plus z khat dotted with a ihat plus b jhat plus c khat.

[00:40:02]This is typically the way you're going to learn it in college. Um, you know what I mean? Actually, yeah.

[00:40:10]Uh, it was very seldom it seldomly did we actually learn this way. Uh, which simplifies it a lot even though the textbooks probably said to do it this way quite often. Okay. Hell yeah. And then the dotproduct is just going to be the multiplication of the x terms um and with the addition of the multiplication of the b terms and then added to the multiplication of the c terms.

[00:40:41]And I didn't leave myself enough room to do the full proof, but I'm just going to jump up to here. So we're going to have or uh I'll do that in a second or maybe I can write it in here. So, we're going to have x - x a plus y - y b + z - z knot c.

[00:41:04]And this is all equal to zero. Sorry about that. It's all equal to zero.

[00:41:10]So, this the inputs are my x y and z inputs. Everything else are um scalers um you know already their coefficients are already chosen. So x knot and a y knot and b and z knot and c.

[00:41:28]And so I'm going to re um well the a b and c are going to be coefficients but we are going to have um you know just like uh just like y is equal to mx + b. uh this is what I'm talking I don't I don't know the formal term for that but we're going to have that u it's not a function of our variable x or y or z it's just kind of an additional um scalar and and and in with lines it's how far you move it from along the y ais in three dimensions it's I mean it's still how far you're moving it from the origin even in three dimensions okay hell yeah but let's uh Let's distribute this through. So, we're going to and we're going to get the a in front. So, in kind of one step, we're going to go ax minus ax knot plus b y - b y + cz minus cz kn. And then we're going to group terms. This, this, and this had our xyz. And then these were the and this is all equal to zero.

[00:42:40]These were those um uh I don't know how to say it kind of the y intercept. they would I don't think you call it the scalers but anyway whatever they are you're going to get to the other side and then you're going to have this final form ax plus b y plus cz is equal to and I'm going to put this in parenthesis because we're going to have ax knot plus b y and it's going to be plus because we got it to the other side of the equation.

[00:43:15]equation and then plus c z knot. Put this in brackets because they're going to say this bracket is just kind of equal to a variable d. And then this is the equation of a plane is that you have ax + b y + cz is equal to um some value.

[00:43:37]Um I guess I'm just going to call it a scalar value.

[00:43:43]And what were our inputs? Our inputs were this normal vector ABC. Well, hey, that's the coefficients on this x, y, and z. This is a lot of jumble in here.

[00:43:56]Um, but we could still use it. If we knew the equation of a if we knew a point on the plane, which even in our problem, we do. You know what I mean? Hey, this plane passes through the point 422.

[00:44:12]And so I could put 422 right there and is parallel um uh to P which means that P in this plane P and P1 are um like this. They're parallel but they're perpendicular vectors. The vectors that are the per the normal vector of both of those are going to be the same value.

[00:44:43]All right. Hell yeah. Uh but yeah, so we're going to use this process to solve this problem. So hell yeah. Let's get to it.

[00:44:53]And because I got to get on with my day, I'm going to try to but I think that was the most important part just going through the fundamentals. I don't think anyone else I think people are just going to say like hey this is you know hey you're going to say hey this is vector A. How do you know it? Because that's going to be vector A. Vector A is 2 31. Um your r for this, you know, your point on the your point on this line. Um which is a point in the plane. So I'm just going to say it's P is going to be 1 3 -2.

[00:45:27]Um cool. And then let's set this uh diagram up. So here's plane P.

[00:45:36]Here is P1 which is parallel and then here is the other plane. So we do know that this normal vector is uh and I'm just going to say this is this is star. So n star this is plain star and then n star is going to be one 2 3 that's the vector that we have this line uh which is this. This is the equation of that line.

[00:46:07]And so what we're going to do is to to get um actually I kind of forget what we're going to do now. This is a transmissible vector. So we can put this right here. This is our equation of a line.

[00:46:26]And uh we we do know a point on this line.

[00:46:32]um we know a vector on this line. So this could be this could represent vector a and then so actually so if I cross a with n I'm going to get a perpendicular. So here's vector a here's vector n and because we have two vectors a and n you know in a plane when we cross them we're going to get the normal to that. So then and if we got the normal to this plane we would get these values um of those you know what I mean we just wouldn't know what that value is uh but we would get those so let's do that first the the simplest way to do cross productduct is going to be you know take vector a 2 3 1 and stack it on top of vector you know n star 1 2 three.

[00:47:27]Put your I J K up here and take the determinant and um well um yeah is take the yeah take the determinant where these are vectors.

[00:47:46]Yeah, that that confused me but yeah these are not scalers these are vectors.

[00:47:49]So you're going to take I and you're going to cover up this column and then you're going to take the the determinant of the remainder. And so this is Yeah.

[00:47:59]So how you would do it is it's the co-actors. You're um anyway, let's just do it. So you're going to cover up this column and you're going to take the determinant of the remainder, which is 9 - 2 is 7. And that's your i value. For this middle one though, because you're at this, you're uh um it's going to be a negative here.

[00:48:23]You're at a one one location in the matrix. One, row one, column one.

[00:48:30]If you were at row two, column 2, row 3, column 3, or row two, row um one, row one, column three, you know, rows plus columns. If it adds up to an even number is going to be a positive. If it's negative, 1 + 2, that's a negative. So, this is going to be a negative. Cover up this row and column. 2 * 3 is 6 - 1. So, that's five, but it's going to be a negative 5g 5 jhat. Khat, that's going to be a positive value. And so, we have 4 - 3 is 1 + 1 k hat. So there you go. So that that is going to be the normal. Um yes.

[00:49:19]So those crossed this is going to be normal to P. So we're just find we're finding NP.

[00:49:27]So we took um Yep. So hell yeah. So that's going to be NP.

[00:49:32]And hey 751.

[00:49:34]Hey, we're looking pretty good from that other formal definition. the you're going to have ax + b y + c z is equal to ax kn + b y + c z kn. So we need to find a point on this plane. The equation of a line gives us this point on the plane. So let's throw in those values. We know that this is our ABC.

[00:50:05]This is our ABC of NP and this is our ABC of A. And so I'm I might even call this A prime, B prime, C prime. Um because ABC for this normal is probably going to be more crucial to this problem.

[00:50:22]But this better equal, you know, the equation of a plane. And so this part, this part satisfied this side of that equation. How about the -10? So we're going to go seven. And then we have our point one um plus b is -5 plus our three and then plus c is um 1 and then z knot is -2. So we got -2 -15 and 7. And even like right here, sometimes my mind melts. Like even this weekend, it was very difficult for me to see this.

[00:51:04]I mean, does this just scream to you -10? Cuz it didn't for me. Like I was like, seven and two is nine. You know what I mean? And I was just like, oh, don't do that. 7 and -2 is is five or -15 and -2 is -17 and seven. So for sure I made a huge error here. I could not do this mental math. I got tripped up in not seeing that negative and I just saw 7 and 2 is 9 and how the hell is that -10 instead of 17 and 7 is going to be 10. You know what I mean? Um or 7 and two is going to be five. You know um stuff like that. Uh yeah, negatives are brutal. So sweet. So this is going to be -10 and that's going to be true.

[00:51:54]So that one is true. Cool. The distance between planes is a little bit trickier.

[00:52:02]So we have P1 over here uh which is a 422 and I'm just going to say that's P1 KN.

[00:52:11]And then we have we have a we have P over here. And then so we do have we can make you know based on taking we can find a line we can find this vector which is just P1 KN you know P which is P1 uh sorry I don't want to write it that way I want to go P1 KN and so this vector is going to be P1 KN minus uh P which is going to be 4 2 2 that vector minus um this point. So 1 3 -2 1 3 -2. So we have 4 - 1 is 3. 2 - 3 is - 1. And then 2 - a -2 is 2 + 2 is 4. So sweet. So there's that vector.

[00:53:10]And look like guess look look look look what its length could be you know like how off could be the length has to be shorter. So even just right there is like holy Toledo if you know um Yeah.

[00:53:24]Yeah. Like what's the magnitude of this?

[00:53:27]The magnitude you know it can't be uh even if you have even if you have four 4 and 4 the magnitude can't be more than 1.7 * 4. And so let's do that in our calculator. 42 2 + 4 2 + 4 2 square root that. And then 4 * 1.7. Yeah, 1.7. You know what I mean? And so even if all these numbers, pick the biggest one, are the same, and you multiply by two, that gives you a very comfortable idea of the magnitude of that vector. This vector is a skew hopefully or probably. And we're just actually learning we just want the perpendicular distance, which is always going to be shorter. So this vector can't be more than eight. and they're saying it could be 30. And so this is just obviously wrong.

[00:54:24]But let's go through how you would actually solve for the actual distance here. We have this vector and we know um and we have np we have uh oh we have n star and n star is transmissible here and transmissible there. Um no what do we want? Oh no no no we we want uh np and so so the perpendicular distance is going to be a perpendicular vector from plane p and work with unit vectors. So I'm going to go little np which is that and then this is the vector that we just found. And when we dot this vector with little np which is going to be 1.

[00:55:12]What is np? NP is over here. Sorry about that. NP is 7 -5 1. And then we're going to divide it by the magnitude of this.

[00:55:20]7^ squar is 49. 5^2 is 25. 1^2 is 1. So we have 50 + 25 is<unk> 75.

[00:55:30]You should probably, you know, 75 / 5.

[00:55:34]Um, I do this 75 * 2 is 150. So it's 5 * 15. You know what I mean? So that's that's kind of a fast way to do fives is that 75 * 2 is 150. 150 / 10 is 15. And that's going to be the same thing as 5* 15 is 75 half that. So that's kind of a fast way, but anyway, 5 * 15 and this is going to be 5* 5 * 3. So we're going to have five and then it's all the square root, which you can put a square root over just each of the individual ones.

[00:56:06]Um, those are going to add up to five and we got five square roots of three.

[00:56:10]So it's good to be fast at that. Um, sorry I burned through that, but even just mentioning it and talking through it is good. Hell yeah. Um, so this is going to be 5 <unk>s 3. When we do the dotproduct, we're going to have 21 + 5 + 4. So we have 21 + 5 + 4. Um, this is 2530. So we're going to have 30 /<unk> 75. So that was the trick. You did get the 30, but you're obviously going to square it by um, you know<unk> 75, which is what? square<unk> 64 is 8 square<unk> uh 9 81. So you know your 30 divided by 9 you're um yeah 30 divid by yeah you're three or four yeah so your your distance is three or four which we thought you know three or four um and it's not 30. Cool.

[00:57:09]All right. I might fold up a sheet of paper but we're almost done with this.

[00:57:13]The distance of the plane from the origin. So the distance from this plane to the origin. So now we have to pick an origin and then uh this plane you know kind of goes in in the two dimensions that it goes in and then so here's a perpendicular to that and so we have a known point we called it a p and we're going to go p to the origin and so we're going to go um so we have origin to p which is p minus so it's vector p minus the origin and so So that's going to be 1 sorry 1.

[00:57:52]P is 1 3 -2 - 0 0. Obviously that's 1 3 -2. And then what are we going what do we need? Well, we have this n we have um yeah this vector is np sorry this is n star.

[00:58:14]This plane is star. This is np. And so here is also NP going that direction. So we have or just visually NP the unit vector of NP is that direction. I found that and so I'm going to dot these together. Or I could I could show this vector this way. So this vector which has and check out its magnitude.

[00:58:37]You know what I mean? Hey, it's maybe it's definitely not 3 * 2 is 6. So it's going to be less than that. This is 2 * 1.7. This is almost 2 * 2. This is 4.

[00:58:51]So, hey, we're pretty close. This could be four. This could be a Czech Ola.

[00:58:55]Let's see if it is.

[00:58:58]And so, we're going to take this vector 1 3 -2 and we're going to dot it with the unit vector NP. So that's going to be 7 -51 divided by its magnitude 75 which is also 5 roots of 3. So the dot product is is 1 * 7. We'll write this out cuz we're almost done. Plus 3 * -5 + -2 * 1 um and then this would be 7 / that 7 divided by that one divided by that. You know what I mean?

[00:59:35]And so overall it's going to be just everything divided by that. So you can factor that out. So we're going to have 21 minus 15 minus 2. This is going to be -7 um 21.

[00:59:51]I'm just going to I'm just going to add my thing. We're going to be a positive 17. 18 19 20 21. And now I'm going to see well yeah it was 3 to 20 and then one more. So it's going to be positive4.

[01:00:01]So we're going to have four square roots of 75. So that's going to be four five square roots of three.

[01:00:09]Four fth squares. I feel like I may have made an error on that, but let's let's do this. So the distance from the plane P to the origin. The the point on plane P is going to be 1 3 -2. So we got 1 3 -2. That's from the origin. So we think its magnitude is fairly close. we're going to dot it with a perpendicular vector um a perpendicular unit vector to plane P which is this 751 / the<unk> 75.

[01:00:41]And so when we dot these together we're going to have 7 um yeah so that's where I I probably messed up. Yeah. Where did I get 21 here? I'm just losing my mind. I haven't eaten anything this morning and I wanted to make this video first because I went on a little vacation and people are just like make more J and um yeah. Isn't that crazy? You know what I mean? And and we're going to get that same -10 which we should have got. Is that I I I don't know. Anyway, um sorry about that error. So, we're going to have five minus 15 is -10. So, we're going to have -10 / 5<unk> 3. We're close. And this is just the absolute value. So we're going to have 2 2 / <unk>3. And you're like, hey, this is 2 * <unk>3. And let's just see if this would have made sense. You know, I mean, so the answer is no.

[01:01:36]That's wrong.

[01:01:42]And 2 / 1.7 is almost like 2 / two. So it's it's a little bit more than one. Let's say it's 1.3.

[01:01:52]And would we have gotten that? Because vector P is 1 3 -2. But yeah, we just don't know how far it is skewed from that plane. You know what I mean? So there is actually no telling. Um even though the you know all we know is that it can't be bigger than this. You know if this is the vector from the origin to that plane it just can't be bigger you know. So that you know it could it could be zero. This plane could literally go through the origin and slice right through it. Um all right cool. All right last one. Now the acute angle between plane P and the plane star is uh no no no no just just any random plane. So we do have this plane which has this as our normal vector. And so we're going to have this plane with this as our normal vector. and then just an acute plane. And so we're just going to, you know, put that right there. And then this is that normal vector. And what's going to happen is that um um we can due to the transmissibility of these. And so this angle right here is the same thing as putting this normal with this normal, you know, tail to tail. And so if we move that one up to there, then those are going to be like that. And guess what? That's the same angle.

[01:03:24]And so what we can do is we can take um um and then so here's the normal of one of the planes and then here's here's normal of plane one, normal plane two.

[01:03:38]we can take the crossroduct to get a perpendic but basically we're just trying to find um um yeah let's see if these were inline oh shoot let's do it I mean the fact that you can do it two different ways with the dot product and the cross productd that that I think melts my mind. If I go n1 cross n2 divided by the magnitude of n1 cross n2 um this is going to give me the cosine of the angle between them.

[01:04:22]Also n1 dotted with n2 um divided by the magnitude of n1 times n2.

[01:04:35]See, I did this right wrong actually. Uh the magnitude of n1 n2. So let's do that again. Magnitude of n1 crossed n2.

[01:04:47]Um if it was 90, if those two vectors were 90, then it's going to be a straight multiplication. Booya. And then we're going to and then it would be the uh magnitude of n1 n2. This would be a value of one. And we would have coine of the angle. And because this would be one, cosine of one is 90. And that's wrong actually. Um.

[01:05:16]Oh yeah. Yeah. My bad. Yeah, that's why crossroduct is a sign of the angle and a dotproduct is a cosine of the angle because if these are in line or let's say they are 90, you know what I mean? And so, um, oh, I'm losing it.

[01:05:43]All right. Hell yeah. I'm just going to play around with this for a second because I guess I have just lost my intuition for, you know, um, if two vectors are in line, they're going to multiply and their magnitude is going to be the multiplication of kind of the magnitudes of both those vectors.

[01:06:01]If one of the vectors is 60°, you know what I mean? It's only half.

[01:06:08]And so it would be multiplication of both these vectors, but then by a factor of a half. We're going to we're going to do that by a half. And a half is going to come up because that's going to be and if here's our angle between these, that's going to be this value. That's going to be the cosine. So cosine of 60 is a half.

[01:06:32]Cool. If we have perpendicular vectors u and we take the crossroduct those are going to be 100% multiply the vectors together. If we take those vectors and do a 30° angle then those vectors are only and a 30 is more like that. Honestly a 30 is pretty decent 60. Um, but if we multiply these two together, you know what I mean? Only half is going to be the case. And that's going to and that's going to be this angle and sine of 30 is a half.

[01:07:08]So that's where you get this property that a cross b um that magnitude is going to be the magnitude of a magnitude of b sign of the angle. And then over here you're going to have a dotted with b.

[01:07:26]The magnitude of that well a a dob is just a scalar but yeah that's fine. Um you know you don't have to put the magnitude of it but a dob is the magnitude of a magnitude of b cosine of the angle. And so cool I think that's useful to be like hey 30° and 60. All right, we got about no more juice. But hell yeah, 30 and 60 at some point's going to make a half uh based on signs and cosiness. Okay, and so for what we want, we want cosine. And so we're going to take the dot product. And so we're going to let's see what we're going to dot. All right. Hell yeah. Last push.

[01:08:12]And so we know the normal of P is going to be 7 -51.

[01:08:19]And then we're going to take the dotproduct with the normal of this new plane. 2 2 1 is going to be the normal to that plane.

[01:08:30]Um and then so what we have is this is going to be the magnitude of this vector.

[01:08:39]And so I can do that magnitude of magnitude of that vector magnitude of that that vector cosine of the angle.

[01:08:47]And so we're interested in the cosine of the angle. So let's divide by the magnitude of this. This is going to be 49^2 + 5^ 2 + 1 2 square<unk> that which is I'm sorry I mean it's 7 7 2ar 7 2 5 2 1 2 which is 49 + 25 + 1 which is<unk> 75 which is 5 <unk>s 3. This is going to be 2^ squar + 2^2 + 1^ 2ar. So we have four. This is going to be nine square<unk> 9. This is going to be three. So this is going to be 15 square roots three on this side is equal to uh no I'm sorry that's just the magnitude. So we're we're we're taking the dotproduct dividing by the magnitude and getting the coine of the angle.

[01:09:39]Um uh we want the we want the actual angle.

[01:09:47]And so um yeah. Yeah. And so uh this this shown right here inverse cosine of 1 3 <unk>3 is the same thing as this is the ratio.

[01:10:08]Um, oh, how do you do that?

[01:10:19]Yeah, sorry about that. So, this is the equation. If you get these over here and then um so those are over there and then take the inverse coine of both sides.

[01:10:30]There you go. Then this becomes the angle. So this is the angle um this is the inverse cosine of this ratio. The dotproduct is 7 * 2 is 14 - 10 + 1. So we have 15 minus 10 is 5 divided by 15 <unk>3 is going to be 1 3 <unk>3 and that's going to be 1 3 <unk>3 which is correct and that's cool because I messed up. That's why I actually got into a huge deal right here is because I was using the crossroduct. I said the crossroduct is equal to the magnitudes times the cosine and I wasn't getting the right or something like that. And so I'm going to use this idea that 30° um 30 and 60, you know what I mean? These are in line 100%. If it's perpendicular, when you multiply this using the cross productduct, it's 100% of the magnitudes multiplied together. As I do this, it's going to get less. You know what I mean?

[01:11:47]But if I only do this to 60, hey, it's it's still almost 100%. So that's like this. And if I did this, if this is 30, hey, it's almost 100%. So that is also the kicker. Inline 100%.

[01:12:03]One's 30, hey, almost 100%. If it's cross productd perpendicular 100%. If 160, hey, it's pretty much perpendicular, almost 100%. It's only until you get to 30 that now it's it's a half. Like you really reduced it to half. And then the dot product in line to 60. Oo, now you reduced it to half.

[01:12:25]Now these are more perpendicular. Okay.

[01:12:28]Hell yeah. All right. So we did it. And we got uh and I redid that answer. So that was pretty sweet. And so we got A and D. I have not looked up the solutions for these. These are just, you know, and this hopefully is a good video because people wanted be like, "Hey, just struggle your way through it.

[01:12:47]We need more of these videos that are pure and uh show your weaknesses." Hell yeah. All right. Thanks for joining me on this one. Booya.