Schnittpunkt Geraden im Raum – Vektorrechnung

Hello everyone! Today I would like to show you step by step how to determine the intersection of two straight lines in space. We have already given our two straight equations here, the straight G and the straight H, and now we are going to determine the intersection of these two straight lines, which we will simply call S. And no matter what your straight equations look like, you can always proceed in the same way. We are now going to set these two straight lines in the same way, namely in terms of coordinates. So we start with everything that is in the first coordinate, also here if we mark it, and now we set equations. So 2 plus S times 2 is the left side, S times 2 we can also turn around to 2S, which is more common, is equal. And now we do exactly the same here with the second straight line, 3 plus T times 1. So 3 plus, and this 1 times T is simply T. That was the first coordinate and we get a first equation from it. You are probably guessing correctly, exactly the same now with the second coordinate for a second equation. 2 plus S times 1 we have on the left side, S times 1, so 1 times S is just S. And on the right side 0 plus T times minus 2, so 0 minus 2T. You don't have to write 0 down, you can leave it out. Second equation, and of course there is a third one, because we have three coordinates here, so we can mark it nicely. And on the left side we have minus 3 plus S times minus 1, so minus 3 minus 1S, so minus S. And on the right side we have minus 1 plus T times 2, so plus 2T. So now we have created an equation system with three equations, but only two unknowns. That's why this equation system is relatively easy to solve. So we want to find out which values come out for S and T here, and we can then use them for the S or the T here to determine our intersection. But for that we have to get S and T out first, and let's take a look at that here. If we really only have two unknowns for three equations, then we can ignore one of the equations first. We still have to check them, but for the calculation until we get S and T out, we can first erase the third equation. That means we look at the first two equations, we can solve one of them relatively easily for S or T. For example, here S comes out on its own, so we could quickly convert the second one to S and then use it in the first one. That's the plan. If we calculate minus 2 here, then S is directly on the left side in the second equation, and on the right side we have minus 2T and plus minus 2. So we immediately converted the second equation to S, and I wrote that down for the second equation here. Here is our S. And we now insert that into the first equation for S. So we insert the second equation into the first one. Then it says 2 plus 2 times, and for S we now insert the part here, a nice bracket around it, because all of this is supposed to be multiplied by 2. So everything that was our S goes into the bracket. And on the right side we only had 3 plus T, so we write that down. Now we have an equation in which only T is present. We can convert that to T, and then we have already found our T. So, solve the bracket. 2 plus 2 times minus 2T. That's minus 4T. Okay, so no plus, but minus 4T. And 2 times minus 2 is minus 4. On the right side we just write that down. Then maybe calculate the numbers here. 2 minus 4 is minus 2 up here. And then bring all T's together on the left side. So let's calculate minus T here. Then we have minus 2 here. Minus 4T minus another T. Then we end up with minus 5T. And on the right side only 3 is left. Then minus 2 over by calculating plus 2. Then only minus 5T is left on the left side. On the right side the numbers are calculated together. 5. And so that T is left alone, the minus 5 has to go. We divide by the minus 5. And then we did it. That our T, 5 divided by minus 5, results in minus 1. So we have a value for T. However, that doesn't mean anything yet. We are still missing the value for S. And we haven't used the third equation yet. So just because you have something for T here, you shouldn't use it yet at this point. So be patient and see what else happens. Let's calculate S first. We can do that with the second equation if we simply use our found value for T. Then we have our S directly in the second equation. Minus 2 times for T. We use our found value of minus 1. We write the rest like this. And then we have minus 2 times minus 1 is 2. The minus 2 at the back. And if we calculate that, we get 0 for S. Okay, that's not bad. It can happen. It's okay, we have two values. But still be patient. We haven't used the third equation yet. Just because we have found two values for S and T, it still doesn't mean anything. It may still be that there is no intersection at all. So don't use the values yet. Let's see in the third equation whether these values are even questioned. We put them in. Simply put them in the equation for S and T and see if the equation is correct. That means we have minus 3 minus on the left. For S we put in 0. And on the right we have minus 1 plus 2 times. And for T we put in our found minus 1. Let's see if the equation is correct. On the left we have minus 3. On the right we have minus 1. 2 times minus 1 is minus 2. So if we calculate the right side, minus 1 minus 2 is minus 3. And yes, on both sides it says the same. That means the equation is correct, the third one. And with that we can use these values here. There is an intersection. Should there be a wrong statement here, for example, minus 3 equals 4, then there is actually no intersection. So you are not allowed to use these values here. They simply have no statement, because something wrong came out here in the third one. But as long as that fits, you are allowed to use these values to calculate the intersection. And for that you can now at the very end either use S here in your equation for S. So either you put the zero in here. A zero is always very practical. Or if you say, no, I like the minus 1 better, you can also use it for the T here in this equation. Very important, S belongs to S, T belongs to T. So don't assign it wrongly at the end. Let's do both so that you can see that the same comes out for both calculations. So we are looking for our intersection in the end and for that we simply insert our found S here. The zero in there for the S is of course very practical. The zero times the vector here all becomes zero. This means that nothing more is added to the vector up here. This is the final vector that we can simply write off and we have already found our intersection. So very fast when you work with a zero. But if you say, no, minus 1, I want the challenge, then we put the minus 1 here for the T and calculate and see if we get to the same intersection. 3 plus minus 1 times 1, so now again coordinate-wise. 3 minus 1 in total is 2. Okay, very good, that's right. Second coordinate, 0 plus minus 1 times minus 2. Minus 1 times minus 2 is 2. And then the last coordinate, minus 1 plus minus 1 times 2. So minus 1 minus 2 is minus 3. And there we also see this intersection. There is only one, so you can choose at the end what you want, which number you like better. Then I hope it helped you. If you have any questions, feel free to write them in the comments. Otherwise, have a wonderful day and we'll see you in the next video. Take care!