Peterlohnes, sometimes the point is not what is the shortest avenue, but the best way to get to a destination based on what you are given. If you noticed Jensen perpetually repeats himself at every question that , THERE ARE MANY WAYS TO SOLVE THESE, the lesson starts with scalar form of a line stated in different ways, however the point of this lesson is to explore a newish way ....Also this is a class lecture, as Jenen always has done, he prepares us for what is to come. I do not know what topics will be taught in the next 5 lessons but the way he teaches, has enabled me to predict how to solve the examples with the new information he is providing me in each lesson while connecting these new information to the old knowledge , I think that is the goal of pedagogy. To teach new stuff and to enable your students to connect them with the old ...Also if you do his homework, question 8 beautifully designed to use exactly what you are saying.
Thank you for another great lesson! This one was really interesting to me because suddenly everything I have learnt from grade 9 and 10 about lines made perfect sense. You brilliantly showed me how mathematics approaches to solve a problem and how the solution must be chosen to fit the problem at hand! How very cool!
Another way to do this is to realize the 4t from the y equation and the 2t from the x equation is actually rise over run: 4/2 = 2. You have a point on this line given by the 3 in the x equation (3,y), and the -5 in the y equation (3,-5). You can use this point to find b by substituting : y=mx + b (put in your m that you found from rise/run)---> y=2x + b (put in your point to solve for b) -5=6 + b, b=-11. Rewrite your line equation using m and b: y=2x -11. Or in standard form: 2x-y-11=0. Saves you from the fractions. Mind you if your rise/run ends up being a fraction, you're back to solving fractions.
Again a faster method at 22:10. Anything is parallel if they have the same m...m is right in the vector equation. In line 2 the rise is 12, the run is 3, so m=4. In line 1, the rise is 4, the run is 2 , so m=2. Not parallel. You should learn the scalar trick for 3 dimensional vectors, so thats why its taught this way.
Yes. You should.... most of the concepts link to previous concepts. All of math is like that, if you skip it can make it really hard to understand what you are doing, and why you are doing it.
Peterlohnes, sometimes the point is not what is the shortest avenue, but the best way to get to a destination based on what you are given. If you noticed Jensen perpetually repeats himself at every question that , THERE ARE MANY WAYS TO SOLVE THESE, the lesson starts with scalar form of a line stated in different ways, however the point of this lesson is to explore a newish way ....Also this is a class lecture, as Jenen always has done, he prepares us for what is to come. I do not know what topics will be taught in the next 5 lessons but the way he teaches, has enabled me to predict how to solve the examples with the new information he is providing me in each lesson while connecting these new information to the old knowledge , I think that is the goal of pedagogy. To teach new stuff and to enable your students to connect them with the old ...Also if you do his homework, question 8 beautifully designed to use exactly what you are saying.
Thank you for another great lesson! This one was really interesting to me because suddenly everything I have learnt from grade 9 and 10 about lines made perfect sense. You brilliantly showed me how mathematics approaches to solve a problem and how the solution must be chosen to fit the problem at hand! How very cool!
You are legendary, keep it up mate
Another way to do this is to realize the 4t from the y equation and the 2t from the x equation is actually rise over run: 4/2 = 2. You have a point on this line given by the 3 in the x equation (3,y), and the -5 in the y equation (3,-5). You can use this point to find b by substituting : y=mx + b (put in your m that you found from rise/run)---> y=2x + b (put in your point to solve for b) -5=6 + b, b=-11. Rewrite your line equation using m and b: y=2x -11. Or in standard form: 2x-y-11=0. Saves you from the fractions. Mind you if your rise/run ends up being a fraction, you're back to solving fractions.
Again a faster method at 22:10. Anything is parallel if they have the same m...m is right in the vector equation. In line 2 the rise is 12, the run is 3, so m=4. In line 1, the rise is 4, the run is 2 , so m=2. Not parallel. You should learn the scalar trick for 3 dimensional vectors, so thats why its taught this way.
Do I need to watch every unit 6 & 7 lesson go understand 8? Our teacher split vectors into 3 units
Yes. You should.... most of the concepts link to previous concepts. All of math is like that, if you skip it can make it really hard to understand what you are doing, and why you are doing it.