Hi, if we have more teachers like you, math would be accepted easier. My university math professor tells us "memory this and that", i cant im sorry. hahah
Gosh, I hope not! I think all my playlists are still here: www.youtube.com/@bevinmaultsby/playlists I don't have anything called Calc 4, you might be looking for Calc 3? ua-cam.com/play/PLBEl4BT8wUgOqJCBijQxClMIuumgtMinc.html&feature=shared
My approach was we have point (1,2,3) and we have normal vector (3,-1,5) so i did this line (x,y,z)=(1,2,3)+k(3,-1,5) intersected with the plane then calculated the distance between points.
It isn't clear to me how you made the assumption that vector n had coordinates 3, -1, 5 . If you are using some 3D coordinate reference system , the coordinates 3, -1 , 5 is just a point on the plane , not a vector. To determine vector n , surely one would need the coordinates of the tip end of vector n as well as the coordinates of its tail end (which is 3,-1,5). Further , sq root 35 x sq root 35 = 35
Hi Jack, when a plane has the general form ax + by + cz = d, then the vector (a,b,c) is perpendicular to the plane -- in this case, the vector (3,-1,5). That is not a point on the plane (notice it does not make the plane equation true). Remember that vectors do not live in prescribed places in space--the vector which begins at the origin and goes to the point (3,-1,5) is the vector (3,-1,5), but you can slide it around in space wherever you'd like. So I can slide that vector into the position where I sketched it on the diagram. For your last remark, when I broke the projection into v.n/||n|| * n/||n||, the idea was that the length of that (which is what we want) is | v.n/||n|| * n/||n|| | = | v.n/||n|| | (because n/||n|| has unit length). Therefore, there is only one copy of sqrt(35) in the denominator, as the other one was used to normalize n.
Excellent video! 😊
Thank you! 😃
Thank you!!
You're welcome :)
Hi, if we have more teachers like you, math would be accepted easier. My university math professor tells us "memory this and that", i cant im sorry. hahah
Happy to help!
Honestly, it took me a couple of minutes to see that one does not need to find the point where “l” and the plane intersect. Thanks for the insight.
You’re welcome! You can also find that point if you’d like, there are often multiple ways to proceed.
Hey there. For some reason, I can no longer find your Cal 4 series of videos. Am I overlooking something, or are they no longer available? Thank you!
Gosh, I hope not! I think all my playlists are still here: www.youtube.com/@bevinmaultsby/playlists
I don't have anything called Calc 4, you might be looking for Calc 3? ua-cam.com/play/PLBEl4BT8wUgOqJCBijQxClMIuumgtMinc.html&feature=shared
My approach was we have point (1,2,3) and we have normal vector (3,-1,5) so i did this line (x,y,z)=(1,2,3)+k(3,-1,5) intersected with the plane then calculated the distance between points.
Perfect! That would land right on that point I sketched at the beginning.
@@bevinmaultsby but your approach shows us there are many more approaches and something we forget is we can find many points in a plane and a line.
It isn't clear to me how you made the assumption that vector n had coordinates 3, -1, 5 . If you are using some 3D coordinate reference system , the coordinates 3, -1 , 5 is just a point on the plane , not a vector. To determine vector n , surely one would need the coordinates of the tip end of vector n as well as the coordinates of its tail end (which is 3,-1,5). Further , sq root 35 x sq root 35 = 35
Hi Jack, when a plane has the general form ax + by + cz = d, then the vector (a,b,c) is perpendicular to the plane -- in this case, the vector (3,-1,5). That is not a point on the plane (notice it does not make the plane equation true). Remember that vectors do not live in prescribed places in space--the vector which begins at the origin and goes to the point (3,-1,5) is the vector (3,-1,5), but you can slide it around in space wherever you'd like. So I can slide that vector into the position where I sketched it on the diagram. For your last remark, when I broke the projection into v.n/||n|| * n/||n||, the idea was that the length of that (which is what we want) is | v.n/||n|| * n/||n|| | = | v.n/||n|| | (because n/||n|| has unit length). Therefore, there is only one copy of sqrt(35) in the denominator, as the other one was used to normalize n.
Sorry for bothering you, then. I thought I remembered seeing it before I left Africa. Probably just wishful thinking.
No worries!