For those wondering where the formula comes from, it comes from the fact that, if we have two vectors, u and v, the area of the parallelogram they form will be | u x v |. If we have a line passing through some point, O, and another point not on our line, P, we can form two vectors. Firstly, d, the direction vector along the line, and OP, the vector from O to P. They will form a parallelogram who's height is the distance between the line and the point. The height of a parallelogram is just the area divided by the base-length. The base-length of our parallelogram is | d |. This gives us the final formula; | d x OP | -------------- | d | An alternative, maybe more intuitive, formula is given by the vector orthogonal to the projection. This gives the same answer as the above formula; Orth d ᴾᴼ
The formula at 4:30 was all I needed to solve and understand this problem. Not sure what the other 8 and a half minutes of this video covered, but that formula was genius! (using cross product to find the area of a parallelogram drawn using the line and point and then dividing out the base of the parallelogram, which leaves you the height, which is what you were solving for)
Thank's so very much for going step by step in your videos! You've helped me a lot! I like how you teach us by breaking down the parts in the formula into definitions...Sometimes I have to pause and replay a few parts because your notation is unorthodox :P
You neglected to tell us WHY you use that formula. It's some kind of vector projection . But I don't remember which, or how you know to use that formula. Which I what I was watching the video for. :(
A much faster approach would be through perpendicular projection of vector b on vector a, with that calculating the closest point on the line to the point of the line, and then finishing by calculating the distance between the two points.
I've noticed that when deriving the vector B, you subtracted by . Is there any particular reason that you did it in this order, and would it be OK if I did it as - ?please someone to make it clear
thanks mam Krista , your way of teaching helps me alot , i wish i could be ur student in university or i could gift u chocolates when every time attending ur lectures there.
I've noticed that when deriving the vector B, you subtracted by . Is there any particular reason that you did it in this order, and would it be OK if I did it as - ?
i have a confusion. i think the formula is not correct according to the recent research it should be bxa instead of axb bcz axb!=bxa. if you think i'm wrong then explain me plz
all the work up to 2:40 seems unnecessary.... i can get the point from the constants and the vector components from the coefficients of the parametric equations...
Does this find the shortest distance between a point and a line? Because I would assume that's the only reason you'd need to answer this question, and I don't understand how finding the distance between a point and a random point on the line counts as achieving that goal. Unless somehow that point is the closest point on the line, which you didn't really explain either way.
My confusion came from |axb|/|a| because my textbook uses the point on the line as the initial point, so I was thrown off thinking it should’ve been |bxa|/|a|.
Math is never a big problem to me (noy that i like it, i detest it), but this vector unit's truly been a massive headache generator, a pain in the bum.
Very poor presentation.. I watched your all videos. In most of the videos, presentation is poor.. You are not giving enough explanation .. So it will be complicated for viewers
I've noticed that when deriving the vector B, you subtracted by . Is there any particular reason that you did it in this order, and would it be OK if I did it as - ?
For those wondering where the formula comes from, it comes from the fact that, if we have two vectors, u and v, the area of the parallelogram they form will be | u x v |.
If we have a line passing through some point, O, and another point not on our line, P, we can form two vectors. Firstly, d, the direction vector along the line, and OP, the vector from O to P. They will form a parallelogram who's height is the distance between the line and the point.
The height of a parallelogram is just the area divided by the base-length. The base-length of our parallelogram is | d |. This gives us the final formula;
| d x OP |
--------------
| d |
An alternative, maybe more intuitive, formula is given by the vector orthogonal to the projection. This gives the same answer as the above formula;
Orth d
ᴾᴼ
I love you. I think you’re beautiful, and I want you to know that.
Thanks a million times
Solid explanation sir
dude that cleared up everything so well
Bros a genius
The formula at 4:30 was all I needed to solve and understand this problem. Not sure what the other 8 and a half minutes of this video covered, but that formula was genius! (using cross product to find the area of a parallelogram drawn using the line and point and then dividing out the base of the parallelogram, which leaves you the height, which is what you were solving for)
love this
your concept is far way better than my lecture notes, super short ,sweet and straight to the point and of course the answer :) thank you!
:D
You are absolutely amazing! Seriously! After 5 years...I am still coming to you! You are a wonderful soul!
Wow, thank you so much, Dee! I'm honored! :D
Thank you so much for the step by step explanation!! It's so much more easier to understand and it helped me through my tutorials!
Where did you get this problem? because on my assignment I literally have the exact same values... the math gods favor me this day!
This is Early Transcendentals. A bit late but I have the same one rn lol.
Thank's so very much for going step by step in your videos! You've helped me a lot! I like how you teach us by breaking down the parts in the formula into definitions...Sometimes I have to pause and replay a few parts because your notation is unorthodox :P
Thanks, you explained it better in 8 mins than an hour lecture
You neglected to tell us WHY you use that formula. It's some kind of vector projection . But I don't remember which, or how you know to use that formula. Which I what I was watching the video for. :(
Your videos help me so much with having a better understanding in my Calc class
Oh good! I'm so glad the videos have been helping! :D
I appreciate your time in this! thank you!
You're welcome!
Can you explain how you got the formula at 4:30
Thank you for making this video! God bless you and your family!
You're so welcome, Candy! :)
Thank you very muchhhh... the steps are clearly explained in the video ❤❤👍
Thank you so much! This makes so much more sense and wasn't explained in my textbook.
So glad it could help!
Thanks, this was helpful :)
You're welcome, Anthony! I'm so glad it helped! :)
A much faster approach would be through perpendicular projection of vector b on vector a, with that calculating the closest point on the line to the point of the line, and then finishing by calculating the distance between the two points.
Thank you so much, you're a lot of help with my Linear Algebra course! =D
You're welcome, I'm so glad I can help! :)
I've noticed that when deriving the vector B, you subtracted by . Is there any particular reason that you did it in this order, and would it be OK if I did it as - ?please someone to make it clear
No issue , as u are taking the cross product of both , it doesnt or shouldnt matter
- =
- =
You will get an inverse answer and there is a perfect explnation for that u still will get the same answer but its just a negative value
Thanks for the explaining
You're welcome, valeria, I'm happy to help! :)
Wow, you are Amazing.
Thank you, Arkan! 😃
thanks mam Krista , your way of teaching helps me alot , i wish i could be ur student in university or i could gift u chocolates when every time attending ur lectures there.
You're very welcome, Faisal, I'm so happy to help!
Is this the shortest distance between the point and the line?
Thank you very much!!! Your videos help me a lot!!!!!!!!!!!
You're welcome, I'm so glad they're helping!
thank you so helpful!
I've noticed that when deriving the vector B, you subtracted by . Is there any particular reason that you did it in this order, and would it be OK if I did it as - ?
Really helpful thank you!👍👍
You're welcome, Kinjal, I'm so glad it helped! :)
i have a confusion. i think the formula is not correct according to the recent research it should be bxa instead of axb bcz axb!=bxa.
if you think i'm wrong then explain me plz
please i want to ask that what if the line is given by 2x-3y-5=0 and a point
all the work up to 2:40 seems unnecessary.... i can get the point from the constants and the vector components from the coefficients of the parametric equations...
she does that because she wants to explain it to people who did not learn the basis of calc 3 I think!
Not everybody is like you tho
Thank you teacher
You’re welcome! ❤️
I understand it now. Thanks!
Mark Lin You're welcome, I'm so glad it makes sense!
So, the distance between a point and a line is equal to the height of the parallelogram?
d = h = area/base ?
Thank you! This is exactly what I needed to work out my Webwork problem lmao.
You're welcome, I'm so glad it helped! :)
Same problem 8 1.5 lol
I also want to know how to understand this formula?
God bless you!
Does this find the shortest distance between a point and a line? Because I would assume that's the only reason you'd need to answer this question, and I don't understand how finding the distance between a point and a random point on the line counts as achieving that goal. Unless somehow that point is the closest point on the line, which you didn't really explain either way.
I have the same question
Thank you Miss
+Misha Nemov You're welcome, Misha!
great video helped a lot!
Thank you so much if I didn't find this video I may lose my score :D
is this the same method to find the shortest distance?
in what way we label vector a and b?
I mean which one is supposed to be a or b?
holy fuck, i figured it out. thank you for making this simple to follow.
Awesome! You rock!! 😊💪👊
شكراً
Which software is she using for writing?
how d u get that formula for distance??
Does using the projection method work for this problem?
how did you know the direction of the vector and the point on it... there are any proof of this please...?
how do you have all the answers thanks so much
My confusion came from |axb|/|a| because my textbook uses the point on the line as the initial point, so I was thrown off thinking it should’ve been |bxa|/|a|.
thank you!
You're welcome! :)
Thank u so much
what are these t ? and can you explain the form of the line ? or tell me where i can look up stuff about it? x=1+5t seem new to me.
t is just any real number
Math is never a big problem to me (noy that i like it, i detest it), but this vector unit's truly been a massive headache generator, a pain in the bum.
thank you
+Azwan Shari You're welcome!
Thanks
You're welcome! Glad you liked it!
Very good
Thank you
what's the software u're using for presenting ?
Hey, Pedram! It's called Sketchbook.
you over complicated this problem so much!
She did it aweesome bro, respect others work.
@TANGENT You prefer speed run teaching ?
To be honest I am bad at math but I understand each thing she teach
Turns what I was doing wrong was not taking the magnitude of the cross product. I forgot that the result was a scalar. Classic me!
Glad this helped clarify! :)
word
god bless you
is this the same as me finding the point on a line closest to another point
first three mins are waste of time you dont have to do all that adding and distributing
Very poor presentation.. I watched your all videos. In most of the videos, presentation is poor.. You are not giving enough explanation .. So it will be complicated for viewers
Have you had a chance to check out any of my latest videos? I've been diving into the details even further, so you may like those better. :)
I've noticed that when deriving the vector B, you subtracted by . Is there any particular reason that you did it in this order, and would it be OK if I did it as - ?
thank you
thank you
You're welcome, Gabe! :)
thank you
You're welcome, Raymond! :)
thank you