Hey professor Dave! you are my hero and I would love to be as cool as you someday. It tickles me the way you combat flat eathers while also helping me learn the material my engineering professors can not. You have made my school life much better since i have stumbled upon your videos and i can not thank you enough. Please keep up what you are doing and know that you are a more than a professor you are a HERO!
Hello. I have a qn on Vector Cross Product. 1) Is there a reason why we have to subtract first followed by adding the next vector? 2) Is there an order as to whether or not I should take -5 x 3 first, and then 7 x 4? 3) May I confirm that in each of the vector after solving its multiplication, I have to subtract it? For example, (-5x3) - (7x4)
Multiplication of ordinary numbers is commutative, but this is not the case for the cross product. For the cross product, a cross b is not the same thing as b cross a, because a cross b is the negative of b cross a. You can multiply the individual numbers in any order you want, that carry out the cross product. But to set up the multiplication, you need to keep the vectors in the correct order.
1) yes firstly we took + sign because of the position of that i cap i.e; a11 (1+1=2)(even) so positive then j cap belongs to a12 (1+2=3)(odd) so negative and so on... 2)That's the way of doing determinants 3)that's the whole way of solving determinants
Cross product: (1) Only in 3D space (2) the value of a determinant is defined to be a scalar quantity, not a vector; the rubric for the cross product is only a mnemonic. Do not study only with this channel unless you wish to treat it simply as prep for exams that will include only very rudimentary "calculate this and give us a number" exercises. In other words, you're not learning linear algebra from Professor Dave in any depth. It's quick, thorough prep for next day's exam if you're only asked to calculate.
I had been struggling for several months with cross products. I never thought your explanation could help me clear the topics smoothly! Thank you so much professor dave
i just want to learn vector cross product to find area of triangle but as i opened your channel i found ocean of knowledge .omg im shocked...thnak you so much prof deck from this learner from nepal
Resolve b into two perpendicular components, one parallel to a, and one perpendicular to a; then: magnitude of cross product = [magnitude of a] X [magnitude of the component of b which is at right angles to a (which is "b sin theta") ]
The second coordinate (j) is the negative of the multiplication. Basically, once you have your result, you just negate the second coordinate. Check 03:22
it's just a dumb thing i made up! if you go to my "just for fun" playlist there is a five hour loop of it in there, just in case you fell like listening for longer!
I do have a question about this part 1:51. How come the answers of 4x7, 4x2, and 3x2 are all negative? Is it always negative when in fact they are all positive?
I’m always told that the cross product is perpendicular to the 2 vectors but never given a explanation so I have to just accept as fact but Professor Dave could you please provide the reason for this.
Set up a general case of a cross product: cross |i j k| |a b c| |d e f| i*(b*f - c*e) + j*(c*d - a*f)+ k*(a*e -b*d) = Then take the dot product with the above, and the original vectors dot = a*(b*f - c*e) + b*(c*d - a*f) + c*(a*e - b*d) a*b*f - a*c*e + b*c*d - a*b*f + c*a*e - c*b*d Cancel equal and opposite terms: a*b*f - a*c*e - a*b*f + c*a*e a*b*f - a*b*f 0 Do the same with the other original vector, and you will see that the dot product of any vector and its cross product with a second vector, will always be zero. This indicates that the cross product is perpendicular to both original vectors.
@@carultch Good explanation I didn’t think of using the dot product on a general crossproduct vector to check if the resulting crossproduct vector was perpendicular to the original vectors ,but this only proof only applies to 2d crossproducts is there a more general proof that extends to general (n dimension) crossproducts.
@@burningsilicon149 The cross product proof I gave, applies to crossing two 3-dimensional vectors in general, and it shows that no matter what the vectors are, the dot product of each source vector and the cross product resultant is zero. The proof depends on it already being established that a dot product of zero means perpendicular vectors. The N-dimensional cross product doesn't really exist, as it is a concept which is only defined in 3-dimensional vector spaces. You can cross two 2-d vectors, and the result is perpendicular to the plane of both of them, requiring 3-d space to work with all three vectors. You can cross two 1-d vectors, and the result is guaranteed to be zero. But for 4-d vectors, there is no cross product defined. There are other types of vector multiplication in higher dimensional vector space, that I know nothing about.
@@carultch I wasn’t aware that above 3 dimensions the crossproduct isn’t defined.I’ve taken linear algebra and they usually extend say matrix vector multiplication to a arbitrary number of dimensions (n) I assumed that applied to crossproducts as well.
True. The input vectors don't necessarily need to be perpendicular. When they are not perpendicular, the output cross product vector will still be perpendicular to both of them, and in a direction determined by the right hand rule. Simply let your middle finger rotate to a position other than perpendicular to your index finger. Your thumb will still identify the direction of the cross product resultant. The magnitude of the cross product will be the area of a parallelogram, defined by the two vectors as two of the adjacent sides of the shape. You can use this as a shortcut for finding the area of a triangle among three points in space in general. Find a vector between point A and point B, and another vector between point A and point C. Take the cross product, find its magnitude, and divide by two.
If cross product of two vectors is orthogonal to both of them doesn't that mean new dimension is added? so if there exists two vectors in 2-d plane if we apply right hand rule does the thumb point upward introducing new 3rd dimension Z-axis?
Yes. A cross product of two vectors in 2-d space, requires a third dimension for the resultant. If you cross unit vectors i-hat and j-hat, in that order, the resultant is k-hat, which produces the z-axis. By convention, a standard right-handed coordinate system, puts the x-axis (i-hat unit vector) on the pointer finger, the y-axis (j-hat unit vector) on the middle finger, and the z-axis (k-hat unit vector) on the thumb. The cross product in a purely 2-dimensional space, is just a scalar, rather than a vector. If we lived in flatland, applications of the cross product like torque, would just be scalars.
Why can't my university prof just put up these videos and call it a day. That way I won't be binge-watching these videos the day before the test
You do know how to find them. Nothing is stopping you from watching them earlier than the day before the test.
Y'all are studying this in uni?
@@yugagalaxa98 Yeah!
@@ConceptualCalculus Woah man you might have a point! Thanks for pointing that out idk what I'd do without you!
@@chenchoon8751 oh. We study it in 11th grade...
this man is single handedly saving my engineering career 😭
us 😭
Omg samee here 😭🥲
@@Sam-em9zy im not a uni student, but the way everyone in these comments sections describe uni, Im cooked
Him and The Chemistry Tutor guy🔥
same bro😂😂
Hey professor Dave! you are my hero and I would love to be as cool as you someday. It tickles me the way you combat flat eathers while also helping me learn the material my engineering professors can not. You have made my school life much better since i have stumbled upon your videos and i can not thank you enough. Please keep up what you are doing and know that you are a more than a professor you are a HERO!
I would completely agree....i feel my highschool year has been easier since i found u❤️
Hello. I have a qn on Vector Cross Product.
1) Is there a reason why we have to subtract first followed by adding the next vector?
2) Is there an order as to whether or not I should take -5 x 3 first, and then 7 x 4?
3) May I confirm that in each of the vector after solving its multiplication, I have to subtract it? For example, (-5x3) - (7x4)
Multiplication of ordinary numbers is commutative, but this is not the case for the cross product. For the cross product, a cross b is not the same thing as b cross a, because a cross b is the negative of b cross a.
You can multiply the individual numbers in any order you want, that carry out the cross product. But to set up the multiplication, you need to keep the vectors in the correct order.
1) yes firstly we took + sign because of the position of that i cap i.e; a11 (1+1=2)(even) so positive then j cap belongs to a12 (1+2=3)(odd) so negative and so on...
2)That's the way of doing determinants
3)that's the whole way of solving determinants
I've come across your videos just recently and am so happy I did; thank you SO MUCH for you crystal clear understanding of these concepts!!
Cross product: (1) Only in 3D space (2) the value of a determinant is defined to be a scalar quantity, not a vector; the rubric for the cross product is only a mnemonic. Do not study only with this channel unless you wish to treat it simply as prep for exams that will include only very rudimentary "calculate this and give us a number" exercises. In other words, you're not learning linear algebra from Professor Dave in any depth. It's quick, thorough prep for next day's exam if you're only asked to calculate.
My Sir explained this topic many times but I couldn't relate, when prof. explained understood very clearly ...Thanks...
I had been struggling for several months with cross products.
I never thought your explanation could help me clear the topics smoothly!
Thank you so much professor dave
3 years later I'm watching this and it still explains it so easily. Keep it up 😊😊😊😊
Thanks prof, I passed the physics paper with ur help.....ur tutorials are awesome
Mate, thank you so much for these videos, I wouldn't have been able to pass my midterms without you.
i just want to learn vector cross product to find area of triangle but as i opened your channel i found ocean of knowledge .omg im shocked...thnak you so much prof deck from this learner from nepal
I used to watch him as I studied for ap's now I watch him at Stanford. thank you goat
Resolve b into two perpendicular components, one parallel to a, and one perpendicular to a; then:
magnitude of cross product =
[magnitude of a] X [magnitude of the component of b which is at right angles to a (which is "b sin theta") ]
You're a life saver Professor Dave !
Thank you, professor Dave! English is not my mother tongue but I understand Linear Algebra better than I am learning in my university
Very clearly explained! Thank you Professor Dave!
professor Dave is just the best. Now are understand more about the topic
Thank you so much sir your explainations are superb and works like a one shot before exams.... By the way, love from India ❤️
Nice tutorials. It helps me alot.
@@RajKapoor-ix4mk sum indian ned pusi
Excellent presentation with explanations that get right to the point.
Came for the flat earth wreckage, stayed for my Master's degree.
Well I guess that's one advantage of flat Earthers existing. Getting you introduced to people like Dave to help you earn your degree.
This man is saving LIVES 🙌🏻
You're currently my favorite math TH-camr!
because of ur teaching i got excellent marks in this chapter thank you sir thank u very much......
Thank you professor dave, I hope i pass my exam later. You are very helpful!!
did you pass?
PROFESSOR YOU ARE TRULY THE BEST. LOVE YOU, FROM KENYA
i have an exam tomorrow. this is a great crash course
I have mine tomorrow as well
i such love the introduction of this channel it is so shiny
This is way easier to understand than memorizing a formula.
Make sure you put the arrow above to indicate that it's a vector!
Thanks for the video!
In comprehension,
a×b= -3i + 19j + 10k
So,
|a×b| = √(-3)^2 i + (19)^2 j + (10)^2 k = √470
Is √470 = |a| |b| sin⊙ ?
I LOVE EDUCATION TH-cam THANK YOU SO MUCH PROFESSOR DAVE, YOU ARE A STEPPING STONE TO MY CAREER AS AN AEROSPACE ENGINEER. Sorry I really love science.
Wow🎉🎉 I really love your way to explains, it's so easy to understand clearly. Thank you so much
Hi Dave. I love your videos. Since we all went online abruptly, I have been using them a lot in my classes. Thank you.
Could you do a playlist on dynamics?
check my classical physics playlist
Thank you sir☺️☺️
Love from India🇮🇳🇮🇳♥️
Bobs and vegan (ʘᴗʘ✿)
I am from India....
State :telangana
2:23 right-hand rule
You are the best in the mond bro
dude u are greatt you are freaking great
i wish indian schools would have teachers like you!!!
Great sir, i understood concept very well . Thanks for being there sir
I learned more from this video than what I learned during my entire degree
you always make me feel physics is easy
Thanks sir ur vedios are short and very helpful🙂🙂🙂
Thanks man I have exam tomorrow and I understand it
Why a x b as a result is 19 j positive? I think is -19j
The second coordinate (j) is the negative of the multiplication. Basically, once you have your result, you just negate the second coordinate.
Check 03:22
Why isn't the Vector Dot Product video not in the Linear Algebra playlist?
Professor Dave!!! Every thing is perfect except the audio. Please be loud!!!!
it's such a wonderful session thank you, sir!!
Great explanation, thank you!
U r the best professor Dave, thanks
you explain so well! thank you!
Very clear,thank you.
The right hand rule remimds me of Poyntings vector
saving me at 1am the night before my mid semester
What is the tune during comprehension...someone plrase tell! I really like it....it is relaxing and satisfying!
it's just a dumb thing i made up! if you go to my "just for fun" playlist there is a five hour loop of it in there, just in case you fell like listening for longer!
Thank you professor!!!
Thank god i got this video... thank you sir....
thank you!! this is informative
thank you prof dave
aint no way i came back 3 years later cos I forgot
you're a life saver 🙏🏻
I really understand the concept ....thanku sir
I wish you would have shown the easier way to set up the multiplication within the vectors set up.
Thank u so much sir .....u made me understand so clearly😊😊
Thank you professor jave
Is the cross product a resulting vector? Or is that term only used when adding vectors?
I love that introduction song
Thank you professor!
Carrying me through calc by day. Crushing flat earther's by night
Amazing. Thank you.
Thank you!
Well, that's the fastest I understood anything in linear algebra
I like the intro so much.and sama as professor dave
I do have a question about this part 1:51. How come the answers of 4x7, 4x2, and 3x2 are all negative? Is it always negative when in fact they are all positive?
That's the way the determinant of a 2*2 matrix is defined: [a b | c d] is ad-bc. So the one that is [3 4 | 7 -5] above yields 3*5-4*7.
tnx Mr Dave u rock 😚
at 6:00 *a x b* is (27 i + 19 j + 10 k) and not (-3 i + 19 j + 10 k)
When I included the negatives : (-12-15)i-(24-(-5))j+12-(-2))k , I got -27i-19j+14k.
Helps a lot! Thanks Professor Dave.
Me too! I use the voice search and sing "Cross Products of Vectors by Professor Dave explains"
Thank you
This right hand rule is the most bull tip i've ever seen in my while life
Do you have a video like this for addition with the same amount set of numbers?
so helpful once again. thanks.
I wish this man knows how he saves our lives in campus
So any advices for electrostats ?😅
How to find the direction of the vector product of 2 vectors : 2:25 to 3:05
Wait, in which previous video did you discuss vector dot product? I can't seem to find it in any previous video in the linear algebra series.
Earlier in the big math playlist
The first one shouldn't be -3i+19j-10k?
Yes it should be -27i+19j+10k
yes you are right i find the same things
the best to ever exist
Superb sir thank you ❤🙏
My University teaches us the cross product in the form i+j+k not as you have as i-j+k, do they give the same results, do you know why it's different?
I think there is a misunderstanding
When you expand a determinant along any row or column you will always get atleast one negative co factor
I’m always told that the cross product is perpendicular to the 2 vectors but never given a explanation so I have to just accept as fact but Professor Dave could you please provide the reason for this.
Set up a general case of a cross product:
cross
|i j k|
|a b c|
|d e f|
i*(b*f - c*e) + j*(c*d - a*f)+ k*(a*e -b*d) =
Then take the dot product with the above, and the original vectors
dot =
a*(b*f - c*e) + b*(c*d - a*f) + c*(a*e - b*d)
a*b*f - a*c*e + b*c*d - a*b*f + c*a*e - c*b*d
Cancel equal and opposite terms:
a*b*f - a*c*e - a*b*f + c*a*e
a*b*f - a*b*f
0
Do the same with the other original vector, and you will see that the dot product of any vector and its cross product with a second vector, will always be zero. This indicates that the cross product is perpendicular to both original vectors.
@@carultch Good explanation I didn’t think of using the dot product on a general crossproduct vector to check if the resulting crossproduct vector was perpendicular to the original vectors ,but this only proof only applies to 2d crossproducts is there a more general proof that extends to general (n dimension) crossproducts.
@@burningsilicon149 The cross product proof I gave, applies to crossing two 3-dimensional vectors in general, and it shows that no matter what the vectors are, the dot product of each source vector and the cross product resultant is zero. The proof depends on it already being established that a dot product of zero means perpendicular vectors.
The N-dimensional cross product doesn't really exist, as it is a concept which is only defined in 3-dimensional vector spaces. You can cross two 2-d vectors, and the result is perpendicular to the plane of both of them, requiring 3-d space to work with all three vectors. You can cross two 1-d vectors, and the result is guaranteed to be zero. But for 4-d vectors, there is no cross product defined. There are other types of vector multiplication in higher dimensional vector space, that I know nothing about.
@@carultch I wasn’t aware that above 3 dimensions the crossproduct isn’t defined.I’ve taken linear algebra and they usually extend say matrix vector multiplication to a arbitrary number of dimensions (n) I assumed that applied to crossproducts as well.
Thanks dude
Thank you , from thai student
At 2:59, the right right-hand diagram seems to show that a & b are perpendicular to each other, but they don't have to always be perpendicular, right?
True. The input vectors don't necessarily need to be perpendicular. When they are not perpendicular, the output cross product vector will still be perpendicular to both of them, and in a direction determined by the right hand rule. Simply let your middle finger rotate to a position other than perpendicular to your index finger. Your thumb will still identify the direction of the cross product resultant.
The magnitude of the cross product will be the area of a parallelogram, defined by the two vectors as two of the adjacent sides of the shape. You can use this as a shortcut for finding the area of a triangle among three points in space in general. Find a vector between point A and point B, and another vector between point A and point C. Take the cross product, find its magnitude, and divide by two.
at 5:37 are a,b and c all vectors (in the distributive property)?
Thanks
Brilliant!!!❤️❤️❤️
It would have been a little better if you would've shown the matrix...
Ty this helped a lot
Why is it minus J on the very first step, explain please
If cross product of two vectors is orthogonal to both of them doesn't that mean new dimension is added? so if there exists two vectors in 2-d plane if we apply right hand rule does the thumb point upward introducing new 3rd dimension Z-axis?
Yes. A cross product of two vectors in 2-d space, requires a third dimension for the resultant. If you cross unit vectors i-hat and j-hat, in that order, the resultant is k-hat, which produces the z-axis. By convention, a standard right-handed coordinate system, puts the x-axis (i-hat unit vector) on the pointer finger, the y-axis (j-hat unit vector) on the middle finger, and the z-axis (k-hat unit vector) on the thumb.
The cross product in a purely 2-dimensional space, is just a scalar, rather than a vector. If we lived in flatland, applications of the cross product like torque, would just be scalars.