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.
Hi is there anyone that can explain for the first example at 1:20 why it is i - j + k ? and not i+j+k or something like that? How do we know whether it is addition/ subtracting i/j/k ? for some reason i always had trouble with vector math, so please forgive me if the answer is obvious
I(-1)^a+b + j(-1)^a+b + k(-1)^a+b. a and b are position of the values I, j and k. I is in position 1,1 j is in position 1,2 and k is in position 1,3 and if we solve that: = I(-1)^2 + j(-1)^3 + k(-1)^4 = I-j+k. Hope that helps.
Look up cofactor expansion. I like to think of cofactor expansion by covering the current row and column with my finger(s), putting the rest of the (visible) numbers in the smaller determinant You can think of the signs as a matrix of only signs where each adjacent sign is alternating. The row or column you choose for calculating the determinant is what decides the sign of each smaller determinant. The top left is *always* a +. You can continue the pattern infinitely. [ + - ] 2x2 determinant |1 2| = (Using top row, which is [+ -]): +1*(4) - 2*(3) [ - +] |3 4| = (Using bottom row, which is [- +]): -3*(2) + 4*(1) For the top row example equation above: 1 is current #, sign is +, cover 2,3, smaller determinant=4. 2 is current #, sign is -, cover 1,4, smaller determinant=3. You can also use the "teacher-sponsored memorization" approach which is a*d minus b*c: (multiply the main diagonal) and subtract (multiply the minor diagonal). |a b| = ad-bc |c d| [ + - + ] 3x3 determinant | 1 2 3| [ - + - ] |4 5 6| = (Using top row, which is [+ - +]): +1*|5 6| -2*|4 6| +3*|4 5| [ + - + ] |7 8 9| |8 9| |7 9| |7 8| For the top row example above: 1 is current #, sign is +, cover 1's row and column (1,2,3,4,7), smaller determinant=|5 6| |8 9| 2 is current #, sign is -, cover 2's row and column (1,2,3,5,8), smaller determinant=|4 6| |7 9| 3 is current #, sign is +, cover 3's row and column (1,2,3,6,9), smaller determinant=|4 5| |7 8| [ + - + - ] 4x4 determinant [ - + - + ] [ + - + - ] [ - + - + ] etc If I choose the top or bottom row (or left or right column) for a 3x3 determinant, my sign coefficients will be +,-,+. If I choose the middle row or column for a 3x3 determinant, my sign coefficients will be -,+,-.
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
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
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
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 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.
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!
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.
I don't. I have a different way of remembering how to do the cross product. I imagine a copy of the matrix to the left, and a copy to the right. I then multiply along the down/right positive diagonals, and then along the down/left negative diagonals. Add up the products along positive diagonals. Then subtract the products along the negative diagonals. You get the same answer, and you don't need to think about a negative sign on the j-term. As for in general, it comes from Matrix cofactor expansion, as part of the procedure for finding determinants through sub-determinants. If there were a 4th term, to take a 4x4 matrix's determinant, there would be a negative sign assigned to the 4th term as well. There is a checkerboard of positive and negative signs that applies for determinants in general, when calculating them through sub-determinants. My method of positive and negative diagonals, only works for 2x2 and 3x3 matrix determinants. But that is all we'll ever need to do with cross products in 3D space.
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...
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❤️
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🔥
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.
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!!
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.
Hi is there anyone that can explain for the first example at 1:20 why it is i - j + k ? and not i+j+k or something like that? How do we know whether it is addition/ subtracting i/j/k ?
for some reason i always had trouble with vector math, so please forgive me if the answer is obvious
I think it is just always set up as i - j + k. I'm not saying it will always end up that way, but that's how it is set up at least
Its why because in an individual determinant we have a1b2-a2b1
I(-1)^a+b + j(-1)^a+b + k(-1)^a+b.
a and b are position of the values I, j and k.
I is in position 1,1 j is in position 1,2 and k is in position 1,3 and if we solve that:
= I(-1)^2 + j(-1)^3 + k(-1)^4
= I-j+k.
Hope that helps.
Look up cofactor expansion. I like to think of cofactor expansion by covering the current row and column with my finger(s), putting the rest of the (visible) numbers in the smaller determinant
You can think of the signs as a matrix of only signs where each adjacent sign is alternating.
The row or column you choose for calculating the determinant is what decides the sign of each smaller determinant. The top left is *always* a +. You can continue the pattern infinitely.
[ + - ] 2x2 determinant |1 2| = (Using top row, which is [+ -]): +1*(4) - 2*(3)
[ - +] |3 4| = (Using bottom row, which is [- +]): -3*(2) + 4*(1)
For the top row example equation above:
1 is current #, sign is +, cover 2,3, smaller determinant=4.
2 is current #, sign is -, cover 1,4, smaller determinant=3.
You can also use the "teacher-sponsored memorization" approach which is a*d minus b*c: (multiply the main diagonal) and subtract (multiply the minor diagonal).
|a b| = ad-bc
|c d|
[ + - + ] 3x3 determinant | 1 2 3|
[ - + - ] |4 5 6| = (Using top row, which is [+ - +]): +1*|5 6| -2*|4 6| +3*|4 5|
[ + - + ] |7 8 9| |8 9| |7 9| |7 8|
For the top row example above:
1 is current #, sign is +, cover 1's row and column (1,2,3,4,7), smaller determinant=|5 6|
|8 9|
2 is current #, sign is -, cover 2's row and column (1,2,3,5,8), smaller determinant=|4 6|
|7 9|
3 is current #, sign is +, cover 3's row and column (1,2,3,6,9), smaller determinant=|4 5|
|7 8|
[ + - + - ] 4x4 determinant
[ - + - + ]
[ + - + - ]
[ - + - + ] etc
If I choose the top or bottom row (or left or right column) for a 3x3 determinant, my sign coefficients will be +,-,+.
If I choose the middle row or column for a 3x3 determinant, my sign coefficients will be -,+,-.
@@abdiladifmohamud5957 Oh now it makes sense. Thanks and I appreciate your help. A lot of kids at my age won’t understand any of this at all.
Very clearly explained! Thank you Professor Dave!
Nice tutorials. It helps me alot.
@@RajKapoor-ix4mk sum indian ned pusi
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 😊😊😊😊
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
You're a life saver Professor Dave !
My Sir explained this topic many times but I couldn't relate, when prof. explained understood very clearly ...Thanks...
Thanks prof, I passed the physics paper with ur help.....ur tutorials are awesome
Carrying me through calc by day. Crushing flat earther's by night
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
Thank you so much sir your explainations are superb and works like a one shot before exams.... By the way, love from India ❤️
Thank you, professor Dave! English is not my mother tongue but I understand Linear Algebra better than I am learning in my university
2:23 right-hand rule
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⊙ ?
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.
professor Dave is just the best. Now are understand more about the topic
Thank you professor dave, I hope i pass my exam later. You are very helpful!!
did you pass?
A random boy of india from 11th grade... 😎
Not cool lol
Yep
Same👊🏻
Random boy from Canada here, we exist too
Ya bro
Excellent presentation with explanations that get right to the point.
Thank you sir☺️☺️
Love from India🇮🇳🇮🇳♥️
Bobs and vegan (ʘᴗʘ✿)
I am from India....
State :telangana
I LOVE EDUCATION UA-cam THANK YOU SO MUCH PROFESSOR DAVE, YOU ARE A STEPPING STONE TO MY CAREER AS AN AEROSPACE ENGINEER. Sorry I really love science.
This man is saving LIVES 🙌🏻
because of ur teaching i got excellent marks in this chapter thank you sir thank u very much......
Could you do a playlist on dynamics?
check my classical physics playlist
You're currently my favorite math UA-camr!
Very clear,thank you.
Thanks sir ur vedios are short and very helpful🙂🙂🙂
i such love the introduction of this channel it is so shiny
Amazing. Thank you.
I learned more from this video than what I learned during my entire degree
i have an exam tomorrow. this is a great crash course
I have mine tomorrow as well
Great sir, i understood concept very well . Thanks for being there sir
Wow🎉🎉 I really love your way to explains, it's so easy to understand clearly. Thank you so much
thank you!! this is informative
Hi Dave. I love your videos. Since we all went online abruptly, I have been using them a lot in my classes. Thank you.
PROFESSOR YOU ARE TRULY THE BEST. LOVE YOU, FROM KENYA
thank you prof dave
aint no way i came back 3 years later cos I forgot
Thank you professor jave
tnx Mr Dave u rock 😚
Thank god i got this video... thank you sir....
dude u are greatt you are freaking great
i wish indian schools would have teachers like you!!!
Thanks man I have exam tomorrow and I understand it
U r the best professor Dave, thanks
Thank u so much sir .....u made me understand so clearly😊😊
Do you have a video like this for addition with the same amount set of numbers?
This is way easier to understand than memorizing a formula.
Great explanation, thank you!
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.
you explain so well! thank you!
I love that introduction song
Thank you professor!
The right hand rule remimds me of Poyntings vector
it's such a wonderful session thank you, sir!!
Make sure you put the arrow above to indicate that it's a vector!
I really understand the concept ....thanku sir
Is the cross product a resulting vector? Or is that term only used when adding vectors?
This right hand rule is the most bull tip i've ever seen in my while life
saving me at 1am the night before my mid semester
Professor Dave!!! Every thing is perfect except the audio. Please be loud!!!!
Well, that's the fastest I understood anything in linear algebra
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.
Superb sir thank you ❤🙏
Thank you
bXc must be -27i -42j -16k am i right?
nope, its correct
It should be -43j
you're a life saver 🙏🏻
I like the intro so much.and sama as professor dave
Thanks :)
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
so helpful once again. thanks.
Why isn't the Vector Dot Product video not in the Linear Algebra playlist?
Helps a lot! Thanks Professor Dave.
Hi Mr Dave.
Thank you!
Me too! I use the voice search and sing "Cross Products of Vectors by Professor Dave explains"
So any advices for electrostats ?😅
Brilliant!!!❤️❤️❤️
Great ,you are great
How to find the direction of the vector product of 2 vectors : 2:25 to 3:05
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!!!
why did you subtract at 1:50
Why is it minus J on the very first step, explain please
Sir what if i am been given a number or component like
a=(1,2, -3) and b=(2,-3,4) How can i solve it
I love your intro
Ty this helped a lot
Thank you sir.
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.
why do we use minus for j?
I don't. I have a different way of remembering how to do the cross product. I imagine a copy of the matrix to the left, and a copy to the right. I then multiply along the down/right positive diagonals, and then along the down/left negative diagonals. Add up the products along positive diagonals. Then subtract the products along the negative diagonals. You get the same answer, and you don't need to think about a negative sign on the j-term.
As for in general, it comes from Matrix cofactor expansion, as part of the procedure for finding determinants through sub-determinants. If there were a 4th term, to take a 4x4 matrix's determinant, there would be a negative sign assigned to the 4th term as well. There is a checkerboard of positive and negative signs that applies for determinants in general, when calculating them through sub-determinants.
My method of positive and negative diagonals, only works for 2x2 and 3x3 matrix determinants. But that is all we'll ever need to do with cross products in 3D space.
Thank you , from thai student
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.
Well explained. Chemistry Jesus 😊👌🏽
the hell??
Thank you. 💯