Question 1

Its difficult to imagine how useful the Dot and Cross products are in real world applications. Some examples would shed a lot of light

Accepted Answer

First, anything with forces/vectors in different directions can use the dot product. The physics video www.khanacademy.org/video/the-dot-product? uses an example where you pull an object along the ground. How much work do you do?
Work = (Force)·(Distance). 

Secondly, if you normalize the vectors, then |a| and |b| are both 1, and the dot product |a||b|cosθ is now just cosθ.   This is useful any time you want the angle between two vectors. 

For example: computers multiply and add much faster than they do trig functions, so dot and cross products are used all the time in 3D appplications. To a video game engine, the player 'camera' is a vector. When you walk around, the orientation of objects in relation to one another, and to you, are computed by a series of dot and cross products. 

Cross products? Those too! Say light hits a surface. Is the light reflected to you? Can you see through the surface, or is the angle too steep? You want to know what direction the surface is facing. You want the vector orthogonal to the surface: the one sticking out into the world: You need its normal, i.e. the cross product. 

Generalize inner products are only necessary when your problem requires a more abstract tool. The dot product is defined. That definition holds; so its definition is absolutely rigorous. You need only secure those properties you require. 

One example in the direction crabhand is hinting, is signal processing. Say you break apart a signal into an arbitrary number of streams. If you want each stream to carry data which does NOT already exist in any of the others, you might call this property, 'Orthogonality.' You could say the streams are mutually orthogonal (all of 'em!). The metaphor sounds nice, but it's just a story. To make it mathematically valid, we must define these ideas in a way that can be carried into mathematics. In general, it may or may not be possible to obtain the desired analogy. In this particular case, it can be done. 
Does that help at all? 

(Also, Linear Algebra is abstract mathematics. The physics videos have concrete examples of vector products)

Question 2

When handling the area of a parallelogram with the cross product, do units become arbitrary? Obviously, the magnitude (length) of a vector is a one-dimensional unit, whereas area is measured as a two-dimensional unit.

Furthermore, when dealing with vectors outside of R^3, does the same analogy still hold, or does the cross product simply not exist in those cases?

Accepted Answer

Yes units are always arbitrary. Generally in most of your math classes, even advanced classes, dimensional analysis are even mentioned. It's up to you to decide if your working with a micro meter or a mile. As for the second question, the cross product is not defined for spaces greater than R3. But the operation is still used in finding a determinate of a matrix which can be far greater that R3. Just augment the matrix and treat is as as many 2x2 matrices as are needed. You can find examples in just about any Linear Algebra book.

Question 3

if both dot and cross product of a vector is zero(0) does it imply that one of these vectors must be null vectors?

Accepted Answer

_Note that the null vector is the only vector with a magnitude of 0_
The dot product of vectors a and b is defined as:
`a.b = |a||b|cos(p)`
The cross product magnitude of vectors a and b is defined as:
`|a x b| = |a||b|sin(p)`
Where |a| and |b| are the magnitudes of the vector and p is the angle between the vectors.
The dot product can be 0 if:
`The magnitude of a is 0`
`The magnitude of b is 0`
`The cosine of the angle between the vectors is 0, cos(p)`
The magnitude of the cross product can be zero if:
`The magnitude of a is 0`
`The magnitude of b is 0`
`The sine of the angle between the vectors is 0, sin(p)`

In order for the dot and cross product magnitude to both be zero, the two angle related requirements cannot both be valid!
If the dot product requirement for a dot product of 0 is true:
`The cosine of the angle between the vectors is 0, cos(p)` 
Then the cross product requirement for a magnitude of 0:
`The sine of the angle between the vectors is 0, sin(p)`
cannot be true, because sin and cos are not equivalent functions.
Therefore, if a dot and cross product are both equal to 0 for the same vectors a and b, then either a or b (or both) must be a null vector.

Question 4

Can I use cross product of two vectors to figure out angle between them ?
I know cross product will give us a vector but that doesn't matter ( If I'm right ). The magnitude of the cross product will give the measure of the angle and direction will signify that relative to which side we are defining the angle.
If I'm wrong, then sorry in advance !

Accepted Answer

Yes! Let `v` and `w` be vectors in `R^3` and let `theta` be the angle between them, The exact formula you are looking for is `|v x w| = |v| |w| Sin(theta)`. This should remind you of the dot product formula which has `|v . w| = |v| |w| Cos(theta)`. Either one can be used to find the angle between two vectors in R^3, but usually the dot product is easier to compute. If you are not in 3-dimensions then the dot product is the only way to find the angle. A common application is that two vectors are orthogonal if their dot product is zero and two vectors are parallel if their cross product is zero.

Question 5

at 12:22 and earlier, could we have constructed the triangle to have the right angle at the tip of a? ..and would this have any other effect besides flipping the projection from a→b to b→a?

Accepted Answer

"a·b" and "b·a" yield the same result, so switching the vectors is fine as long as you evaluate them consistently.

For simplicity, I would call the shorter vector "a" so I'm doing most of my work with smaller values. Sal was also unclear on what happens when the projection of "a" is _shorter_ than the length of "b." I doubt it would be an issue but am having trouble visualizing the "area" in that situation.

Question 6

can someone tell me why we cant find cross product between two parallel vectors? why is it zero?
i mean if there are two lines parallel in xy plane........ we can get a perpendicular to both of the lines................. which will be along z axis..
then why cross product between two parellel lines is zero?

Accepted Answer

The simple answer is: because the angle between them is zero.  That means the sine of that angle is zero, and therefore the product of ||a|| ||b|| sin(<ab) is zero (here I use "<ab" to mean "the angle between vectors a and b).

You are right that there is a vector along the z axis which is perpendicular to both - and that is exactly what the cross product is: it's that vector perpendicular to both a and b.  But the *length* of that vector is defined by the formula ||a|| ||b|| sin(<ab), and that length is zero.  So yes, there is such a vector, but its length is zero.

Question 7

I understand that the dot product gives you an indicator of our similar the two vectors are in direction. But not the part when it is explained by |b| adj. can someone explain what that product means intuitively?

Accepted Answer

To find how much the vectors go together, we multiply their parallel parts together. By dropping a perpendicular line from *a*, Sal forms a right triangle. The adjacent side of the right triangle is how much of *a* goes parallel with *b*. So, we can say that the dot product is the product of *b* with that adjacent side of the right triangle made from *a*.

Question 8

Sir,
When to use Dot product and Cross Product and how can we multiply  ( -6i+8k ) X ( 8j + 6k ) /   | -6i+8k  | | 8j + 6k  |   . Note vector symbol is on top on every i,j,k multiples

Accepted Answer

You might be able to get the zero vector that is orthogonal to all vectors and parallel to all. But the magnitude of the cross product uses sine of theta. Because the sine of zero is zero. And zero times anything is zero.

Question 9

Seeing that one is proportional to sine and the other to cosine made me think of derivatives. Would it be correct to say that the derivative of a function that outputs the cross product of 2 constant vectors when the angle between them varies is the dot product of the same thing?

Accepted Answer

Close. The derivative of a function that outputs _the magnitude_ of the cross product of two vectors of fixed magnitude is the same function that returns their dot product.

We need the function to give the magnitude, because the derivative of a vector-valued function is another vector-valued function.

Question 10

Hi, I have a question I hope someone can help me with.  
Take any arbitrary vector like A = ( 2i + 2j +2k), if I was to dot that with just ' i ', would I change ' i ' into a vector ( i + 0j + 0k) and dot that with A or should all the componants of ' A ' be dotted with ' i '  to get something like ( 2i.i + 2j.i + 2k.i) ?  Or what should be done?  Please and thank you

Accepted Answer

î already is a vector, and yes it is equal to 1î + 0ĵ + 0ƙ. So, you can dot it with *A* normally.

Linear algebra

Course: Linear algebra > Unit 1

Dot and cross product comparison/intuition

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