amistre64
  • amistre64
What does the scalar result of dotting 2 vectors signify? I know a 0 means they are perps, but what is the scalar product actually mean?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
<1,1>.<0,2sqrt(2)> = 0 + 2sqrt(2) = 2sqrt(2). That doesnt make sense to me a whole lot.
amistre64
  • amistre64
that just seems to give the magnitude of the bottom vector
anonymous
  • anonymous
<1,1>.<0,2sqrt(2)> = 0 + 2sqrt(2) = 2sqrt(2). how?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

nowhereman
  • nowhereman
It gives the cosine of the angle between them. This is the way you can get angles in any vector space with an inner product.
amistre64
  • amistre64
<1 , 1 > <0,2sqrt(2)> ------------ 0 + 2sqrt(2) = 2sqrt(2)
amistre64
  • amistre64
yes, it gives the adjacent part of the cosine angle which is just the magnitude of the bottom vector right? the product of the magnitudes of the top and bottom vectors account for the 'hypotenuse' then
anonymous
  • anonymous
<1 , 1 > <0,2sqrt(2)> ------------ 0 + 2sqrt(2) = 2sqrt(2). iam not geting it amistre
amistre64
  • amistre64
see; it is how you multiply vectors; or at least one way to do it; the other being the cross product
myininaya
  • myininaya
cross product can be found two ways if i remember correctly finding the determinant of a matrix or some kind of trig thing right?
amistre64
  • amistre64
determinate of 2 vectors is usual
amistre64
  • amistre64
the other way is just to equate the zero dot product of 2 vectors inthe plane and equat them
amistre64
  • amistre64
v n ------- vx.a + vy.b = 0 u n -------- ux.a +uy.b = 0 equate them and solve the system of equations :)
amistre64
  • amistre64
geterminate is mor emechanical tho
amistre64
  • amistre64
lol... read thru the typos
amistre64
  • amistre64
i now the scalar projection projects the top onto the bottom at |top|cos(a)
amistre64
  • amistre64
i just dont understand if there is another way to use the scalar product other than associating it to the cos(a) between the 2 vectors :)
myininaya
  • myininaya
example: find u*v if u=2i+3j and v=-2k | i j k | = i|3 0 | -j|2 0| +k|2 3| =i(-2(3)-0)-j(-2-0)+k(0-0)=-6i+2j | 2 3 0 | |0 -2| |0 -2| |0 0| | 0 0 -2|
amistre64
  • amistre64
correct; just keep in mind to do +-+ :) that middle - is to account for being lazy i think lol
myininaya
  • myininaya
=-6j+4k
myininaya
  • myininaya
oops =-6i+4j
myininaya
  • myininaya
i forgot about that othr two up there
amistre64
  • amistre64
2a+3b+0c = 0 0a+0b-2c = 0 lol... thats would be hard to determine that way ; cross product is better in that case; and in most cases prolly
myininaya
  • myininaya
yes that would be hard lol (or impossible) i dont see how we could do it that way

Looking for something else?

Not the answer you are looking for? Search for more explanations.