anonymous
  • anonymous
consider the vector I =[ 1/Square root (2) , 1/Square root (2)] and J= [ -1/Square root (2), 1/Square root (2)]. Write the vector (2,-6) in terms of I and J.
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
james gave you a good starting point. I= [1 1]/sqrt(2) J= [-1 1]/sqrt(2) you want to find two scalars (two simple numbers) that you multiply I and J by to get (2,-6) call the numbers a and b then you must solve (2,-6)= a*I + b*J
phi
  • phi
First rule: multiplying a vector by a scalar uses this rule: a (x,y)= (ax, ay)
phi
  • phi
so, start with (2,-6)= a*I + b*J (2, -6) = a*(1,1)/sqrt(2) + b*(-1,1)/sqrt(2) multiply both sides by sqrt(2) . This means multiply every term by sqrt(2) you get sqrt(2) * (2, -6) = a*(1,1) + b*(-1,1)

Looking for something else?

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

More answers

anonymous
  • anonymous
hmm this one doesn't make sense to me at all
anonymous
  • anonymous
ohh ok
phi
  • phi
Now use the rule First rule: multiplying a vector by a scalar uses this rule: a (x,y)= (ax, ay) on all the terms
anonymous
  • anonymous
so its 2 sqrt(2) , -6 sqrt(2) = a (1,1) + b (-1,1)
phi
  • phi
yes, but apply the rule to the two terms on the right hand side also
anonymous
  • anonymous
how do you do that?
phi
  • phi
a (x,y)= (ax, ay)
anonymous
  • anonymous
so its 2sqrt(2) , -6sqrt(2) = (a1, a1) + (-b, b1)
phi
  • phi
well, I would not write a1, because a*1 is just a.... also, put the left side in parens because we do not want to forget is is a vector
anonymous
  • anonymous
ok. its [2sqrt(2), -6sqrt(2)] = (a,a) + (-b,b) ?
phi
  • phi
yes, now add the two vectors on the right hand side
anonymous
  • anonymous
how do you do that?
phi
  • phi
You were able to subtract two vectors, for example P-Q in the last problem. Adding is the same, except you add instead of subtract.
anonymous
  • anonymous
so its like (1+-1, 1+1) = (0,2) ?
phi
  • phi
yes, but now we are adding variables instead of numbers
anonymous
  • anonymous
what do you mean?
phi
  • phi
(a,a) + (-b,b) = ?? add corresponding entries
anonymous
  • anonymous
(-ab, ab)
phi
  • phi
add (not multiply)
anonymous
  • anonymous
(a+-b, a+b)
phi
  • phi
yes, but normally you would write a + -b as just a-b .
anonymous
  • anonymous
oh ok. now what do i do?
phi
  • phi
You have (2sqrt(2), -6sqrt(2)) = (a-b, a+b)
phi
  • phi
this means you have two vectors that equal each other. If they equal each other, what does that mean about each entry in the vector?
anonymous
  • anonymous
umm no clue
phi
  • phi
if I told you ( 1, 2) = (x, y) what is x and what is y how do you know?
anonymous
  • anonymous
that x = 1 and y =2
phi
  • phi
Now use the same idea on our problem (2sqrt(2), -6sqrt(2)) = (a-b, a+b)
anonymous
  • anonymous
that a-b = 2sqrt(2) and a+b = -6sqrt(2)
phi
  • phi
yes. we have 2 equations and two unknowns. The way to solve these types of equations is write them down like this: a-b = 2sqrt(2) a+b = -6sqrt(2) add the two equations together (we are adding equal things to equal things, so the sum of them is still equal things) 2a-b+b= 2sqrt(2)-6sqrt(2) can you solve for a?
anonymous
  • anonymous
lol thats confusing
anonymous
  • anonymous
would it be a = 2sqrt(2) -6sqrt(2) / 2 ??
phi
  • phi
first, you should put parens around (2sqrt(2) -6sqrt(2)) because you are dividing *both* terms by 2 (you divide the whole right hand side by 2) next, you should simplify
anonymous
  • anonymous
so its [sqrt(2), -3sqrt(2)]?
phi
  • phi
it is not a vector.
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
oh sqrt(2)i , -3sqrt(2) j
phi
  • phi
no. I mean you start with (2*sqrt(2)-6*sqrt(2))/2 which is just a (ugly) number: divide both terms by 2 to get (sqrt(2)-3*sqrt(2)) factor out sqrt(2) to get (1-3)*sqrt(2) simplify 1-3 to get a= -2*sqrt(2)
phi
  • phi
btw, if you want info on how to solve simultaneous equations like this one, watch http://www.khanacademy.org/video/solving-systems-by-elimination?topic=worked-examples-4 and his other examples
anonymous
  • anonymous
ok
anonymous
  • anonymous
but what would be the answer for this?
phi
  • phi
we are getting there.... use a+b = -6sqrt(2) plus your value for a that we just found to get b
anonymous
  • anonymous
so b = -6sqrt(2) - 2sqrt(2) ?
phi
  • phi
well let's check that a + b= -6sqrt(2) -2*sqrt(2) + b= -6sqrt(2) add +2sqrt(2) to both sides to get b = -6sqrt(2)+2sqrt(2) so close, but not exactly correct. can you simplify this?
anonymous
  • anonymous
lol i am so confused
phi
  • phi
can you simplify b = -6sqrt(2)+2sqrt(2)
anonymous
  • anonymous
umm -4sqrt(2) ?
phi
  • phi
yes. we are almost done.
phi
  • phi
we have to answer Write the vector (2,-6) in terms of I and J. which we decided meant: (2,-6)= a*I + b*J we now know a and b, so the final answer is?
anonymous
  • anonymous
2sqrt(2) i + -4sqrt(2) j ?
phi
  • phi
and what is I and J ? It does not hurt to write out the whole solution
anonymous
  • anonymous
what do you mean?
phi
  • phi
I =[ 1/Square root (2) , 1/Square root (2)] J= [ -1/Square root (2), 1/Square root (2)] [2,-6]=2sqrt(2) I + -4sqrt(2) J
phi
  • phi
You can check your answer by multiplying the scalars times the vectors, and then adding the vectors to see if you get [2, -6] btw, if you have time, you might want to brush up on your algebra using http://www.khanacademy.org/#core-algebra
anonymous
  • anonymous
ohh k
anonymous
  • anonymous
thanks a lot

Looking for something else?

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