- 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.

- jamiebookeater

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- 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

First rule: multiplying a vector by a scalar uses this rule: a (x,y)= (ax, ay)

- 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)

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## More answers

- anonymous

hmm this one doesn't make sense to me at all

- anonymous

ohh ok

- 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

so its 2 sqrt(2) , -6 sqrt(2) = a (1,1) + b (-1,1)

- phi

yes, but apply the rule to the two terms on the right hand side also

- anonymous

how do you do that?

- phi

a (x,y)= (ax, ay)

- anonymous

so its 2sqrt(2) , -6sqrt(2) = (a1, a1) + (-b, b1)

- 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

ok. its [2sqrt(2), -6sqrt(2)] = (a,a) + (-b,b) ?

- phi

yes, now add the two vectors on the right hand side

- anonymous

how do you do that?

- 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

so its like (1+-1, 1+1) = (0,2) ?

- phi

yes, but now we are adding variables instead of numbers

- anonymous

what do you mean?

- phi

(a,a) + (-b,b) = ??
add corresponding entries

- anonymous

(-ab, ab)

- phi

add (not multiply)

- anonymous

(a+-b, a+b)

- phi

yes, but normally you would write a + -b as just a-b .

- anonymous

oh ok. now what do i do?

- phi

You have
(2sqrt(2), -6sqrt(2)) = (a-b, a+b)

- 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

umm no clue

- phi

if I told you
( 1, 2) = (x, y)
what is x and what is y
how do you know?

- anonymous

that x = 1 and y =2

- phi

Now use the same idea on our problem
(2sqrt(2), -6sqrt(2)) = (a-b, a+b)

- anonymous

that a-b = 2sqrt(2) and a+b = -6sqrt(2)

- 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

lol thats confusing

- anonymous

would it be a = 2sqrt(2) -6sqrt(2) / 2 ??

- 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

so its [sqrt(2), -3sqrt(2)]?

- phi

it is not a vector.

- anonymous

what do you mean?

- anonymous

oh sqrt(2)i , -3sqrt(2) j

- 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

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

ok

- anonymous

but what would be the answer for this?

- phi

we are getting there....
use
a+b = -6sqrt(2)
plus your value for a that we just found to get b

- anonymous

so b = -6sqrt(2) - 2sqrt(2) ?

- 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

lol i am so confused

- phi

can you simplify
b = -6sqrt(2)+2sqrt(2)

- anonymous

umm -4sqrt(2) ?

- phi

yes.
we are almost done.

- 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

2sqrt(2) i + -4sqrt(2) j ?

- phi

and what is I and J ? It does not hurt to write out the whole solution

- anonymous

what do you mean?

- 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

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

ohh k

- anonymous

thanks a lot

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