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Thread starter 14 years ago
#1
How do you do this?

2. Given that a = 2 + j and b = i + 3 j find a, (cant do the symbol on comp looks like a lambda), if a + lambda b is parallel to the vector i

Thanks
0
14 years ago
#2
(Original post by DOJO)
How do you do this?

2. Given that a = 2 + j and b = i + 3 j find a, (cant do the symbol on comp looks like a lambda), if a + lambda b is parallel to the vector i

Thanks
Sorry why a = 2 + j? You mean 2i + j, or 2j?
0
Thread starter 14 years ago
#3
(Original post by BCHL85)
Sorry why a = 2 + j? You mean 2i + j, or 2j?
opps sorry its a=2i+j
0
14 years ago
#4
Ok, a + l.b = ki
2i + j + li + 3lj = ki
(2+l)i + (1+3l)j = ki
So 1 + 3l = 0 -> l = -1/3
And you want to find lamda (l), don't you?
0
Thread starter 14 years ago
#5
(Original post by BCHL85)
Ok, a + l.b = ki
2i + j + li + 3lj = ki
(2+l)i + (1+3l)j = ki
So 1 + 3l = 0 -> l = -1/3
And you want to find lamda (l), don't you?
Yes, so I think thats what the question is asking me to do (my first time doing M1 and vectors...)

Your answer of -1/3 was right, however where did the k lamda come from?

Thanks.
0
14 years ago
#6
(Original post by DOJO)
Yes, so I think thats what the question is asking me to do (my first time doing M1 and vectors...)

Your answer of -1/3 was right, however where did the k lamda come from?

Thanks.
You confused of the first line a+l.b = ki?
Because if a vector is parallel to another vector, it can be expressed by that vector multiply a real number.
0
Thread starter 14 years ago
#7
(Original post by BCHL85)
You confused of the first line a+l.b = ki?
Because if a vector is parallel to another vector, it can be expressed by that vector multiply a real number.
I understand the a+l.b , its just the kl , where did k come from?

Here was my working before I posted on :

a + lb = (2i + j) + l(i+3j)

and then I did I multiplied out of b with l :

a + lb = (2i + j) + l(i+3j) =
(2i+j) + (li + 3lj)

and then my friend it all went horribly wrong... However by looking at your working I was not far away from getting the right answer... 0
14 years ago
#8
(Original post by DOJO)
I understand the a+l.b , its just the kl , where did k come from?

Here was my working before I posted on :

a + lb = (2i + j) + l(i+3j)

and then I did I multiplied out of b with l :

a + lb = (2i + j) + l(i+3j) =
(2i+j) + (li + 3lj)

and then my friend it all went horribly wrong... However by looking at your working I was not far away from getting the right answer... Right, k is just put there. k can be any real value. (You can find it after solving l = -1/3, k = 2 + l)
You just need to group i and j into 2 groups .. then find l.
0
Thread starter 14 years ago
#9
(Original post by BCHL85)
Right, k is just put there. k can be any real value. (You can find it after solving l = -1/3, k = 2 + l)
You just need to group i and j into 2 groups .. then find l.
Oh right , I now see what you have done, you basically did what I did, but then made lamba in b the subject of the equation to get -1/3
0
Thread starter 14 years ago
#10
(Original post by DOJO)
Oh right , I now see what you have done, you basically did what I did, but then made lamba in b the subject of the equation to get -1/3
Just a quick question, do i have to draw questions like this, or can I just get away with algebra? Because when I started this question, I thought by drawing the vectors would make things better, but it made it more confusing....
0
14 years ago
#11
(Original post by DOJO)
Just a quick question, do i have to draw questions like this, or can I just get away with algebra? Because when I started this question, I thought by drawing the vectors would make things better, but it made it more confusing....
Erhh, I think you just can write because a+l.b is parallel to vector i, so
a + l.b = ki where k E R.
0
Thread starter 14 years ago
#12
(Original post by BCHL85)
Erhh, I think you just can write because a+l.b is parallel to vector i, so
a + l.b = ki where k E R.
Ok, I will just do that in future (add k in there , as it is a parallel line), and just = the equation to 0 and then make lamba the subject of the equation.

Thanks!
0
14 years ago
#13
You're welcome 0
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