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Simple M1 Vector help ?

This is most likely a simple question although im having some trouble and any help would be greatly appreciated.

Given that a = 2i + 5j and b = 3i - j

find a) λ if a + λb is parallel to the vector i

b) μ if μa + b is parallel to the vector j


So far for a) i have :
(2i+5j)+λ(3i-j)=i
2i+5j+3iλ-jλ=i
Then i( 2+3λ ) + j( 5-λ )

Although im unsure about the last line and where to go from there.

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Reply 1
you don't know that what you have written is equal to i, only that it is parallel to it.

your last line makes no sense. there is no equals sign. assuming you meant an equals sign instead of a plus sign.... does what you've written make any sense? can ANY amount of i vectors equal some amount of j vectors?
Reply 2
Yes sorry I intened the last line to = i, I do however think my working out is wrong and im not sure what to do =/.
Reply 3
Well played James. We need a book tbh!
Reply 4
does λ=-2/3?
Reply 5
I don't know, I need help working out λ, I dont have the answers.
Reply 6
draw a diagram. and think about what i said. give me something here.
Reply 7
Hi, Basically if its saying that its parellel to the vector, i, then it means its horizontal right? so the j component must equal 0 (if it equalled anything else, the vector would be at an angle).

so 5-λ=0
λ=5

I think I'm right :smile:
Hope it helps anyway. (apply the same thing to part b with the i component equating to 0)
Reply 8
LifeIsLoud
Hi, Basically if its saying that its parellel to the vector, i, then it means its horizontal right? so the j component must equal 0 (if it equalled anything else, the vector would be at an angle).

so 5-λ=0
λ=5

I think I'm right :smile:
Hope it helps anyway. (apply the same thing to part b with the i component equating to 0)


But how do you know that Vector I is horizontal and not at an angle ?
Reply 9
jhwilliam
But how do you know that Vector I is horizontal and not at an angle ?

because this is the convention used. It just is. The unit vector i is 1 unit in the horizontal direction and j is one unit in the vertical direction.
Reply 10
Thats what defines the i vector, sorry I don't how to better explain it but its the fundamentals of vector mechanics. Go to page 1 on your M1 book! The i component is the horizontal (x) component and the j component is the vertical (y) component.
Reply 11
as per ur working the line before last is
2i+5j+3iλ-jλ=i
3iλ-jλ=i-2i-5j
λ(3i-j)=-i-5j
λ=-i-5j/3i-j
u can then simplify further if u want
this answer is according 2 your working
Reply 12
His working is wrong, the top equation doesn't equal i. It is a multiple of i, but that doesn't really matter. When it says parallel to vector i, it means there is no j component, thats it.
Reply 13
LifeIsLoud
His working is wrong, the top equation doesn't equal i. It is a multiple of i, but that doesn't really matter. When it says parallel to vector i, it means there is no j component, thats it.


I understand now, thank you for your help !
Reply 14
the vector (2i + 5j) + λ(3i - j) is parallel to the unit vector i so it has no j component, so multiply out the brackets and collect the i terms together and the j terms together and set the j terms = 0 (because there is no j component).
Reply 15
Danielllll
the vector (2i + 5j) + λ(3i - j) is parallel to the unit vector i so it has no j component, so multiply out the brackets and collect the i terms together and the j terms together and set the j terms = 0 (because there is no j component).



Yes so I then have, 2i + 5j + 3iλ - jλ

and the j values = 0

so I am left with 2i+3iλ

so is the answer λ = -2/3 ?
Reply 16
Is that correct ? or have I gone down the wrong path yet again?
Reply 17
jhwilliam
Yes so I then have, 2i + 5j + 3iλ - jλ

and the j values = 0

so I am left with 2i+3iλ

so is the answer λ = -2/3 ?

no, you don't need the i values for this question. you collect the j terms so you have 5j - λj which = ( 5 - λ )j and then you set this equal to 0 and solve for λ.
Reply 18
Danielllll
no, you don't need the i values for this question. you collect the j terms so you have 5j - λj which = ( 5 - λ )j and then you set this equal to 0 and solve for λ.


So j=0 there for λ = 5 ?
Reply 19
jhwilliam
So j=0 there for λ = 5 ?

yeppp :smile:

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