# Vectors and proof 3points are a straight line

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Thread starter 1 month ago
#1
Hi, this is a 3 part question. I believe I’m right in my working out, however could someone verify my working out as I believe my QM is incorrect.
Above is the question.

Here is my solution.
Last edited by KingRich; 1 month ago
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1 month ago
#2
Looking good. (How) Did you find the value of t in the end?

Minor things which might help in the end, I'd always express stuff like
()a + ()b
so the coefficients of a and b can be easily seen, even if they're negative.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#3
(Original post by mqb2766)
Looking good. (How) Did you find the value of t in the end?

Minor things which might help in the end, I'd always express stuff like
()a + ()b
so the coefficients of a and b can be easily seen, even if they're negative.
Thank you for confirming.
I was having a little trouble finding the t, so the confirmation helps me to re-evaluate my approach.

Noted: the a and b
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1 month ago
#4
(Original post by KingRich)
Thank you for confirming.
I was having a little trouble finding the t, so the confirmation helps me to re-evaluate my approach.

Noted: the a and b
Draw the line connecting the 3 points on and draw the two corresponding triangles in terms of *a, *b and the lines youve found. Is there anything "similar" between the two (three) triangles?

Edit -
As an alternative way to do it, you could express a point on BM in terms of a parameter s, then find where (value of t (and s)) the two lines meet.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#5
(Original post by mqb2766)
Draw the line connecting the 3 points on and draw the two corresponding triangles in terms of *a, *b and the lines youve found. Is there anything "similar" between the two (three) triangles?

Edit -
As an alternative way to do it, you could express a point on BM in terms of a parameter s, then find where (value of t (and s)) the two lines meet.
I have connected the dots and believe I’ve done as you explained. In this instance do the a and b represent the I and J from the vector form?

Between BQ and QM The only thing I see in common is a (t-1), however one is for a and the other is for b.

I think this is what you meant, although like usual I could be wrong
Edit: BP= positive a
Last edited by KingRich; 1 month ago
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1 month ago
#6
There are two ways to go about this

1) is to construct similar triangles and find the t value for which they are actually similar.
2) to find the intersection point between OP and BM which gives Q and t.

For the first method, you can construct 3 "triangles" involving BQ, QM or BM, then argue that the a:b ratio must be the same for each. The "triangle" for BM is formed starting at B then moving -b/2 and then a. So the side ratio of the coefficients a:b is 1:-1/2. You should be able to do the same for one or both of BQ and QM and the a:b side ratios will involve t. Then for the lines to be in the same direction, all would have the same ratio which means you can solve for t.

It would be worth drawing each triangle with one side being parallel to a, the other parallel to b and the third is one of the sides listed above to make sure you understand it
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#7
(Original post by mqb2766)
There are two ways to go about this

1) is to construct similar triangles and find the t value for which they are actually similar.
2) to find the intersection point between OP and BM which gives Q and t.

For the first method, you can construct 3 "triangles" involving BQ, QM or BM, then argue that the a:b ratio must be the same for each. The "triangle" for BM is formed starting at B then moving -b/2 and then a. So the side ratio of the coefficients a:b is 1:-1/2. You should be able to do the same for one or both of BQ and QM and the a:b side ratios will involve t. Then for the lines to be in the same direction, all would have the same ratio which means you can solve for t.

It would be worth drawing each triangle with one side being parallel to a, the other parallel to b and the third is one of the sides listed above to make sure you understand it
Took me a while but I found my way lol.
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1 month ago
#8
Looks correct, but a qucker way would have been to choose one of the "triangles" to be BM as it does not involve t, so its easier. Using that with your BQ
1/-0.5 = t/(t-1)
t-1 = -0.5t
1.5t = 1
t = 2/3
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#9
I didn’t really understand what you were trying to explain with the triangles.
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1 month ago
#10
(Original post by KingRich)
I didn’t really understand what you were trying to explain with the triangles.
Its really just to think of each line segment in terms of a and b. Give me 10 mins and Ill upload a sketch. But if you had a normal set of basis vectors a=i and b=j, then a point like
x = 3i + 4j
would be 3 along the x-axis and 4 along the y-axis. The line OX would form a 35 right angled triangle with the other two sides 3i and 4j.
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1 month ago
#11
This guy is always worth watching. The first four minutes pretty much cover everything (after that its not relevant) apart from they use v,w for the two (basis) vectors rather than a,b. Obvously you can use whatever names you want
https://www.youtube.com/watch?v=k7RM...el=3Blue1Brown

The lines will be parallel if the multipliers (coefficients) associated with a and b are in the same ratio, just as two "normal" lines are parallel if they have the same gradient (same y:x ratio). The corresponding triangles are similar.
Last edited by mqb2766; 1 month ago
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