Vectors and similar triangles question

Original post by petrus123

Vector [4,-10] is the same as vector [2,-5] (same direction, less magnitude) which is where a similar vector triangle is created. Can you see the link from there?

(edited 7 years ago)

Reply 2

Original post by RDKGames

Vector [4,-10] is the same as vector [2,-5] which is where a similar vector triangle is created. Can you see the link from there?

I understand that much, but how can you explain they are perpendicular (sorry for being dopey)?

EDIT: I know that the fact one of the vectors is reversed makes them perpendicular, but I don't know how to explain it or how I'm meant to answer the question.

(edited 7 years ago)

Reply 3

Original post by petrus123

If you were to join the triangles you can use ratios and angles to prove perpendicularity.

Reply 4

Original post by RDKGames

If you were to join the triangles you can use ratios and angles to prove perpendicularity.

OK, I'll try that. Thanks :-)

Reply 5

Original post by petrus123

OK, I'll try that. Thanks :-)

Does this work?

Original post by petrus123

Does this work?

You haven't shown they're similar so you haven't used that fact.

Reply 7

Original post by RDKGames

You haven't shown they're similar so you haven't used that fact.

How would you 'show they're similar'? Just add that 5/2 = 10/4 = 2.5?

Reply 8

Original post by petrus123

How would you 'show they're similar'? Just add that 5/2 = 10/4 = 2.5?

By showing that the vector

\begin{bmatrix} 4 \\ -10 \end{bmatrix} = k \begin{bmatrix} 2 \\ -5 \end{bmatrix}

where k is a constant. Then proceeding to draw the [2,-5] vector triangle alongside the [5,2] one. Then using angles and ratios to prove what you need.

Reply 9

Original post by RDKGames

By showing that the vector

\begin{bmatrix} 4 \\ -10 \end{bmatrix} = k \begin{bmatrix} 2 \\ -5 \end{bmatrix}

where k is a constant. Then proceeding to draw the [2,-5] vector triangle alongside the [5,2] one. Then using angles and ratios to prove what you need.

Thanks.
Do you need to do all three? Isn't the first sufficient (if the x to y ratio is the same, the angle will be the same).

Is the third step what I did (in the picture) or is that wrong?

Reply 10

Original post by petrus123

Thanks.
Do you need to do all three? Isn't the first sufficient (if the x to y ratio is the same, the angle will be the same).

Is the third step what I did (in the picture) or is that wrong?

It would be enough if they didn't ask for similar triangles as part of the explanation.

Reply 11

Original post by RDKGames

It would be enough if they didn't ask for similar triangles as part of the explanation.

Thanks. I think I get it now. Is this the right sort of answer?

Original post by petrus123

Thanks. I think I get it now. Is this the right sort of answer?

Yeah... "sort of". You need to get the vector right with the correct signs and use that vector's similar triangle in your working out. This is nothing complicated.

Reply 13

Original post by RDKGames

Yeah... "sort of". You need to get the vector right with the correct signs and use that vector's similar triangle in your working out. This is nothing complicated.

I'm aware. I think I'm missing the obvious.

Isn't flipping the triangle necessary to show it's a right angle? Otherwise, what would you try to show with the similar triangles?

Thanks, and sorry this is taking me so frustratingly long!

Reply 14

Original post by petrus123

Of course not. What do you mean? You are missing the obvious because you are not using the similar triangle to [4,-10] when you join up the two distinct vector triangles.

The thing with the similar triangle is that it clearly shows how that vector is related to the other one (from using the same numbers) so from there on in you should be able to prove it without much hassle.

Reply 15

Original post by RDKGames

I'm getting very confused at this point. How would adding a similar triangle help me to prove this, and where and what size would it be? Any chance you could draw a quick sketch (MS Paint)? Again, I have a feeling I'm going to kick myself, but I just don't get what I'm trying to do...

Reply 16

Original post by petrus123

Just literally this alongside your previous working where you show similarity. Here you can show which angles are the same and work from there.

(edited 7 years ago)

Reply 17

Original post by RDKGames

Just literally this alongside your previous working where you show similarity. Here you can show which angles are the same and work from there.

Thanks for the drawing and the help. Is the point of that just to show the angles (which weren't clear on my drawings)?

Like this? (also the question sheet used parantheses for vectors and I know I forgot the arrows on the bottom part)

Sorry for being such a numpty with this problem...

Original post by petrus123

Yes. When working with similar triangles the angles remain the same. One triangle is simply an enlargement and/or rotation of the other.

Reply 19