Need help w/ vector task: Trapeze ABCD, AB parallel w/ CD, Vector AB=a, Vector AD=b, Vector DC=(2/3)a

[john44]

I'll try my best to translate the task, I've done the previous questions, but need help the last
Trapeze ABCD
AB is parallel with CD
Vector AB = a
Vector AD = b
Vector DC = two thirds ( 2/3 ) of a

Point E is in middle of BC, so BE is (1/2)BC
AE and BD intersect eachother at point S
Find AS expressed by a and b
At first i thought i could say AS = t * AE (because point E is just further behind than S, but on same line. But I think I can only have a's and b's in the answer

Last question is:
A point F is given by: CF = 3/8 DB
Show that A, E, and F are in a line
For this I think I need to show that AE is parallel with AF but my answers doesnt match..

 

[topsquark]

You've given a large amount of information to go on. Good for you! But nothing in the problem tells us where F is. It could be anywhere on an arc of radius CF centered at C. I can't see why it would have to be on the extension of the line segment AE.

Is there something missing?

-Dan

 

[Dr.Peterson]

But nothing in the problem tells us where F is. It could be anywhere on an arc of radius CF centered at C. I can't see why it would have to be on the extension of the line segment AE.

Is there something missing?

I think vector CF = 3/8 of vector DB. Then it makes sense.

 

[topsquark]

I think vector CF = 3/8 of vector DB. Then it makes sense.

Ahhhh. I get it now. I guess the OP is using an underline to denote vectors. I haven't seen someone do that for a long time. Thanks for the catch!

-Dan

 

[Dr.Peterson]

I'll try my best to translate the task, I've done the previous questions, but need help the last
Trapeze ABCD
AB is parallel with CD
Vector AB = a
Vector AD = b
Vector DC = two thirds ( 2/3 ) of a

Point E is in middle of BC, so BE is (1/2)BC
AE and BD intersect each other at point S
Find AS expressed by a and b
At first i thought i could say AS = t * AE (because point E is just further behind than S, but on same line. But I think I can only have a's and b's in the answer

Last question is:
A point F is given by: CF = 3/8 DB
Show that A, E, and F are in a line
For this I think I need to show that AE is parallel with AF but my answers doesnt match..
View attachment 11632

There are probably many ways to proceed. To express AS in terms of a and b, I might first express AE, and write AS as mAE, then express AS as b + nDB (in terms of a and b). Setting these equal, you can solve for m and n, and then you have AS.

Once you've done that part, we can discuss the second part. (My first attempt at that didn't work out, but I haven't taken the time to find my error or determine that the problem is wrong.)

 

[john44]

There are probably many ways to proceed. To express AS in terms of a and b, I might first express AE, and write AS as mAE, then express AS as b + nDB (in terms of a and b). Setting these equal, you can solve for m and n, and then you have AS.

Once you've done that part, we can discuss the second part. (My first attempt at that didn't work out, but I haven't taken the time to find my error or determine that the problem is wrong.)

Do you mean: AS = m(AE) = m ( (5/6)a + (1/2)b )
And AS = b + n(DB) = b + n (-b + a)
How do I that solve for m and n?

 

[Dr.Peterson]

Good. Now, one way to go is to note that if a and b are linearly independent, AS can be obtained in only one way as a linear combination of a and b, so when you set these equal, the coefficients of a on each side must be equal, and likewise for b. Solve the resulting system of equations.

Others will probably have alternative methods to suggest.

 

[john44]

Good. Now, one way to go is to note that if a and b are linearly independent, AS can be obtained in only one way as a linear combination of a and b, so when you set these equal, the coefficients of a on each side must be equal, and likewise for b. Solve the resulting system of equations.

Others will probably have alternative methods to suggest.

Alright I ended up with AS = (5/8)a + (3/8)b which I think is correct, but can't confirm - thank you for the help

 

[Dr.Peterson]

That's what I get.

One way to check would be to verify that AS and AE are collinear, and DS and DB are collinear.