Vector help

[alexander9408]

The position vector of points A,B and C are a=4i-9j-k, b=i+3j+5k and c=pi-j+3k respectively.

(A) Find the value of p such that A,B and C are collinear.
(B) If p=2, find the position vector of D so that ABCD is a parallelogram.

Sorry that I don't even know how to start off....

Please help.....Thanks.

 

[pka]

The position vector of points A,B and C are a=4i-9j-k, b=i+3j+5k and c=pi-j+3k respectively.
(A) Find the value of p such that A,B and C are collinear.
(B) If p=2, find the position vector of D so that ABCD is a parallelogram.

For (A) find a \(\displaystyle p\) such for some \(\displaystyle \alpha\) we have \(\displaystyle \alpha(a-b)=c-b\)

For (B) find \(\displaystyle d=d_1{i}+d_2j+d_3k\) so that \(\displaystyle b-a=d-c\).

 

[mmm4444bot]

Sorry that I don't even know how to start off....

Draw a picture, perhaps? :cool:

 

[Deleted member 4993]

Not "perhaps" - it is "essential" for partial grade.

 

[pka]

Not "perhaps" - it is "essential" for partial grade.

I hope not essential. These are 3-d vectors and difficult to draw.

 

[mmm4444bot]

Mere surmise, sir.

But drawing such 3d vectors is easier, when people use graph paper. The components of a and b are Integers near zero, too.

The simple act of attempting a drawing, might bring sufficient visualization to cause something to click. Or, it might generate a specific question. Or, it might be a total waste of time.

Perhaps, the poster could try working a 2d-vector version of this exercise first, and then extend to 3d. Anything, really, to make a start at something, to show some kind of effort.

I have little notion why this poster is stuck, what this poster needs, or what thoughts have been thought.

Here's more surmise, sir. Perhaps, the poster could start by talking with their instructor. :idea:

 

[Yogi]

The position vector of points A,B and C are a=4i-9j-k, b=i+3j+5k and c=pi-j+3k respectively.
(A) Find the value of p such that A,B and C are collinear.

Things to know before we start:
Collinear essentially means "same line".
If you multiply a vector by "2" it becomes twice as long, or "-1" it flips directions(try it in 2D). In both cases the resulting vector ends up on the same line as the original vector(collinear).

From A to B move [3 west, 12 north, 6 up]
From A to C move [?, 8 north, 4 up]
Multiply the first row by something to get the second row and you finish the rest. Essentially its the same thing as the formula stated in a previous post but without having to remember this formula.

(B) If p=2, find the position vector of D so that ABCD is a parallelogram.

Draw a 2D parallelogram ABCD.
The move from A to B will be the same as the move from D to C.
The move from A to B will be opposite the move from C to D.

From a to b move [3 west, 12 north, 6 up]
From d to c move [3 west, 12 north, 6 up] (same as a to b)
From c to d move [3 east, 12 south, 6 down] (opposite d to c)

Starting at c (2,-1,3) we make the above move to get to d which is..._________
Again, the same result as the above mentioned formulas.

 

[Deleted member 4993]

Superb response Yogi...

I teach now - engineering. We breakdown very complicated machine - on paper - using Free Body Diagrams. Trying to draw on paper - like mmmbot said - at least shows your efforts to solve the problem and thought process.