Find Points C Along A(1, -1, 2) to B(2, 0, 1) Line

[Kaede_N9]

Homework Statement

Find all points C on the line through A(1, -1, 2) and B(2, 0, 1) such that vectors llACll= 2 llBCll

Homework Equations

Not sure.

The Attempt at a Solution

I found the equation of the line for vector AB:
(1,2,-1) +t(2,0,1)

Then found the scalar equation:
x=1+2t
y=-1
x=2+t

I found that t is 1/4 from knowing that C is 1/4 from llACll= 2 llBCll , (where the distance of C is 1/4 from B, and 3/4 from A).

Plugging 1/4 = t gives
x=3/2
y=-1
z=9/4

I stopped here and did not bother plugging in 3/4 since the answer in the back of the book says:
C(3,1,0) and C(5/3,-1/3,4/3)

 

[LCKurtz]

Homework Statement

Find all points C on the line through A(1, -1, 2) and B(2, 0, 1) such that vectors llACll= 2 llBCll

Homework Equations

Not sure.

The Attempt at a Solution

I found the equation of the line for vector AB:
(1,2,-1) +t(2,0,1)

That isn't an equation since there is no = sign. And if you put (x,y,z)= on the left, it still isn't the correct equation for the line AB. (2,0,1) is point B, not the direction vector from A to B.

 

[Kaede_N9]

That isn't an equation since there is no = sign. And if you put (x,y,z)= on the left, it still isn't the correct equation for the line AB. (2,0,1) is point B, not the direction vector from A to B.

Okay, I re did it and obtained the direction vector AB which is (2,-1,-1).
Using the same principle of t=1/4, I obtained (x,y,z) through the scalar equations, to be (3/2,-3/2,7/4).

What did I do wrong?

 

[HallsofIvy]

The vector AB has length \sqrt{4+ 1+ 1}= \sqrt{6}. Taking 1/4 of each coordinate gives an line segment of length (1/4)\sqrt{6}, not 1/4.

 

[SammyS]

The vector AB has length \sqrt{4+ 1+ 1}= \sqrt{6}. Taking 1/4 of each coordinate gives an line segment of length (1/4)\sqrt{6}, not 1/4.

Halls,

You're missing a "/" in your final [/itex] tag.

I put it into the above "QUOTE".

(I will remove this post shortly, assuming you edit yours.)

 

[SammyS]

Halls,

You're missing a "/" in your final [/itex] tag.

I put it into the above "QUOTE".

(I will remove this post shortly, assuming you edit yours.)

Well, I fell asleep & failed to remove it.

DUH!