Question

Vectors
Given OA=2i-3j,OB=3i-6j and OC=5i-12j,show that the points A,B and C are collinear.
@ganeshie8

Answer 1

can you find the "direction vector" from OA to OB ?

Answer 2

yes\[AB=\left(\begin{matrix}1 \\ 3\end{matrix}\right)\]

Answer 3

is it correct?

Answer 4

the j component is -6 - (-3)

Answer 5

ops\[AB=\left(\begin{matrix}1 \\ -3\end{matrix}\right)\]

Answer 6

the equation of a line can be written as
\[ p= p_0 + t \vec{v} \]
where v is the direction vector, t is a scalar variable and p0 is a point on the line
so the eq for a line with points A and B is
\[ p= (2, -3) + t ( 1, -3)
\]
collinear means line on the same line.
so we have to show that point (5, -12) is on the line.
I would show there is a t that "works"
i.e.
solve for the i component
2+ t = 5
and show that t=3 works

Answer 7

2i-3j + 3 * (i -3j) = 2i-3j + 3i -9j = 5i -12j

Answer 8

all 3 points satisfy the equation (with t=0, t=1, and t=3 )

Answer 9

can u tell me how to get t=0 and t=1 @phi

Answer 10

when you find the equation
P0 + t * v
if we use point A as P0, (which is what we did), then t=0 gives us P0, i.e. gives us A

but even if you did not know this, you would write
(2, -3) + t (1, -3) = (2, -3)

now pick either dimension (or component)
for example, if we look at the "j" component (2nd dimension)
we have this equation
-3 + t*-3 = -3
add +3 to both sides
t*-3= 0
t= 0

you can do that same thing for the 2nd point
(3, -6) to find t= 1 works

but because we used points A and B to find this equation
it always turns out t=0 and t=1 will work to "get back" A and B

Answer 11

Thank you @phi :)

Answer 12

of course, once we find t, we have to verify that t gives us back the entire point.
for example if we had

(2, -3) + t (1, -3) = (2, -4)
and we solve using i component:
2 + t*1 = 2
we find t=0

now we use t=0 in the equation
(2,-3) + 0* (1,-3) = (2+0, -3+0) or just (2, -3)
and that does not equal (2,-4)
this shows there is no t that works for all of the components at the same time
and that means that the point (2, -4) is not on the line

Answer 13

We could also show all 3 vectors make the same angle with the x axis by using dot product

Answer 14

***(2, -3) + t (1, -3) = (2, -4)***
that is an example to show a point that is not on the line. I picked (2,-4) as an arbitrary point that I know is not on the line.

Answer 15

i c

Answer 16

The book answer show that \[AB=\frac{ 1 }{ 3 }AC\],is it acceptable if i put \[AB=\frac{ 1 }{ 2 }BC\]

Answer 17

yes

Answer 18

Okay,thnx @phi :)