Question

With respect to the origin O, the points A and B have position vectors given by −−→OA = i + 2j + 2k and
−−→OB = 3i + 4j. The point P lies on the line AB and OP is perpendicular to AB.
(ii) Find the position vector of P.

Answer 1

@phi

Answer 2

I like to use <a,b,c> to represent a vector rather than a i + b j + c k. (easier to type)
A is at (1,2,2)
B is at (3,4,0)
the vector from A to B is B-A. I would find B-A

Answer 3

then

Answer 4

write down the equation of the line AB

P (for point on the line) = <1,2,2> + t (B-A)
where B-A is ?

Answer 5

2,2,-2

Answer 6

ok, and because we only care about the direction we can write it as 2<1,1,-1> and ignore the 2
the equation of the line AB is
P= <1,2,2> + t<1,1,-1>
where t is a scalar

Answer 7

and?

Answer 8

I was thinking OP perpendicular to the line means
means P is perpendicular to the direction vector B-A i.e. to <1,1,-1>
and when we dot product P with <1,1,-1> we get 0

Answer 9

and how do we do that

Answer 10

let P= <x,y,z>
then because P is orthogonal to <1,1,-1> we know
<x,y,z> dot <1,1,-1> =0
and
x + y - z =0

Answer 11

using the equation of the line
P= <1,2,2> + t<1,1,-1>
<x,y,z> = <1+t, 2+t, 2-t>
we see
x= 1+t
y= 2+t
z= 2-t
put those into
x + y - z =0

Answer 12

i dont think it is good ?

Answer 13

what do you mean ?

Answer 14

op = 2/3i + 5/3 j + 7/3 k

Answer 15

this is the answer u should be getting

Answer 16

yes, that is what we will get

Answer 17

how ? i am not getting it

Answer 18

do you understand this part:
let P= <x,y,z> then because P is orthogonal to <1,1,-1>
we know <x,y,z> dot <1,1,-1> =0
and x + y - z =0

Answer 19

yes

Answer 20

also, P is on the line so it must be that
P= <1,2,2> + t<1,1,-1>
<x,y,z> = <1+t, 2+t, 2-t>

Answer 21

yes good

Answer 22

and component by component,
<x,y,z> = <1+t, 2+t, 2-t>
means
x= 1+t
y= 2+t
z= 2-t

Answer 23

yes

Answer 24

now put all the variables in terms of "t" in
x + y - z =0

using
x= 1+t
y= 2+t
z= 2-t

and solve for t

Answer 25

t is ?

Answer 26

the equation of a line in n-dim is typically written as
\[ P= P_0 + t \vec{V} \]
where V is the direction we move in starting at some point P0
the "t" is a scalar that we multiply the direction vector by to show how far we move

we are using the info in this problem to find the "t" that gets us to a point that solves the problem

Answer 27

i mean how much t is ?

Answer 28

what answer uve got for t

Answer 29

x + y - z =0 using x= 1+t ,y= 2+t ,z= 2-t
replace x with 1+t, etc

(1+t) + (2+t) - (2-t) = 0

Answer 30

yes i know this

Answer 31

what do you get ?

Answer 32

-1/3

Answer 33

now use
P= <x,y,z> = <1,2,2> + t<1,1,-1> = <1+t, 2+t, 2-t>

to find the answer

Answer 34

is t -1/3 ?

Answer 35

and then write it using i, j and k

yes , t= -1/3

Answer 36

if you do the problem with t= -1/3 you will get the answer you posted earlier.