# Thread: Vector Problem

1. ## Vector Problem

Given 3 points A(0,2,7) , B(5,-3,2) and C(1,1,1). (1)Find the position vector of the point R on AB such that CR is perpendicular to AB. Hence (2)find the perpendicular distance from C to AB and the (3)position vector of the reflection of C in AB.

Here's my solution, i just have trouble finding the "position vector of the reflection of C in AB."

Vector AB = OB - OA
= (5-0,-3-2,2-7)
=(5,-5,-5)

vector equation of line AB l: (0,2,7) + λ(5,-5,-5)

vector OR = (5λ, 2-5λ, 7-5λ)

vector CR= OR - OC = (5λ-1, 1-5λ, 6-5λ)

vector CR.AB=0 , since cos(90) = 0, given by scalar product

(5λ-1, 1-5λ, 6-5λ).(5,-5,-5) = 0

λ=8/15

vector OR= (8/3, -2/3, 13/3)= 1/3(8,-2,13)

vector CR=(5/3,-5/3,10/3)
Magnitude of CR = (50/3)^0.5

now i'm having trouble with the reflection part

2. Consider $\displaystyle OC$ (the position of point $\displaystyle C$)

$\displaystyle = OR + RC$

Now the position vector of the reflection of $\displaystyle C$ in $\displaystyle AB$ ,

it should be $\displaystyle OR - RC$

$\displaystyle = \frac{1}{3} [(8,-2,13) - (5,-5,10) ]$

$\displaystyle = \frac{1}{3} (3 , 3 , 3 ) = (1,1,1)$

3. the answer is 1/3(13,-7,23) but thanks for helping

4. Originally Posted by simplependulum
Consider $\displaystyle OC$ (the position of point $\displaystyle C$)

$\displaystyle = OR + RC$

Now the position vector of the reflection of $\displaystyle C$ in $\displaystyle AB$ ,

it should be $\displaystyle OR - RC$

$\displaystyle = \frac{1}{3} [(8,-2,13) - (5,-5,10) ]$

$\displaystyle = \frac{1}{3} (3 , 3 , 3 ) = (1,1,1)$
Dear simplependulum,

You have made a slight mistake, instead of $\displaystyle \frac{1}{3} [(8,-2,13) - (5,-5,10)]$ it should be $\displaystyle \frac{1}{3} [(8,-2,13) - (-5,5,-10)]$ since $\displaystyle RC=(-5,5,-10)$. (You have accidently taken CR instead of RC).

Dear mephisto50,

The above correction will give you the correct answer.

Thank you.