# Position Vector of Reflection of a Point

• August 10th 2009, 01:18 AM
MathJunction
Position Vector of Reflection of a Point
Hello, I am currently having trouble understanding how to find the reflection point of a point with the coordiantes of that point and the vector equation of the line. I have tried finding the maginitude of the foot of the prependicular, but I still could not get the answer. I hope my doubts can be cleared.

This is the question :
Find the position vector of the image C' of the point C(5,2,-1) under a reflection in the line l joining the points A(3,1,3) and B(-1,-1,-3).
• August 10th 2009, 02:37 AM
Vectors
Hello MathJunction

Welcome to Math Help Forum!
Quote:

Originally Posted by MathJunction
Hello, I am currently having trouble understanding how to find the reflection point of a point with the coordiantes of that point and the vector equation of the line. I have tried finding the maginitude of the foot of the prependicular, but I still could not get the answer. I hope my doubts can be cleared.

This is the question :
Find the position vector of the image C' of the point C(5,2,-1) under a reflection in the line l joining the points A(3,1,3) and B(-1,-1,-3).

Using vectors, suppose that the position vector of $A$ is $\vec{a} = 3\vec{i} + \vec{j} + 3\vec{k}$. Similarly $\vec{b} = -\vec{i} -\vec{j} -3\vec{k},\, \vec{c} = 5\vec{i} + 2\vec{j} -\vec{k}$.

Then if $M$ is the mid-point of $AB$, its position vector is given by $\vec{m} = \tfrac12(\vec{a}+\vec{b}) = -\vec{i}$.

$\Rightarrow \vec{CM} = \vec{m} - \vec{c} = -6\vec{i} -2\vec{j} +\vec{k}$

Now if $C'$ is the reflection of $C$ in $AB$, $\vec{MC'}=\vec{CM}$

$\Rightarrow \vec{c'} = \vec{m} + \vec{MC'} =\vec{m} + \vec{CM}$, where $\vec{c'}$ is the position vector of $C'$.

Can you complete it now?

• August 10th 2009, 04:26 AM
MathJunction
Quote:

Hello MathJunction

Welcome to Math Help Forum!Using vectors, suppose that the position vector of $A$ is $\vec{a} = 3\vec{i} + \vec{j} + 3\vec{k}$. Similarly $\vec{b} = -\vec{i} -\vec{j} -3\vec{k},\, \vec{c} = 5\vec{i} + 2\vec{j} -\vec{k}$.

Then if $M$ is the mid-point of $AB$, its position vector is given by $\vec{m} = \tfrac12(\vec{a}+\vec{b}) = -\vec{i}$.

$\Rightarrow \vec{CM} = \vec{m} - \vec{c} = -6\vec{i} -2\vec{j} +\vec{k}$

Now if $C'$ is the reflection of $C$ in $AB$, $\vec{MC'}=\vec{CM}$

$\Rightarrow \vec{c'} = \vec{m} + \vec{MC'} =\vec{m} + \vec{CM}$, where $\vec{c'}$ is the position vector of $C'$.

Can you complete it now?

Hello Grandad, thank you for taking some time off in helping me. I regret to tell you I am still blurred. Shouldn't vector m be 1/2(a+b) = (1,0,0) since vector AB is ( -4,-2,-6) ? Thus AM is half of AB which is (-2,-1,-3) and then OM would be vector AM + vector OA?

Sorry to trouble you again.
• August 10th 2009, 05:08 AM
Vectors
Hello MathJunction
Quote:

Originally Posted by MathJunction
Hello Grandad, thank you for taking some time off in helping me. I regret to tell you I am still blurred. Shouldn't vector m be 1/2(a+b) = (1,0,0) since vector AB is ( -4,-2,-6) ? Thus AM is half of AB which is (-2,-1,-3) and then OM would be vector AM + vector OA?

Sorry to trouble you again.

You're quite right - I got a sign wrong. Sorry!

$\vec{m} = \tfrac12(\vec{a}+\vec{b})= \vec{i}$

So $\vec{CM}=\vec{m}-\vec{c}=-4\vec{i} -2\vec{j} +\vec{k}$

Then continue as I said before.