# Thread: Supposingly Easy Vector Question

1. ## Supposingly Easy Vector Question

In triangle ABC, the point X divides BC internally in the ration m:n, where m+n = 1. Express (AX)^2 in terms of AB , BC,CA,m and n.

My attempt :

(AX)^2 = (AB)^2 - (BX)^2 ----1
(AX)^2 = (CA)^2 - (XC)^2 ----2

1+2 , 2(AX)^2 = (AB)^2 + (CA)^2 - (m BC)^2 - (n BC)^2
(AX)^2 = [(AB)^2 +(CA)^2 - (m BC)^2 - (n BC)^2]/2

2. Hello,
Originally Posted by ose90
In triangle ABC, the point X divides BC internally in the ration m:n, where m+n = 1. Express (AX)^2 in terms of AB , BC,CA,m and n.

My attempt :

(AX)^2 = (AB)^2 - (BX)^2 ----1
(AX)^2 = (CA)^2 - (XC)^2 ----2

1+2 , 2(AX)^2 = (AB)^2 + (CA)^2 - (m BC)^2 - (n BC)^2
(AX)^2 = [(AB)^2 +(CA)^2 - (m BC)^2 - (n BC)^2]/2
I don't quite understand your 1 & 2 equations...

What exactly do you want ? Is it $\vec{AX}$ ? Or the length $AX^2$ ?

Be careful, $\vec{AX}=\vec{AB} {\color{red}+} \vec{BX}$ and $\vec{AX}=\vec{AC}+\vec{CX}$

3. Originally Posted by Moo
Hello,

I don't quite understand your 1 & 2 equations...

What exactly do you want ? Is it $\vec{AX}$ ? Or the length $AX^2$ ?

Be careful, $\vec{AX}=\vec{AB} {\color{red}+} \vec{BX}$ and $\vec{AX}=\vec{AC}+\vec{CX}$
Sorry for the carelessly constructed post.

I want to find AX^2, thanks.