Let A,B,C be points in the xy-plane with position vectors OA=√3 i+ j, OB= i+√3 j, OC=a i+(…

Question

Let $A,B,C$ be points in the $xy$-plane with position vectors $\vec{OA}=\sqrt3\hat i+\hat j$, $\vec{OB}=\hat i+\sqrt3\hat j$, $\vec{OC}=a\hat i+(1-a)\hat j$ w.r.t. origin $O$. If the distance of $C$ from the line bisecting the angle between $\vec{OA}$ and $\vec{OB}$ is $\dfrac9{\sqrt2}$, the sum of all possible values of $a$ is:

Accepted Answer

$|\vec{OA}|=|\vec{OB}|=2$, so the angle bisector direction is $\vec{OA}+\vec{OB}=(\sqrt3+1)(\hat i+\hat j)$, i.e. the line $y=x$, or $x-y=0$.

Distance from $C=(a,1-a)$ to $x-y=0$: $\dfrac{|a-(1-a)|}{\sqrt2} = \dfrac{|2a-1|}{\sqrt2} = \dfrac9{\sqrt2} \implies |2a-1|=9 \implies a=5	ext{ or }a=-4$

Sum $= 5+(-4)=1$.

Solutions · 1