Let √3 i+ j, i+√3 j, i+(1-) j be position vectors of A,B,C.

Question

Let $\sqrt3\hat i+\hat j$, $\hat i+\sqrt3\hat j$, $\beta\hat i+(1-\beta)\hat j$ be position vectors of $A,B,C$. If the distance of $C$ from the bisector of the acute angle between $\vec{OA},\vec{OB}$ is $\dfrac3{\sqrt2}$, the sum of all possible $\beta$ is:

Accepted Answer

The bisector of the acute angle between $\vec{OA},\vec{OB}$ (both magnitude 2, symmetric about $y=x$) is the line $y=x$, i.e. $x-y=0$.

Distance from $C=(\beta,1-\beta)$: $\dfrac{|\beta-(1-\beta)|}{\sqrt2}=\dfrac3{\sqrt2} \implies |2\beta-1|=3 \implies \beta=2	ext{ or }-1$

Sum $=2+(-1)=1$.

Solutions · 1