JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 13)
Let $A, B, C$ be three points in xy-plane, whose position vector are given by $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $a \hat{i}+(1-a) \hat{j}$ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ is $\frac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is :
2
0
$ \frac{9}{2} $
1
Explanation
$$\begin{aligned}
&\text { Equation of angle bisector : } \mathrm{x}-\mathrm{y}=0\\
&\begin{aligned}
& \left|\frac{\mathrm{a}(1-\mathrm{a})}{\sqrt{2}}\right|=\frac{9}{\sqrt{2}} \Rightarrow \mathrm{a}=5 \text { or }-4 \\
& \text { Sum }=5+(-4)=1
\end{aligned}
\end{aligned}$$
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