JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 22)
Let $$\sqrt 3 \widehat i + \widehat j,$$ $$\widehat i + \sqrt 3 \widehat j$$ and $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \over {\sqrt 2 }}$$, then the sum of all possible values of $$\beta $$ is :
4
1
2
3
Explanation
Angle bisector is x $$-$$ y = 0
$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$
$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1
$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$
$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1
Comments (0)
