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Let \sqrt3\hat{i}+\hat{j},\: \: \hat{i}+\sqrt3 \hat{j} \: \: and\: \: \beta \hat{i}+(1-\beta )\hat{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 \frac{3}{\sqrt2},  then the sum of all possible values of \beta is:

  • Option 1)

    1

  • Option 2)

    3

  • Option 3)

    2

  • Option 4)

    4

Answers (1)

best_answer

 

Angle between vector a and vector b -

\cos \Theta =\frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}

- wherein

Here 0\leq \Theta \leq \pi??????

Angle bisector passes through (\frac{1+\sqrt3}{2},\frac{1+\sqrt3}{2})   and   origin.

Angle bisector is x - y = 0

Perpendicular distance from c =\frac{3}{\sqrt2}

=>\frac{1\cdot \beta -(1-\beta )}{\sqrt2}=\frac{3}{\sqrt2}

=>\left | 2\beta -1 \right |=3

=>\beta = 2\: \: or\: \: -1

Answer is 2+(-1)=1

 


Option 1)

1

Option 2)

3

Option 3)

2

Option 4)

4

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