Question

If in a ∆ABC, A = (0, 0), B = (3, 3 \[\sqrt{3}\]), C = (−3\[\sqrt{3}\], 3), then the vector of magnitude 2 \[\sqrt{2}\] units directed along AO, where O is the circumcentre of ∆ABC is 

 

Options

• \[\left( 1 - \sqrt{3} \right) \hat{i} + \left( 1 + \sqrt{3} \right) \hat{j}\]

   

• \[\left( 1 + \sqrt{3} \right) \hat{i} + \left( 1 - \sqrt{3} \right) \hat{j}\]

   

• \[\left( 1 + \sqrt{3} \right) \hat{i} + \left( \sqrt{3} - 1 \right) \hat{j}\]

   

• none of these

Answer 1

\[\left( 1 - \sqrt{3} \right) \hat{i} + \left( 1 + \sqrt{3} \right) \hat{j}\]

\[\begin{array}{l}| \overrightarrow{AO} | = 2\sqrt{2} \\ | \overrightarrow{AO} | = | \overrightarrow{BO} | = | \overrightarrow{CO} | = 2\sqrt{2} = R \\ \text{ Let the position vector of }\text{ be }x \hat{i} + y \hat{j} . \\ | \overrightarrow{AO} | = \sqrt{x^2 + y^2} \\ \therefore  x^2 + y^2 = 8 . . . . . (1) \\ \text{ Also, }| \overrightarrow{BO} | = | \overrightarrow{CO} | \\ \sqrt{\left( x - 3 \right)^2 + \left( y - 3\sqrt{3} \right)^2} = \sqrt{\left( x + 3\sqrt{3} \right)^2 + \left( y - 3 \right)^2} \\ x^2 - 6x + 9 + y^2 - 6\sqrt{3}y + 27 = x^2 + 6\sqrt{3}x + 27 + y^2 - 6y + 9 \\ y\left( 6 - 6\sqrt{3} \right) = x\left( 6\sqrt{3} + 6 \right) \\ y = \frac{x\left( 1 + \sqrt{3} \right)}{\left( 1 - \sqrt{3} \right)} . . . . . (2)\end{array}\]
Substituting y from (2) in (1) we get,
\[\begin{array}{l}\left( 1 - \sqrt{3} \right)^2 x^2 + \left( 1 + \sqrt{3} \right)^2 x^2 = 8 \left( 1 - \sqrt{3} \right)^2 \\ x^2 \times 8 = 8 \left( 1 - \sqrt{3} \right)^2 \\ x = 1 - \sqrt{3} \\ y = 1 + \sqrt{3} \\ \therefore \text{ The position vector of O is }\left( 1 - \sqrt{3} \right) \hat{i} + \left( 1 + \sqrt{3} \right) \hat{j}  \\ \overrightarrow{AO} = \left( 1 - \sqrt{3} \right) \hat{i} + \left( 1 + \sqrt{3} \right) \hat{j}\end{array}\]