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
(a) $(1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j}$

(b) $(1 + \sqrt{3}) \hat{i} + (1 - \sqrt{3}) \hat{j}$

(c) $(1 + \sqrt{3}) \hat{i} + (\sqrt{3} - 1) \hat{j}$

(d) none of these

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Solution

(a) $(1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j}$

$\begin{array}{l} | \vec{A O} | = 2 \sqrt{2} \\ | \vec{A O} | = | \vec{B O} | = | \vec{C O} | = 2 \sqrt{2} = R \\ Let the position vector of O b e x \hat{i} + y \hat{j} . \\ | \vec{A O} | = \sqrt{x^{2} + y^{2}} \\ ∴ x^{2} + y^{2} = 8 . . . . . (1) \\ Also, | \vec{B O} | = | \vec{C O} | \\ \sqrt{{(x - 3)}^{2} + {(y - 3 \sqrt{3})}^{2}} = \sqrt{{(x + 3 \sqrt{3})}^{2} + {(y - 3)}^{2}} \\ x^{2} - 6 x + 9 + y^{2} - 6 \sqrt{3} y + 27 = x^{2} + 6 \sqrt{3} x + 27 + y^{2} - 6 y + 9 \\ y (6 - 6 \sqrt{3}) = x (6 \sqrt{3} + 6) \\ y = \frac{x (1 + \sqrt{3})}{(1 - \sqrt{3})} . . . . . (2) \end{array}$

Substituting y from (2) in (1) we get,

$\begin{array}{l} {(1 - \sqrt{3})}^{2} x^{2} + {(1 + \sqrt{3})}^{2} x^{2} = 8 {(1 - \sqrt{3})}^{2} \\ x^{2} \times 8 = 8 {(1 - \sqrt{3})}^{2} \\ x = 1 - \sqrt{3} \\ y = 1 + \sqrt{3} \\ ∴ The position vector of O is (1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j .} \\ \vec{AO} = (1 - \sqrt{3}) \hat{i} + (1 + \sqrt{3}) \hat{j} \end{array}$

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