Question

If in a Δ ABC, A ≡ (0, 0), B ≡ (3, 3, √3), C ≡ (–3, √3, 3), then the vector of magnitude units directed along AO, where O is the circumcentre of Δ ABC is

A. \((1-\sqrt3)\hat i+(1+\sqrt3)\hat j\)

B. \((1+\sqrt3)\hat i+(1-\sqrt3)\hat j\)

C. \((1+\sqrt3)\hat i+(\sqrt3-1)\hat j\)

D. none of these

Answer 1

Correct option is  A. \((1-\sqrt3)\hat i+(1+\sqrt3)\hat j\) 

Slope of a line joining two points = \(\cfrac{y_2-y_1}{x_2-x_1}\)

Slope of AC = \(\cfrac{0-3}{0+3\sqrt3}\)

Product of Slopes (AC \(\times\) AB) = \(\big(-\cfrac{1}{\sqrt3}\times\sqrt3\big)\) = -1

As the Product of Slopes (AC \(\times\) AB) = -1, so AC \(\times\) AB, ie..,  ∠CAB = 90°.

Circumcentre (O) of Triangle ABC = Mid-Point of BC

Mid-Point of BC = \(\cfrac{3-\sqrt3}2, \cfrac{3+3\sqrt3}2\)

Vector along \(\vec {OA},\) whose magnitude is