Mark the correct alternative in each of the following:
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

Slope of a line joining two points ![]()
Slope of AC ![]()
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Slope of AB ![]()
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Product of Slopes (AC
AB) = ![]()
= -1
As the Product of Slopes (AC
AB) = -1, so AC ⟘ AB, ie..,
CAB = 90°.
Circumcentre (O) of Triangle ABC = Mid-Point of BC
Mid-Point of BC = ![]()


Now, ![]()

![]()
![]()
Unit Vector ![]()

Vector along
, whose magnitude is ![]()
![]()
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Option (A) is the answer.
Couldn't generate an explanation.
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