KVPY SX SB Exam 2016 Question Paper

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Question : 17 of 120

Marks: +1, -0

Let

v↖{→}

be a vector in the plane such that

| v↖{→}- i↖{→} |= | v↖{→}- 2i↖{→} |= | v↖{→}- j↖{→} |

. Then

| v↖{→} |

lies in the interval -

(0, 1]
(1, 2]
(2, 3]
(3, 4]

Solution:

V is the circumcentre of ΔABC
∀ A ≡ (1, 0), B ≡ (0, 1), C(2, 0)
Let V(x, y)
VA = VB = VC

(x- 1)^2 + y^2

= x^2 + (y- 1)^2

= (x - 2)^2 + y^2

(x,y)=(3/2,3/2)

\V={3i+3j}/2

|\V|=3/√2 ∈ (2,3)

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