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Let sqrt(3)hati+hatj,hati+sqrt(3)hatj" a...

Let `sqrt(3)hati+hatj,hati+sqrt(3)hatj" and "betahati+(1-beta)hatj` respectively be the position vectors of the points A, B and C with respect to the origin O. If distance of C from the bisector of the acute angle between OA and OB is `(3)/(sqrt(2))`, then the sum of all possible values of `beta` is (a) 1 (b) 3 (c) 4 (c) 2

A

1

B

3

C

4

D

2

Text Solution

Verified by Experts

The correct Answer is:
A
According to given information, we have the following figure.

Clearly, angle bisector divides the sides AB in OA:OB,
`i.e., 2 : 2 =1 : 1" "[using angle bisector theorem]`
So, D is the mid-point of AB and hence coordinates of D are `((sqrt(3)+1)/(2),(sqrt(3)+1)/(2))`
Now, eauation of bisector OD is
`(y-0)=(((sqrt(3)+1)/(2)-0)/((sqrt(3)+1)/(2)-0))(x-0)impliesy=x`
`impliesx-y=0`
According to the problem,
`(3)/(sqrt(2))=CM=|(beta-(1-beta))/(sqrt(2 ))|`
[Distance of a point `P(x_(1),y_(1))` from the line `ax+by+c=0" is "|(ax_(1)+by_(1)+c)/(sqrt(a^(2)+b^(2)))|]`
`implies" "|2beta-1|=3implies2beta=+-3+1`
`implies" "2beta=4, -2impliesbeta=2, -1`
Sum of 2 and -1 is 1.
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