Let A, B, C be three points in xy -plane, whose position vectors are given by 3i^+j^, i^+3j^ and ai^+(1−a)j^ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors OA and OB is 2θ, then the sum of all the possible values of a is:
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Step-by-Step Solution
Step 1: Find the unit vectors and angle bisector
First, we find the unit vectors along OA and OB. The magnitude of OA is (3)2+12=3+1=2, and similarly for OB. The vector along the angle bisector of OA and OB is given by the sum of their unit vectors. We denote this bisector vector as d.
Step 2: Determine the equation of the angle bisector line
From the previous step, we see that the angle bisector vector d is proportional to i^+j^. This means the line passes through the origin and has a slope of 1. Therefore, the equation of the line bisecting the angle between OA and OB is x−y=0.
Step 3: Calculate the distance from point C to the line
The position vector of point C is ai^+(1−a)j^, so its coordinates are (a, 1-a). We use the formula for the distance from a point (x1,y1) to a line Ax+By+C=0, which is A2+B2∣Ax1+By1+C∣. Substituting the coordinates of C and the line equation x−y=0, we get the distance as 12+(−1)2∣a−(1−a)∣.
Step 4: Solve for 'a' using the given distance
The problem states that the distance of point C from the line is 23. We equate our calculated distance to this given value. This leads to an absolute value equation, ∣2a−1∣=3, which can be split into two linear equations: 2a−1=3 and 2a−1=−3.
Step 5: Find the possible values of 'a' and their sum
Solving the first equation, 2a−1=3, we get 2a=4, so a=2. Solving the second equation, 2a−1=−3, we get 2a=−2, so a=−1. The sum of all possible values of a is 2+(−1)=1.