Vector Algebra
Position Vectors and Angle Bisectors
Grade 12
Question:
<p><strong>68.</strong> Let \(\sqrt{3}\hat{i} + \hat{j}\), \(\hat{i} + \sqrt{3}\hat{j}\) and \(\beta\hat{i} + (1-\beta)\hat{j}\) respectively be the position vectors of the points A, B and C with respect to the origin O. If the 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 ________.</p>
Step-by-Step Solution
Key Concept: The angle bisector of two vectors is in the direction of their normalized sum. The distance from a point to a line through origin equals |projection perpendicular to bisector direction|, using the formula: distance = |r × û|/|û| where û is the unit bisector direction.
Step 1: Find unit vectors along OA and OB OA = √3î + ĵ, |OA| = √(3+1) = 2, so â = (√3/2)î + (1/2)ĵ OB = î + √3ĵ, |OB| = √(1+3) = 2, so b̂ = (1/2)î + (√3/2)ĵ Step 2: Find bisector direction Bisector direction: â + b̂ = (√3/2 + 1/2)î + (1/2 + √3/2)ĵ = ((√3+1)/2)(î + ĵ) Unit bisector: û = (1/√2)î + (1/√2)ĵ Step 3: Apply distance formula OC = βî + (1-β)ĵ Distance from C to bisector = |OC × û|/(|û|) = |OC · n̂| where n̂ ⊥ û Perpendicular direction: n̂ = (1/√2)î - (1/√2)ĵ Distance = |OC · n̂| = |β(1/√2) + (1-β)(-1/√2)| = |(2β-1)/√2| Step 4: Solve for β |(2β-1)/√2| = 3/√2 |2β-1| = 3 2β-1 = 3 → β = 2 2β-1 = -3 → β = -1 ∴ Sum of all possible values = 2 + (-1) = 1
Correct Answer: 1