Vector Algebra
Vector Algebra
nta_pyq_2025_jan
Grade 12
Question:
Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC , divides the arc AC - -\to - -\to - -\to length of arc AB such that length of arc BC = 1 5 , and OC = \alphaOA + \betaOB, then \alpha + \sqrt2(\sqrt3 - 1)\beta is equal to
2\sqrt3
2 - \sqrt3
5\sqrt3
2 + \sqrt3 ^ \to \to = ^i + 2^j + 3k,
Step-by-Step Solution
Key Concept: Apply the core result for dot product, cross product and projections and simplify using the given constraints.
(2) \to \to \to c = \alpha a + \beta b \ldots . . (1) \to \to \to \to \to \to a ⋅ c = \alpha a ⋅ a + \beta b ⋅ a ∘ 0 = \alpha + \beta cos 15 \ldots . (2) \to \to \to \to \to \to (1) \Rightarrow b ⋅ c = \alpha a ⋅ b + \beta b ⋅ b ∘ ∘ \Rightarrow cos 75 = \alpha cos 15 + \beta \ldots . (3) ∘ 2 ∘ (2)&(3) \Rightarrow cos 75 = -\beta cos 15 + \beta ∘ cos 75 1 2\sqrt2 \beta = = ∘ = 2 ∘ sin 15 sin 15 \sqrt3-1 ∘ - cos 15 -(\sqrt3+1) (2) \Rightarrow \alpha = ∘ = sin 15 (\sqrt3-1) \to \to -(\sqrt3+1) \to 2\sqrt2 \therefore c = a + ( ) b (\sqrt3-1) \sqrt3-1 Now -(\sqrt3 + 1) \sqrt2(\sqrt3 - 1) ⋅ 2\sqrt2 \alpha + \sqrt2(\sqrt3 - 1)\beta = + (\sqrt3 - 1) \sqrt3 - 1 2 -(\sqrt3 + 1) = + 4 2 -3 - 1 - 2\sqrt3 + 8 = 2 = 2 - \sqrt3 \to \to
Correct Answer: 2