Vector Algebra
Expression of a radius vector as a linear combination of two others
nta_pyq_2025_apr
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 $AC$ such that $\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac{1}{5}$, and $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$, then $\alpha+\sqrt{2}(\sqrt{3}-1)\beta$ is equal to:
$2\sqrt{3}$
$2-\sqrt{3}$
$5\sqrt{3}$
$2+\sqrt{3}$
Step-by-Step Solution
Key Concept: Arc $AC$ subtends $90°$, divided in ratio $1:5$, so arc $AB=15°$ and arc $BC=75°$; the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is $15°$, and between $\overrightarrow{OB}$ and $\overrightarrow{OC}$ is $75°$. Dot the relation $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$ with $\overrightarrow{OA}$ and $\overrightarrow{OB}$ to get a $2\times2$ system.
Unit vectors: $|\overrightarrow{OA}|=|\overrightarrow{OB}|=|\overrightarrow{OC}|=r$ (radius). Angles: $\overrightarrow{OA}\cdot\overrightarrow{OB}=\cos15°$, $\overrightarrow{OA}\cdot\overrightarrow{OC}=\cos90°=0$, $\overrightarrow{OB}\cdot\overrightarrow{OC}=\cos75°$.
Dot $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$ with $\overrightarrow{OA}$: $0=\alpha+\beta\cos15°$ ... (1)
Dot with $\overrightarrow{OB}$: $\cos75°=\alpha\cos15°+\beta$ ... (2)
From (1): $\alpha=-\beta\cos15°$. Sub in (2): $\cos75°=-\beta\cos^215°+\beta=\beta\sin^215°$.
$\beta=\dfrac{\cos75°}{\sin^215°}=\dfrac{2\sqrt{2}}{\sqrt{3}-1}$, $\alpha=\dfrac{-(\sqrt{3}+1)}{\sqrt{3}-1}$.
$\alpha+\sqrt{2}(\sqrt{3}-1)\beta=\dfrac{-(\sqrt{3}+1)}{\sqrt{3}-1}\cdot\dfrac{\sqrt{3}-1}{1}+\sqrt{2}(\sqrt{3}-1)\cdot\dfrac{2\sqrt{2}}{\sqrt{3}-1}$
$=-(\sqrt{3}+1)+4=\dfrac{-3-1-2\sqrt{3}+8}{2}\ldots$
Simplifying: $=-\dfrac{(\sqrt{3}+1)^2}{\sqrt{3}-1}\cdot\dfrac{\sqrt{3}-1}{2}+4=\dfrac{-(4+2\sqrt{3})}{2}+4=2-\sqrt{3}$.
Correct Answer: 2