Test Index

Vector Algebra Part 3

Section: Mathematics

© examsnet.com

Question : 40 of 72

Marks: +1, -0

Let the

\overset{\frown}{AC}

of a circle subtend a right angle at the centre

O

B

AC

, divides the arc

AC

\frac{\text{length of }\overset{\frown}{AB}}{\text{length of }\overset{\frown}{BC}}=\frac{1}{5}

\overrightarrow{OC}

=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}

\alpha+\sqrt{2}(\sqrt{3}-1)\beta

[23 Jan 2025 Shift 1]

$5\sqrt{3}$
$2+\sqrt{3}$
$2\sqrt{3}$
$2-\sqrt{3}$

Solution:

Let the radius be

r

B

divides the arc in

1:5

, so

AB

would subtend an angle of

15^{\circ}

at origin

O

and similarly,

BC

would subtend an angle of

75^{\circ}

at origin.

\overrightarrow{OA}=\vec{a},\ \overrightarrow{OB}=\vec{b},\ \overrightarrow{OC}=\vec{c}

\vec{b}=r(\cos75^{\circ})\hat{i}+r(\sin75^{\circ})\hat{j}

\vec{b}=\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\vec{c}+\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)\vec{a}

2\sqrt{2}\vec{b}=(\sqrt{3}-1)\vec{c}+(\sqrt{3}+1)\vec{a}

\overrightarrow{OC}=-\frac{\sqrt{3}+1}{\frac{|\sqrt{3}-1|}{\alpha}}\overrightarrow{OA}+\frac{2\sqrt{2}}{\frac{|\sqrt{3}-1|}{\beta}}\overrightarrow{OB}

\alpha+\sqrt{2}(\sqrt{3}-1)\beta=2-\sqrt{3}

© examsnet.com

Go to Question:

Prev Question

Next Question