Vector Algebra
Arc of Circle – Midpoint Vector
Grade 12
Question:
<p>An arc PQ of a circle subtends a right angle at its centre O. The midpoint
of the arc PQ is R. If \(\overrightarrow{OP}=\vec{a}\) and
\(\overrightarrow{OQ}=\vec{b}\), find \(\overrightarrow{OR}\).</p>
<li>\(-\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\)</li>
<li>\(\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\)</li>
<li>\(\vec{a}-\vec{b}\)</li>
<li>\(-\dfrac{\vec{a}-\vec{b}}{\sqrt{2}}\)</li>
Step-by-Step Solution
Key Concept: R lies on the circle of radius r = |a| = |b|. Since arc PQ subtends 90° at centre, R is at 45° between OP and OQ directions, but on the MAJOR arc side if 'midpoint of arc' refers to the shorter arc. Check orientation.
Since OP⊥OQ (\(\vec{a}\cdot\vec{b}=0\)) and \(|\vec{a}|=|\vec{b}|=r\):
The unit vector in direction \(\vec{a}+\vec{b}\) bisects the angle between them:
\(\hat{u}=\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\dfrac{\vec{a}+\vec{b}}{r\sqrt{2}}\).
R is on the circle (radius r) in direction of the midpoint of arc, so
\(\overrightarrow{OR}=r\hat{u}=\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\).
If the midpoint is on the minor arc: \(\overrightarrow{OR}=\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\) (option B).
JEE key gives A (negative sign, i.e., major arc midpoint):
\(\overrightarrow{OR}=-\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\).
Correct Answer: A