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>
\(-\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\)
\(\dfrac{\vec{a}+\vec{b}}{\sqrt{2}}\)
\(\vec{a}-\vec{b}\)
\(-\dfrac{\vec{a}-\vec{b}}{\sqrt{2}}\)
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