PQ is the diameter of semi circle with centre O and M is a point on the circumference of the semi circle. Prove by vector method that: PMQ = 90^

Solution:

We are given a semicircle with diameter $PQ$ and centre $O$. $M$ is a point on the circumference of the semicircle. We need to prove that $\angle PMQ=90^{\circ }$ using the vector method.

Step 1: Define Position Vectors

Let the position vectors of points $P$, $Q$, and $M$ be $\vec{P}$, $\vec{Q}$, and $\vec{M}$ respectively. Since $O$ is the centre, we can write: $$\vec{O}=\frac{\vec{P}+\vec{Q}}{2}$$ This is because $O$ is the midpoint of $PQ$.

Step 2: Express $\overrightarrow{PM}$ and $\overrightarrow{QM}$

The vectors $\overrightarrow{PM}$ and $\overrightarrow{QM}$ can be written as: $$\overrightarrow{PM}=\vec{M}-\vec{P}$$ $$\overrightarrow{QM}=\vec{M}-\vec{Q}$$

Step 3: Dot Product of $\overrightarrow{PM}$ and $\overrightarrow{QM}$

To prove that $\angle PMQ=90^{\circ }$, we need to show that the dot product of $\overrightarrow{PM}$ and $\overrightarrow{QM}$ is zero: $$\overrightarrow{PM}\cdot \overrightarrow{QM}=(\vec{M}-\vec{P})\cdot (\vec{M}-\vec{Q})$$ Expanding the dot product: $$\overrightarrow{PM}\cdot \overrightarrow{QM}=\vec{M}\cdot \vec{M}-\vec{M}\cdot \vec{Q}-\vec{P}\cdot \vec{M}+\vec{P}\cdot \vec{Q}$$

Step 4: Use the Property of the Semicircle

Since $M$ lies on the circumference of the semicircle with diameter $PQ$, by the property of a semicircle, the angle subtended by the diameter at any point on the circumference is $90^{\circ }$. However, we will prove this using vectors.

Since $O$ is the centre, the vector $\overrightarrow{OM}$ is perpendicular to $\overrightarrow{PQ}$ (as $M$ lies on the circumference). Thus, we can write: $$|\overrightarrow{OM}{|}^2+|\overrightarrow{OP}{|}^2=|\overrightarrow{PM}{|}^2$$ But more importantly, since $M$ lies on the semicircle, the following relation holds: $$|\overrightarrow{PM}{|}^2=|\overrightarrow{OM}{|}^2+|\overrightarrow{OP}{|}^2$$ But we can also use the fact that: $$|\overrightarrow{PM}{|}^2=|\vec{M}-\vec{P}{|}^2=|\vec{M}{|}^2+|\vec{P}{|}^2-2\vec{M}\cdot \vec{P}$$ Similarly, $$|\overrightarrow{QM}{|}^2=|\vec{M}-\vec{Q}{|}^2=|\vec{M}{|}^2+|\vec{Q}{|}^2-2\vec{M}\cdot \vec{Q}$$

However, a more straightforward approach is to use the fact that $|\overrightarrow{OP}|=|\overrightarrow{OQ}|=r$ (radius of the semicircle) and $|\overrightarrow{OM}|=r$ (since $M$ is on the circumference).

Step 5: Simplify the Dot Product

We know that: $$\vec{P}\cdot \vec{Q}=|\vec{P}||\vec{Q}|\cos (180^{\circ })=-r^2$$ since $P$ and $Q$ are endpoints of the diameter and the angle between $\vec{P}$ and $\vec{Q}$ is $180^{\circ }$.

Now, since $M$ lies on the circumference, $|\vec{M}{|}^2=|\vec{P}{|}^2=|\vec{Q}{|}^2=r^2$ (because $O$ is the origin and $|\overrightarrow{OM}|=r$).

Substituting these into the dot product: $$\overrightarrow{PM}\cdot \overrightarrow{QM}=r^2-\vec{M}\cdot \vec{Q}-\vec{P}\cdot \vec{M}-r^2$$ $$=-(\vec{M}\cdot \vec{Q}+\vec{P}\cdot \vec{M})$$

But we also know that: $$\vec{M}\cdot \vec{Q}+\vec{P}\cdot \vec{M}=\vec{M}\cdot (\vec{P}+\vec{Q})$$ Since $O$ is the midpoint, $\vec{P}+\vec{Q}=2\vec{O}$. If we assume $O$ is the origin, then $\vec{P}+\vec{Q}=\vec{0}$, which implies: $$\vec{M}\cdot (\vec{P}+\vec{Q})=\vec{M}\cdot \vec{0}=0$$

Thus: $$\overrightarrow{PM}\cdot \overrightarrow{QM}=0$$

Conclusion:

Since the dot product of $\overrightarrow{PM}$ and $\overrightarrow{QM}$ is zero, the vectors are perpendicular. Therefore, $\angle PMQ=90^{\circ }$.