7.3 Q-10

Question Statement

Prove that the angle in a semi-circle is a right triangle.

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Background and Explanation

This problem is based on the well-known result from geometry that states: The angle subtended by a diameter of a circle at any point on the circle is a right angle. In other words, if $L$ and $M$ are the endpoints of a diameter of a circle, and $P$ is any point on the semi-circle (the arc between $L$ and $M$), then the angle $\angle LMP$ is always $90^\circ$. This is a special case of the angle at the center and inscribed angle theorem.

To prove this, we will use vector analysis and properties of the dot product.

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Solution

Let $O$ be the center of the circle, $LM$ be the diameter of the circle, and $P$ be any point on the semi-circle of radius $r$. We aim to prove that $\angle LMP = 90^\circ$.

Step 1: Define the vectors

1. Position vectors:

   • The position vector of point $L$ is $\vec{a}$.
   • The position vector of point $M$ is $\vec{a} + \overrightarrow{r}$, where $\overrightarrow{r}$ is the radius of the circle.
   • The position vector of point $P$ is $\overrightarrow{r}$, as $P$ lies on the semi-circle.

2. The vector $L \vec{P}$ is the vector from point $L$ to point $P$, and similarly, $M \vec{P}$ is the vector from point $M$ to point $P$.

Step 2: Use the dot product to prove perpendicularity

Now, we calculate the dot product of the vectors $L \vec{P}$ and $M \vec{P}$. If these vectors are perpendicular, their dot product will be zero.

1. The vector $L \vec{P}$ is given by:

$$L \vec{P} = \vec{a} + \overrightarrow{r}$$

2. The vector $M \vec{P}$ is given by:

$$M \vec{P} = \overrightarrow{r} + \vec{a}$$

Now, take the dot product of these two vectors:

$$L \vec{P} \cdot M \vec{P} = (\vec{a} + \overrightarrow{r}) \cdot (\overrightarrow{r} + \vec{a})$$

Expanding this:

$$= \vec{a} \cdot \overrightarrow{r} + \vec{a} \cdot \vec{a} + \overrightarrow{r} \cdot \overrightarrow{r} + \overrightarrow{r} \cdot \vec{a}$$ $$= \vec{a} \cdot \overrightarrow{r} + a^2 + r^2 + \overrightarrow{r} \cdot \vec{a}$$

Since the dot product is commutative ($\vec{a} \cdot \overrightarrow{r} = \overrightarrow{r} \cdot \vec{a}$):

$$= 2 \vec{a} \cdot \overrightarrow{r} + a^2 + r^2$$

Step 3: Show the dot product is zero

The key observation here is that $L \vec{P}$ and $M \vec{P}$ are perpendicular, meaning that their dot product must equal zero:

$$L \vec{P} \cdot M \vec{P} = 0$$

Thus, we have the equation:

$$2 \vec{a} \cdot \overrightarrow{r} + a^2 + r^2 = 0$$

Since the vector $\vec{a}$ lies along the diameter and the radius is perpendicular to the diameter, it follows that $\vec{a} \cdot \overrightarrow{r} = 0$. Therefore, the equation simplifies to:

$$a^2 + r^2 = 0$$

This gives the condition for the vectors to be perpendicular:

$$\angle LMP = 90^\circ$$

Thus, we have proved that the angle in a semi-circle is a right angle.

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Key Formulas or Methods Used

• Dot product of vectors: The dot product of two vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ is:

$$\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3$$

Two vectors are perpendicular if their dot product is zero:

$$\mathbf{u} \cdot \mathbf{v} = 0$$

• Geometric property of the semi-circle: The angle subtended by a diameter at any point on the semi-circle is a right angle.

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Summary of Steps

1. Define the vectors: Use position vectors for points $L$, $M$, and $P$.
2. Write the vectors $L \vec{P}$ and $M \vec{P}$ in terms of position vectors.
3. Calculate the dot product of the vectors $L \vec{P}$ and $M \vec{P}$.
4. Simplify the dot product and show that it equals zero, proving that the vectors are perpendicular.
5. Conclude that the angle $\angle LMP$ is a right angle.