An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If O

Q: Wait, why do we need to take the dot product with $\vec{u}$ and $\vec{v}$? Can't we just use components?

You absolutely could use components! However, vector algebra is often more elegant. By taking the dot product with the basis vectors $\vec{u}$ and $\vec{v}$, we are essentially projecting the equation onto those directions. It is a powerful, general technique that works even if you don't want to set up a coordinate system.

Q: Is it okay that the radius $r$ cancels out in the equations?

Yes, it is! The problem is scale-invariant. The relationship between the vectors $\vec{u}$, $\vec{v}$, and $\vec{OQ}$ depends only on the angles between them, not on the size of the circle. The fact that $r^2$ cancels out is a beautiful confirmation that our geometric intuition is correct.

Q: Why do we need $\beta^2$ instead of just $\beta$ for the quadratic equation?

That is a classic trap! The question specifically asks for the roots to be $\alpha$ and $\beta^2$. Always read the question carefully to see exactly what the roots are. In this case, $\beta$ is $\sqrt{2}$, so $\beta^2$ is $2$. Don't let a simple oversight ruin your hard work!

Question

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The mid point of the arc $PQ$ is $R$. If $\vec{OP} = \vec{u}$, $\vec{OR} = \vec{v}$ and $\vec{OQ} = \alpha\vec{u} + \beta\vec{v}$, then $\alpha, \beta^2$, are the roots of the equation

Options:

(A) $x^2 + x - 2 = 0$

(B) $x^2 - x - 2 = 0$

(C) $3x^2 - 2x - 1 = 0$

(D) $3x^2 + 2x - 1 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the center $O$ of a circle. You look out at two points, $P$ and $Q$, on the circumference. The arc $PQ$ subtends a right angle at your position.

We have vectors $\vec{OP} = \vec{u}$ and $\vec{OQ}$, which are perpendicular because the angle between them is $90^\circ$. Now, consider the midpoint $R$ of the arc $PQ$.

The vector $\vec{OR} = \vec{v}$ acts as a bisector, splitting that $90^\circ$ angle into two perfect $45^\circ$ slices. This geometric setup is the foundation of our analysis.

### The Power of the Dot Product

Since $P$, $Q$, and $R$ all lie on the same circle, the magnitudes of vectors $\vec{u}$, $\vec{v}$, and $\vec{OQ}$ are all equal to the radius $r$. We know that $\vec{u} \cdot \vec{OQ} = 0$ because they are perpendicular.

For the other pairs, we use the definition $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos 	heta$. Thus, we calculate the following dot products:

$$\vec{u} \cdot \vec{v} = r^2 \cos 45^\circ = \frac{r^2}{\sqrt{2}}$$

$$\vec{v} \cdot \vec{OQ} = r^2 \cos 45^\circ = \frac{r^2}{\sqrt{2}}$$

These dot products are the keys that will unlock the values of $\alpha$ and $\beta$.

### The Algebraic Dance

We are given the linear combination $\vec{OQ} = \alpha\vec{u} + \beta\vec{v}$. To find $\alpha$ and $\beta$, we perform a two-step algebraic process.

First, we take the dot product of the entire equation with $\vec{u}$:

$$\vec{u} \cdot \vec{OQ} = \alpha(\vec{u} \cdot \vec{u}) + \beta(\vec{u} \cdot \vec{v})$$

Substituting our known values, we get $0 = \alpha r^2 + \beta \frac{r^2}{\sqrt{2}}$. Dividing by $r^2$, we find the elegant relationship:

$$\alpha = -\frac{\beta}{\sqrt{2}}$$

Next, we take the dot product with $\vec{v}$, yielding:

$$\vec{v} \cdot \vec{OQ} = \alpha(\vec{v} \cdot \vec{u}) + \beta(\vec{v} \cdot \vec{v})$$

This simplifies to:

$$\frac{r^2}{\sqrt{2}} = \alpha \frac{r^2}{\sqrt{2}} + \beta r^2 \Rightarrow \frac{1}{\sqrt{2}} = \frac{\alpha}{\sqrt{2}} + \beta$$

By substituting our expression for $\alpha$ into this second equation, we solve for $\beta$:

$$\frac{1}{\sqrt{2}} = \frac{-\beta/\sqrt{2}}{\sqrt{2}} + \beta \Rightarrow \frac{1}{\sqrt{2}} = -\frac{\beta}{2} + \beta \Rightarrow \frac{1}{\sqrt{2}} = \frac{\beta}{2}$$

This yields $\beta = \sqrt{2}$. Consequently, $\alpha = -1$.

### The Final Construction

We have our roots: $\alpha = -1$ and $\beta^2 = (\sqrt{2})^2 = 2$. The problem asks for a quadratic equation with these roots.

Using the sum and product of roots, the sum is $-1 + 2 = 1$, and the product is $(-1)(2) = -2$. The standard form $x^2 - (	ext{Sum})x + (	ext{Product}) = 0$ leads us directly to:

**$x^2 - x - 2 = 0$**

This is the final result of our journey. You have successfully navigated the geometry, the vector algebra, and the final construction.

Animated Solution for Mathematics - Vector Algebra: An arc $P Q$ of a circle subtends a right angle at its centre $O$ . The mid point of the arc $P Q$ is $R$ . If $O P = u$ , $O R = v$ and $O Q = α u + β v$ , then $α, β^{2}$ , are the roots of the equation

Select Answer:

Visualized Solution

Visualizing the Arc $P Q$

Defining Magnitudes and Dot Products

Using the Vector Equation $O Q = α u + β v$

Finding Relation Between $α$ and $β$

Dot Product with $v$

Substituting Values into Second Equation

Solving for $β$

Finding the Roots $α$ and $β^{2}$

Constructing the Final Equation

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Power of the Dot Product

The Algebraic Dance

The Final Construction

Similar Questions

Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that $\frac{length of arc A B}{length of arc B C} = \frac{1}{5}$ and $O C = α O A + β O B,$ then $α + 2 (3 - 1) β$ is equal to

Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively, with respect to $O$ and $∣ p ∣ = p, ∣ q ∣ = q$ . The points $R$ and $S$ divide $P Q$ internally and externally in the ratio $2 : 3$ respectively. If $O R$ and $O S$ are perpendicular then

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that $O P \cdot O Q + O R \cdot O S = O R \cdot O P + O Q \cdot O S = O Q \cdot O R + O P \cdot O S$ Then the triangle PQR has S as its

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Let $a$ and $b$ be the vectors of the same magnitude such that $\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1$ . Then $\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}}$ is :

Let $a, b$ and $c$ be three unit vectors such that $∣ a - b ∣^{2} + ∣ a - c ∣^{2} = 8$ . Then $∣ a + 2 b ∣^{2} + ∣ a + 2 c ∣^{2}$ is equal to

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

Select Answer:

Visualized Solution

Visualizing the Arc .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Defining Magnitudes and Dot Products

Using the Vector Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ​=αu+βv

Finding Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v