Let √3i + j, i + √3j and βi + (1-β)j respectively be the position vectors of the points A, B and C w

Q: Why is the bisector $y=x$?

The angle of $OA$ is $30^\circ$ and $OB$ is $60^\circ$. The bisector is at $(30^\circ + 60^\circ)/2 = 45^\circ$. A line at $45^\circ$ passing through the origin is $y = \tan(45^\circ)x$, which simplifies to $y=x$.

Q: Why does $C$ lie on $x+y=1$?

The position vector of $C$ is $\beta\hat{i} + (1-\beta)\hat{j}$. This means $x = \beta$ and $y = 1-\beta$. Adding them gives $x+y = \beta + 1 - \beta = 1$.

Q: Why are there two values for $\beta$?

The distance formula involves a modulus. $|2\beta - 1| = k$ means $2\beta - 1 = k$ or $2\beta - 1 = -k$. Geometrically, this means there are two points on the line $x+y=1$ that are at a distance of $3/2$ from the line $x-y=0$, one on each side.

Question

Let $\sqrt{3}\hat{i} + \hat{j}$, $\hat{i} + \sqrt{3}\hat{j}$ and $\beta\hat{i} + (1-\beta)\hat{j}$ respectively be the position vectors of the points $A$, $B$ and $C$ with respect to the origin $O$. If the distance of $C$ from the bisector of the acute angle between $OA$ and $OB$ is $\frac{3}{2}$, then the sum of all possible values of $\beta$ is:

Options:

(A) 3

(B) 4

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Angular Landscape

We begin by identifying the positions of points $A(\sqrt{3}, 1)$ and $B(1, \sqrt{3})$. The slope of $OA$ is $\frac{1}{\sqrt{3}}$, corresponding to an angle of $30^\circ$ with the $x$-axis. Similarly, the slope of $OB$ is $\sqrt{3}$, corresponding to an angle of $60^\circ$.

The angle between vectors $OA$ and $OB$ is $60^\circ - 30^\circ = 30^\circ$. The angle bisector of these two vectors must lie exactly in the middle, at an angle of $45^\circ$ from the $x$-axis.

A line passing through the origin at an angle of $45^\circ$ is defined by the equation $y = x$, or:
$$x - y = 0$$

### The Mystery of Point C

Point $C$ is defined by the position vector $\beta\hat{i} + (1-\beta)\hat{j}$, which gives it the coordinates $(\beta, 1-\beta)$. Observing these coordinates, we see that the sum of the $x$ and $y$ components is constant:
$$\beta + (1 - \beta) = 1$$

This implies that point $C$ is constrained to lie on the line $x + y = 1$. We are given that the perpendicular distance of point $C$ from the bisector $x - y = 0$ is $\frac{3}{2}$.

### The Distance Calculation

The distance $d$ of a point $(x_1, y_1)$ from a line $ax + by + c = 0$ is given by:
$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

For our line $x - y = 0$, we have $a=1$, $b=-1$, and $c=0$. Substituting the coordinates of point $C(\beta, 1-\beta)$ into the formula, we obtain:
$$d = \frac{|\beta - (1 - \beta)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2\beta - 1|}{\sqrt{2}}$$

Setting this distance equal to the given value of $\frac{3}{2}$, we arrive at the equation:
$$\frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{2}$$

### Solving for $\beta$

We simplify the modulus equation as follows:
$$|2\beta - 1| = \frac{3\sqrt{2}}{2} = \frac{3}{\sqrt{2}}$$

This yields two possible cases for $\beta$:
$$2\beta - 1 = \frac{3}{\sqrt{2}} \quad 	ext{or} \quad 2\beta - 1 = -\frac{3}{\sqrt{2}}$$

Solving for $\beta$ in each case:
$$\beta_1 = \frac{1}{2} + \frac{3}{2\sqrt{2}}, \quad \beta_2 = \frac{1}{2} - \frac{3}{2\sqrt{2}}$$

To find the sum of all possible values of $\beta$, we add $\beta_1$ and $\beta_2$:
$$\beta_1 + \beta_2 = \left( \frac{1}{2} + \frac{3}{2\sqrt{2}} ight) + \left( \frac{1}{2} - \frac{3}{2\sqrt{2}} ight) = 1$$

The sum of all possible values of $\beta$ is **1**.

Select Answer:

Visualized Solution

Position Vectors of $A$ and $B$

Angles of $O A$ and $O B$

The Angle Bisector

Equation of the Bisector

Position Vector of $C$

Distance from Point to Line

Applying the Distance Formula

Simplifying the Equation

Solving the Modulus

Finding the Values of $β$

Sum of All Possible Values

The Sigma Insight: Components of a Vector

Analyzing the Angular Landscape

The Mystery of Point C

The Distance Calculation

Solving for $β$

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