Let A, B, C be three points in xy-plane, whose position vector are given by √3i+j, i+√3j and ai+(1-a

Q: Why can we just add the vectors to get the bisector?

Because the vectors $\vec{OA}$ and $\vec{OB}$ have equal magnitude, the parallelogram formed by them is a rhombus. In a rhombus, the diagonal bisects the angle between the sides. Thus, $\vec{OA} + \vec{OB}$ is the direction of the bisector.

Q: What if the magnitudes were different?

If the magnitudes were different, the sum vector would not bisect the angle. You would need to normalize the vectors first: $\hat{OA} + \hat{OB}$ would give you the bisector direction.

Q: Why do we get two values for $a$?

The distance condition $d = \frac{9}{\sqrt{2}}$ defines two lines parallel to the bisector, one on either side. Point $C$ can lie on either of these lines, leading to two possible values for $a$.

Question

Let A, B, C be three points in xy-plane, whose position vector are given by $\sqrt{3}\hat{i}+\hat{j}$, $\hat{i}+\sqrt{3}\hat{j}$ and $a\hat{i}+(1-a)\hat{j}$ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors $\vec{OA}$ and $\vec{OB}$ is $\frac{9}{\sqrt{2}}$ then the sum of all the possible values of a is :

Options:

(A) 2

(B) $9/2$

(C) 1

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the $xy$-plane, looking at two vectors, $\vec{OA}$ and $\vec{OB}$. The points $A(\sqrt{3}, 1)$ and $B(1, \sqrt{3})$ are not randomly placed.

If you calculate their magnitudes, you find that:

$$|\vec{OA}| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$$

$$|\vec{OB}| = \sqrt{1^2 + (\sqrt{3})^2} = 2$$

Because the magnitudes are identical, the parallelogram formed by these vectors is a rhombus. In a rhombus, the diagonal—which is the vector sum $\vec{OA} + \vec{OB}$—perfectly bisects the angle between the sides. This geometric intuition simplifies our entire path.

### Finding the Path: The Bisector Line

Now that we know the bisector lies along the sum of the vectors, let us calculate it. Adding $\vec{OA} = \sqrt{3}\hat{i} + \hat{j}$ and $\vec{OB} = \hat{i} + \sqrt{3}\hat{j}$, we get:

$$\vec{V} = (\sqrt{3}+1)\hat{i} + (\sqrt{3}+1)\hat{j}$$

The direction is clearly proportional to $\hat{i} + \hat{j}$. Since the line passes through the origin $(0,0)$ and has a direction vector $(1, 1)$, its slope is $m = 1$.

The equation of our bisector line is $y = x$, or in standard form:

$$x - y = 0$$

This line is the central axis of our problem, the reference point from which we will measure the distance to point $C$.

### The Distance Challenge: Bridging Geometry and Algebra

We are introduced to point $C(a, 1-a)$. We need to find the perpendicular distance from this point to our line $x - y = 0$.

We reach for our trusty perpendicular distance formula:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substituting $x_1 = a$, $y_1 = 1-a$, $A = 1$, $B = -1$, and $C = 0$, we get:

$$d = \frac{|a - (1-a)|}{\sqrt{1^2 + (-1)^2}}$$

Simplifying the numerator, we get $|a - 1 + a| = |2a - 1|$. The denominator is $\sqrt{2}$. So, the distance is:

$$d = \frac{|2a - 1|}{\sqrt{2}}$$

### The Final Resolution: Solving for $a$

The problem tells us that this distance is $\frac{9}{\sqrt{2}}$. Setting our expression equal to this value, we have:

$$\frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}}$$

The $\sqrt{2}$ terms cancel out beautifully, leaving us with the absolute value equation:

$$|2a - 1| = 9$$

This splits into two cases:

1. $2a - 1 = 9 \Rightarrow 2a = 10 \Rightarrow a = 5$

2. $2a - 1 = -9 \Rightarrow 2a = -8 \Rightarrow a = -4$

The sum of these values is $5 + (-4) = 1$. We have navigated the geometry, applied the algebra, and arrived at the **final answer of 1**.

Select Answer:

Visualized Solution

Visualize Vectors $O A$ and $O B$

Calculate Magnitudes

Find the Bisector Vector

Equation of the Bisector Line

Point $C$ and Perpendicular Distance

The Distance Formula

Substitute Coordinates of $C$

Simplify the Distance Expression

Equate to Given Distance

Solve Absolute Value: Case $1$

Solve Absolute Value: Case $2$

Final Sum of Values

The Sigma Insight: Addition of Vectors

Analyzing the Setup

Finding the Path

The Distance Challenge

The Final Resolution

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