Question ID: #474
Let A, B, C be three points in the xy-plane, whose position vectors 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:
- (1) 1
- (2) $\frac{9}{2}$
- (3) 0
- (4) 2
Solution:
Given position vectors:
$\vec{OA} = \sqrt{3}\hat{i} + \hat{j}$
$\vec{OB} = \hat{i} + \sqrt{3}\hat{j}$
$\vec{OC} = a\hat{i} + (1-a)\hat{j}$
First, we find the slopes of lines OA and OB.
Slope of OA ($m_1$) = $\frac{1}{\sqrt{3}} \Rightarrow \theta_1 = 30^\circ$
Slope of OB ($m_2$) = $\frac{\sqrt{3}}{1} \Rightarrow \theta_2 = 60^\circ$
The angle bisector of OA and OB will make an angle equal to the average of $30^\circ$ and $60^\circ$:
$\theta_{bisector} = \frac{30^\circ + 60^\circ}{2} = 45^\circ$
The slope of the angle bisector is $\tan(45^\circ) = 1$.
The equation of the angle bisector passing through the origin is:
$y = x \Rightarrow x – y = 0$
We are given that the perpendicular distance of point C $(a, 1-a)$ from the line $x – y = 0$ is $\frac{9}{\sqrt{2}}$.
Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:
$$\frac{|a – (1-a)|}{\sqrt{1^2 + (-1)^2}} = \frac{9}{\sqrt{2}}$$
$$\frac{|2a – 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}}$$
$$|2a – 1| = 9$$
$$2a – 1 = 9 \quad \text{or} \quad 2a – 1 = -9$$
$$2a = 10 \Rightarrow a = 5$$
$$2a = -8 \Rightarrow a = -4$$
The sum of all possible values of $a$ is:
$$5 + (-4) = 1$$
Ans. (1)
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