SATHEE: 3D Geometry 2 Question 3

3D Geometry 2 Question 3

####3. Let $\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}$ and $β \hat{i} + (1 - β) \hat{j}$ 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 $O A$ and $O B$ is $\frac{3}{\sqrt{2}}$ , then the sum of all possible values of $β$ is

(2019 Main, 11 Jan II)

(a) 1

(b) 3

(d) 2

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Answer:

Correct Answer: 3. (a)

Solution:

According to given information, we have the following figure.

Clearly, angle bisector divides the sides $A B$ in $O A : O B$ , i.e., $2 : 2 = 1 : 1$ [using angle bisector theorem] So, $D$ is the mid-point of $A B$ and hence coordinates of $D$ are $[\frac{\sqrt{3} + 1}{2}, \frac{\sqrt{3} + 1}{2}]$

Now, equation of bisector $O D$ is

$\begin{aligned} (y - 0) = \frac{\frac{\sqrt{3} + 1}{2} - 0}{\frac{\sqrt{3} + 1}{2} - 0} (x - 0) \Rightarrow y = x \\ \Rightarrow x - y & = 0 \end{aligned}$

According to the problem,

$\frac{3}{\sqrt{2}} = C M = | \frac{β - (1 - β)}{\sqrt{2}} |$

[Distance of a point $P (x_{1}, y_{1})$ from the line

$\begin{aligned} a x + b y + c = 0 is | \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} | \\ \Rightarrow | 2 β - 1 | = 3 \Rightarrow 2 β = \pm 3 + 1 \\ \Rightarrow 2 β = 4, - 2 \Rightarrow β = 2, - 1 \end{aligned}$

Sum of 2 and -1 is 1 .

3D Geometry 2 Question 3

Answer:

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3D Geometry 2 Question 3

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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