SATHEE: 3D Geometry 3 Question 51

3D Geometry 3 Question 51

####51. In $R^{3}$ , consider the planes $P_{1} : y = 0$ and $P_{2} : x + z = 1$ . Let $P_{3}$ be a plane, different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point $(0, 1, 0)$ from $P_{3}$ is 1 and the distance of a point $(α, β, γ)$ from $P_{3}$ is 2 , then which of the following relation(s) is/are true?

(2015 Adv.)

(a) $2 α + β + 2 γ + 2 = 0$

(b) $2 α - β + 2 γ + 4 = 0$

(d) $2 α - β + 2 γ - 8 = 0$

Show Answer

Answer:

Correct Answer: 51. (b, d)

Solution:

Here, $P_{3} : (x + z - 1) + λ y = 0$

i.e. $P_{3} : x + λ y + z - 1 = 0$

whose distance from $(0, 1, 0)$ is 1 .

$\begin{aligned} ∴ \frac{| 0 + λ + 0 - 1 |}{\sqrt{1 + λ^{2} + 1}} = 1 \\ \Rightarrow | λ - 1 | = \sqrt{λ^{2} + 2} \\ \Rightarrow λ^{2} - 2 λ + 1 = λ^{2} + 2 \Rightarrow λ = - \frac{1}{2} \end{aligned}$

$∴$ Equation of $P_{3}$ is $2 x - y + 2 z - 2 = 0$ .

$∵$ Distance from $(α, β, γ)$ is 2 .

$\begin{array}{lc} ∴ & \frac{| 2 α - β + 2 γ - 2 |}{\sqrt{4 + 1 + 4}} = 2 \\ \Rightarrow & 2 α - β + 2 γ - 2 = \pm 6 \\ \Rightarrow & 2 α - β + 2 γ = 8 and 2 α - β + 2 γ = - 4 \end{array}$

3D Geometry 3 Question 51

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

3D Geometry 3 Question 51

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu