SATHEE: 3D Geometry 3 Question 63

3D Geometry 3 Question 63

####63. Find the equations of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{\sqrt{6}}$ from the point $(2, 1, - 1)$ .

$(2005, 2 M)$

Show Answer

Answer:

Correct Answer: 63. $2 x - y + z - 3 = 0; 62 x + 29 y + 19 z - 105 = 0$

Solution:

Equation of plane containing the lines

$\begin{aligned} 2 x - y + z - 3 = 0 and 3 x + y + z = 5 is \\ (2 x - y + z - 3) + λ (3 x + y + z - 5) = 0 \\ \Rightarrow (2 + 3 λ) x + (λ - 1) y + (λ + 1) z - 3 - 5 λ = 0 \end{aligned}$

Since, distance of plane from $(2, 1, - 1)$ to above plane is $1 / \sqrt{6}$ .

$∴ \frac{6 λ + 4 + λ - 1 - λ - 1 - 3 - 5 λ}{\sqrt{(3 λ + 2)^{2} + (λ - 1)^{2} + (λ + 1)^{2}}} = \frac{1}{\sqrt{6}}$

$\begin{aligned} \Rightarrow & 6 (λ - 1)^{2} & = 11 λ^{2} + 12 λ + 6 \\ \Rightarrow & λ & = 0, - \frac{24}{5} \end{aligned}$

$∴$ Equations of planes are

$2 x - y + z - 3 = 0 and 62 x + 29 y + 19 z - 105 = 0$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

3D Geometry 3 Question 63

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu