Q. 295.0( 1 Vote )

Find the equation

Answer :

Given: Planes




The equation of the planes passing through the line of intersection of these planes is given by;


2x − 3y + 4z − 1 + λ(x − y + 4) = 0


(2 + λ)x + (−3 − λ)y + (4)z − 1 + 4λ = 0


((2 + λ), (−3 − λ), (4)) ……(i)


Perpendicular plane is


In Cartesian form (2, −1, 1) ……(ii)


According to the given condition product of (i) and (ii) should be zero.


2(2 + λ) − (−3 − λ) + 4 = 0



The required equation is


The given line x − 1 = 2y − 4 = 3z − 12



(1, 2, 4) is a point on the line




the plane contains the line x − 1 = 2y − 4 = 3z − 12


OR


Given: The point (1, 2, −4) and lines



The line perpendicular to these lines will be parallel to the line obtained by the cross product of these lines.





The vector equation of the required line is



The Cartesian equation of the required line is



Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Find the equationMathematics - Board Papers

Find the vector aMathematics - Board Papers

Write the vector Mathematics - Board Papers

Find the value ofMathematics - Board Papers

<span lang="EN-USMathematics - Board Papers