Answer :

Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:


2x – 3y + 5z + θ = 0


According to question,


The plane passes through (3, 4, –1)


2(3) – 3(4) +5(–1) + θ = 0


θ = 11


So, the equation of the plane is as follows:


2x – 3y + 5z + 11 = 0


Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):


We know, the distance of point (x1,y1,z1) from the plane


is given by:



Putting the necessary values,




the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1) is


units


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 equationMathematics - Board Papers

Find the length oMathematics - Board Papers

Find the coordinaMathematics - Board Papers

Find the equationMathematics - Board Papers

Show that the linMathematics - Board Papers