Q. 5

# Find the equation of the plane passing through the point (1, 4, - 2) and parallel to the plane 2x – y + 3z + 7 = 0.

Formula : Plane = r . (n) = d

Where r = any random point

n = normal vector of plane

d = distance of plane from origin

If two planes are parallel , then their normal vectors are same.

Therefore ,

Parallel Plane 2x – y + 3z + 7 = 0

Normal vector = (2i - j + 3k)

Normal vector of required plane = (2i - j + 3k)

Equation of required plane r . (2i - j + 3k) = d

In cartesian form 2x - y + 3z = d

Plane passes through point (1,4, - 2) therefore it will satisfy it.

2(1) - (4) + 3( - 2) = d

d = 2 – 4 – 6 = - 8

Equation of required plane 2x - y + 3z = - 8

2x - y + 3z + 8 = 0

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
The Embryonic Development62 mins
NCERT Exemplar Special | The Living World Part 225 mins
Nernst Equation - Learn the Concept33 mins
Understanding the Combination of cells54 mins
Revising the basics of Organic Chemistry35 mins
Types of solution on the basis of Raoult's Law44 mins
Ray Optics | Getting the basics46 mins
Nucleophilic Substitution Reaction | Getting the basics39 mins
Understant the concept of Development of pollen grains55 mins
Getting into the world of coliisionsFREE Class
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
view all courses