148k views
2 votes
Find an equation of the plane that passes through the point (-1,1/2,3) with normal vector [1,4,1]

User TOvidiu
by
8.5k points

1 Answer

4 votes

Answer:

x + 4y + z = 4

Explanation:

The standard equation is of the form, Ax + By + Cz = 0

The equation of the plane that passes through the point(-1, 1/2, 3) and is perpendicular to [1, 4, 1] is

A = 1, B = 4, C = 1

x = -1; y = 1/2; z = 3

A(x-(-1)) + B(y-1/2) + C(z-3) = 0

(x+1) + 4(y-1/2) + (z-3) = 0

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

x + 1 + 4y -2 + z -3 = 0

x + 4y + z - 4 = 0

x + 4y + z = 4

User Dennis Sakva
by
7.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories