Find an equation of the plane. The plane through the point (5,3,5) and with normal vector 2i + j - k?

Answer 1

Greetings! Lets solve this shall we ?

So, we must find the equation of a plane through the point (5,3,5) with a normal vector of 2i + j - k. Now, we yield the general equation for an equation of a plane with a given normal vector n such that,

n * <x - x₀, y - y₀, z - z₀> = 0

Let n = 2i + j - k = <2, 1, -1> and (x₀, y₀, z₀) = (5, 3, 5).

Then,

n * <x - x₀, y - y₀, z - z₀> = 0

=| <2, 1, -1> * <x-5, y-3, z-5> = 0................Multiply the normal vector inside the equation of a plane

=| 2(x-5) + 1(y-3) -1(z-5) = 0...............Then distribute ans simplify such that

=| 2x - 10 + y - 3 - z + 5 = 0

=| 2x + y - z - 10 - 3 + 5 = 0

=| 2x + y - z - 8 = 0

=| 2x + y - z = 8

I hope this helped!

Answer 2

Let n = 2i +j -k To find the plane through (5,3,5) that has this normal do the following.

2(x-5) +1(y-3) -1(z-5) = 0 hence 2x -10 +y -3 -z +5 =0 2x +y -z -8 =0 is the equation of the plane.