Calculus III question

Question

Find an equation of the plane. the plane that passes through the point (2, 4, −1) and contains the line x = 4 − t, y = 2t − 1, z = −3t

 

 

Answer 1

x = 4-t, y = 2t-1, z = -3t

When t = 0, we get (x, y, z) = (4, -1, 0) is a point on the plane.

Since (2, 4, -1) is also on the plane, v = <4-2, -1-4, 0-(-1)> = <2, -5, 1> is parallel to the plane.

<x, y, z> = t<-1, 2, -3> + <4, -1, 0>

w = <-1, 2, -3> is also parallel to the plane

n = v x w is perpendicular to the plane

n = <13, 5, -1>

Let (x, y, z) be a point on the plane. Then, <x-2, y-4, z+1> is a vector parallel to the plane.

So, <x-2, y-4, z+1> • <13, 5, -1> = 0

13(x - 2) + 5(y - 4) -1(z + 1) = 0

13x + 5y - z = 47

Answer 2

17x−7y−9z−15=0

Answer 3

Writing the line in vector form:

L: P₁ + t*v

where P₁ is a point on the line, setting t = 0, P₁ = (4, -1, 0)

and v is the direction vector of the line v = <-1, 2, -3>

L: (4, -1, 0) + t<-1, 2, -3>

Call our given point P₂ = (2, 4, -1)

Find vector from P₂ to P₁, call it u = <4-2, -1-4, 0-(-1)> = <2, -5, 1>

Normal vector for the plane, n, is the cross product of u and v (calculate or use your calculator)

n = u x v = <13, 5, -1>

Equation of plane , using n and P₂:

13(x-2) + 5(y-4) -(z+1) = 0

simplify to:

13x + 5y -z = 47

Geogebra is a good tool for this type of problem. Here is a link to an illustrated solution to this problem:

https://www.geogebra.org/calculator/sfzfjqa5