find the linear equation of the plane through the point (1,3,7) and parallel to the plane x + 5y +2z +4 = 0

Answer 1

Find the linear equation of the plane through the point (1,3,7) and parallel to the plane

x + 5y +2z +4 = 0

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Helpful Information:

Plane form of an equation: ax + by +cz +d = 0 where (a,b,c) is the normal vector.

Step 1:

Find the normal vector of the given plane.

The normal vector of the plane:

x + 5y +2z +4 = 0 is n vector = (1,5,2)

Step 2:

The equation of a plane with normal vector n = (a,b,c) passing through point (xo,yo,zo) is given by:

a(x-xo)+ b(y-yo) + c(z-zo) = 0

Substitute the normal vector n = (1,5,2) and the point (1,3,7):

a=1,xo=1 1(x-1) = x-1

b=5,yo=3 5(y-3) = 5y-15

c=2,zo=7 2(z-7) = 2z-14

Group Information:

x-1+5y-15+2z-14 =0

Expand and simplify the equation:

x-1 +5y-15+2z-14 = 0

1. Add all numbers to right side of the equation 1+15+14 = 30
2. cancelling out the numbers on the left side
3. 
   

Now you have the final plane equation: x+5y+2z = 30

Solution:

The equation of the plane is x + 5y +2z = 30

I hope the mathematical calculations (step by step approach) was helpful. Please let me know if you require any more assistance. If anyone in my neighborhood is interested in setting up an in-person math tutoring session. I look forward to hearing from them. Have an amazing day. Doris H.