# Lines and Planes Constants

A) If a and b are constants that are not both zero then an equation of the form ax+by+c=0

represents a line in R^2 with normal nbar=(a,b)

B) If a, b and c are constants that are not all zero, then an equation of the form ax+by+cz+d=0

represents a plane in R^3 with normal nbar=(a,b,c)

Please demonstrate how to prove the theorem parts (B and C) in detail.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Theorem

A) If a and b are constants that are not both zero then an equation of the form ax+by+c=0

represents a line in R^2 with normal nbar=(a,b)

B) If a, b and c are constants that are not all zero, then an equation of the form ax+by+cz+d=0

represents a line in R^3 with normal nbar=(a,b,c)

First of all, I wanted to point out the part B) should be:

B) If a, b and c are constants that are not all zero, then an equation of the form ax+by+cz+d=0

represents a plane in R^3 with normal nbar=(a,b,c)

Proof:

A) As the solutions to the equation ax+by+c=0 are ordered pairs (x, y), which are points in . Take and a special point satisfying

....................(1)

Then take an arbitrary solution (x, y) to the equation

ax+by+c=0 ..................(2)

So, by (1)(2), we have

....................(3)

Let . Then (3) can be rewritten as

..............................(4)

The equation (4) implies that the two vectors and are orthogonal (perpendicular). In other words, the solutions to the equation ax+by+c=0 are all points (x, y) on the line passing through and perpendicular to the normal vector .

B) As the solutions to the equation ax+by+cz+d=0 are triples (x, y, z), which are points in (3-dimensional space). Take and a special point satisfying

....................(4)

Then take an arbitrary solution (x, y,z) to the equation

ax+by+cz+d=0 ....................(5)

So, by (4)(5), we have

....................(6)

Let . Clearly, it has as initial point and (x,y,z) as its terminal point.

The equation (6) implies that the two vectors and are orthogonal (perpendicular), as the dot product of and is equal to zero from (6).

Note that the vector has its initial point at and terminal point at (x, y,z). So, we conclude that for any solution (x,y,z) to the equation ax+by+cz+d=0, the vector is perpendicular to the vector . In other words, the solutions to the equation ax+by+cz+d=0 are all points (x,y,z) that are on the plane passing through and having as the normal vector.

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