find equations for the sets of pts (x,y,z) that are equidistant from the planes 2x-y-z+6=0 and x+2y-7z+12=0?

Answer 1

The two sets of points satisfying the question are the two planes that bisect the angles between the given planes.

To find the equations of the bisecting planes we need to find normal vectors of the bisecting planes and one point on the planes. If you recall from geometry the diagonals of a rhombus bisect the angles. So if we take the unit normal vectors of the two given planes and add them we get the normal vector of one bisecting plane. If we subtract them we get the normal vector of the other bisecting plane. We are using unit normal vectors to do this calculation because they are the same length, as are the sides of a rhombus. So the angles of the unit normal vectors of the given planes will be bisected.

The two given planes are:

2x - y - z + 6 = 0
x + 2y - 7z + 12 = 0

The normal vectors of the two given planes are:

n1 = <2, -1, -1>
n2 = <1, 2, -7>

Calculate the magnitude of the normal vectors.

| n1 | = √[2² + (-1)² + (-1)²] = √6
| n2 | = √[1² + 2² + (-7)²] = √54 = 3√6

Add the unit normal vectors to get the normal vector N1, of one bisecting plane.

N1 = n1/|n1| + n2/|n2| = <2, -1, -1>/√6 + <1, 2, -7>/(3√6)
N1 = (<6, -3, -3> + <1, 2, -7>)/(3√6)
N1 = <7, -1, -10>/(3√6)

Any non-zero multiple of N1 is also a normal vector of the bisecting plane. Multiply by 3√6.

N1 = <7, -1, -10>

Subtract the unit normal vectors to get the normal vector N2, of the other bisecting plane.

N2 = n1/|n1| - n2/|n2| = <2, -1, -1>/√6 - <1, 2, -7>/(3√6)
N2 = (<6, -3, -3> - <1, 2, -7>)/(3√6)
N2 = <5, -5, 4>/(3√6)

Any non-zero multiple of N2 is also a normal vector of the bisecting plane. Multiply by 3√6.

N2 = <5, -5, 4>

Find a point P on the line of intersection of the two given planes. If it is on the line of intersection it will also be a point on both bisecting planes.

The equations of the given planes are:

2x - y - z + 6 = 0
x + 2y - 7z + 12 = 0

Let z = 1 and solve for x and y.

2x - y - 1 + 6 = 0
x + 2y - 7 + 12 = 0

2x - y + 5 = 0
x + 2y + 5 = 0

Add twice the first equation to the second.

5x + 15 = 0
x = -3

Substitute back into the first equation.

2x - y + 5 = 0
2(-3) - y + 5 = 0
-6 - y + 5 = 0
-y - 1 = 0
y = -1

So a point on the line of intersection is P(-3, -1, 1).
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With the normal vector N1 and a point P that lies in the plane we can write the equation of one of the bisecting planes.

N1 = <7, -1, -10>
P(-3, -1, 1)

7(x + 3) - 1(y + 1) - 10(z - 1) = 0
7x + 21 - y - 1 - 10z + 10 = 0
7x - y - 10z + 30 = 0
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With the normal vector N2 and a point P that lies in the plane we can write the equation of the other bisecting plane.

N2 = <5, -5, 4>
P(-3, -1, 1)

5(x + 3) - 5(y + 1) + 4(z - 1) = 0
5x + 15 - 5y - 5 + 4z - 4 = 0
5x - 5y + 4z + 6 = 0

Answer 2

It's a sphere. Say:

Sqrt( (2-x)^2 + (-1-y)^2 + (-1-z)^2) = Sqrt( (1-x)^2 +(2-y)^2 + (-7-z)^2)

Just solve that. Start out by squaring both sides, then expand the squared terms in the parentheses, then get everything to one side. I had this question on my calc final. It's a pain in the butt.