Find the distance of the centre of the circle x^2+y^2+z^2+x-2y+2z=3,2x+y+2z=1 from the plane ax+by+cz=d, where a,b,c,d are constants.?

Answer 1

See belo.

Given the Sphere

#S -> x^2+y^2+z^2+x-2y+2z-3=0#

and the plane

#Pi_1 -> 2x+2z-3=0# or #Pi_1-> << p-p_1, vec n >> = 0#

with

#p = (x,y,z)# #p_1 = (0,0,3/2)# #vec n = (2,2,0)#

The circle

#C-> S nn Pi_1# is centered at the orthogonal projection of the center of #S# over #Pi_1#

but

#S -> (x+1/2)^2+(y-1)^2+(z+1)^2 = 21/4#
has as center the point #p_0 = (-1/2,1,-1)#
now calling #p_2# the projection of #p_0# onto #Pi_1# we know that #p_2# is at the intersection
#Pi_1 nn L#
with #L-> p = p_0 + lambda vec n# or
#<< p_0-p_1+lambda vec n, vec n >> = 0#
giving #lambda = - (<< p_0-p_1, vec n >>)/norm(vec n)^2# and
#p_2 = p_0-(<< p_0-p_1, vec n >>)/norm(vec n)^2 vec n = (-3/4,3/4,-1)#

now given

#Pi_2 -> ax+by+cz-d=0# or #Pi_2-> << p-p_3, vec v >> = 0#

with

#vec v = (a,b,c)# #p_3 = (0,0,d/c)#
we have #p_4# as the projection of #p_2# onto #Pi_2# as
#p_4 = p_2 -(<< p_2-p_3, vec v >> )/norm(vec v)^2 vec v#
and the distance #d = norm(p_2-p_4)# is
#d = 1/4((3a-3b+4(c+d))/sqrt(a^2+b^2+c^2))#
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Answer 2

To find the distance between the center of the circle and the plane, we need to follow these steps:

  1. Rewrite the equations of the circle and the plane in standard form.
  2. Find the center of the circle.
  3. Use the formula for the distance between a point and a plane to find the distance.

Given equations: Circle: (x^2 + y^2 + z^2 + x - 2y + 2z = 3) Plane: (ax + by + cz = d)

Step 1: Rewrite equations in standard form: Circle: (x^2 + y^2 + z^2 + x - 2y + 2z - 3 = 0) Plane: (ax + by + cz - d = 0)

Step 2: Complete the square for the circle equation to find the center. ((x^2 + x) + (y^2 - 2y) + (z^2 + 2z) = 3) ((x + \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 + (z + 1)^2 - 1 = 3) ((x + \frac{1}{2})^2 + (y - 1)^2 + (z + 1)^2 = \frac{15}{4}) Center of the circle: ((- \frac{1}{2}, 1, -1))

Step 3: Use the distance formula for a point and a plane. Distance (= \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}})

Substitute the values: Distance (= \frac{|a(-\frac{1}{2}) + b(1) + c(-1) - d|}{\sqrt{a^2 + b^2 + c^2}})

Thus, the distance between the center of the circle and the plane is given by the above expression.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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