Respuesta :
Answer:
Its center = ([tex]\frac{a_{1}+b_{1}}{2}[/tex],[tex]\frac{a_{2}+b_{2}}{2}[/tex],[tex]\frac{a_{3}+b_{3}}{2}[/tex])
Its radius = (|a-b|)/2
Step-by-step explanation:
Using the vector equation, we have:
(r-b).(r-a) = 0
r.r - a.r - b.r + a.b = 0
We are also given the following:
r = [x,y,z], a = [[tex]a_{1},a_{2},a_{3}[/tex]], b = [[tex]b_{1},b_{2},b_{3}[/tex]]
Using the expressions above for the given vector equation, we have:
(x,y,z)*(x,y,z) - ([tex]a_{1},a_{2},a_{3}[/tex])*(x,y,z) - ([tex]b_{1},b_{2},b_{3}[/tex])*(x,y,z) + ([tex]a_{1},a_{2},a_{3}[/tex])*([tex]b_{1},b_{2},b_{3}[/tex]) = 0
[tex]x^{2} +y^{2} +z^{2} - (a_{1}+b_{1})x -(a_{2}+b_{2})y - (a_{3}+b_{3})z + a_{1} b_{1} + a_{2}b_{2}+a_{3}b_{3}[/tex]
Therefore, using completing the square method, we have:
[tex](x-\frac{a_{1}+b_{1}}{2})^{2}[/tex] + [tex](y-\frac{a_{2}+b_{2}}{2})^{2}[/tex]
+[tex](z-\frac{a_{3}+b_{3}}{2})^{2}[/tex] + [tex]a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}[/tex] - (a1+b1)^2/2 - (a2+b2)^2/2 - (a3+b3)^2/2= 0
Further simplification and rearrangement lead to:
= [tex]\frac{-4a_{1}b_{1}}{4}[/tex] - [tex]\frac{-4a_{2}b_{2}}{4}[/tex] - [tex]\frac{-4a_{3}b_{3}}{4}[/tex] + [tex]\frac{a_{1}^{2}+2a_{1}b_{1}+b^{2}_{2}}{4} +[/tex][tex]\frac{a_{2}^{2}+2a_{2}b_{2}+b^{2}_{2}}{4}[/tex] +[tex]\frac{a_{3}^{2}+2a_{3}b_{3}+b^{2}_{3}}{4}[/tex]
= [tex]\frac{(a_{1}-b_{1})^{2}}{4}[/tex] +[tex]\frac{(a_{2}-b_{2})^{2}}{4}[/tex] +[tex]\frac{(a_{3}-b_{3})^{2}}{4}[/tex] = [tex]\frac{(|a-b|)^{2} }{4}[/tex]
Thus,
Its center = ([tex]\frac{a_{1}+b_{1}}{2}[/tex],[tex]\frac{a_{2}+b_{2}}{2}[/tex],[tex]\frac{a_{3}+b_{3}}{2}[/tex])
Its radius = (|a-b|)/2
You can use the dot product to expand the equation (r-a).(r-b) and then can compare if with the equation of the sphere.
Center of the sphere = [tex]( \frac{a_1+b_1}{2} , \frac{a_2+b_2}{2},\frac{a_3+b_3}{2} )[/tex]
Radius of the considered sphere [tex]\dfrac{|a-b|}{2}[/tex] units.
What is the equation of sphere with radius at (h,k,l) coordinate and length of radius as r units?
The equation of sphere with center at (h,k,l) coordinate and radius of length r units is given by:
[tex](x-h)^2 + (y-k)^2 + (x-l)^2 = r^2[/tex]
How to find the dot product for given vectors using given expression?
[tex](r -a).(r-b) &=0\\([x-a_1, y-a_2, z-a_3]).([x-b_1,y-b_2,z-b_3]) = 0\\(x-a_1)(x-b_1) + (x-a_2)(x-b_2) + (x-a_3)(x-b_3) = 0\\x^2 + y^2 + z^2 +a_1b_1 + a_2b_2 + a_3b_3 -[ (a_1+b_1)x + (a_2 + b_2)y + (a_3 + b_3)z] = 0\end{aligned}[/tex]
Using completing the squares method, we will complete the squares containing variables, and the constants left will be the radius squared.
Thus,
[tex](x-(\frac{a_1+b_1}{2}) )^2 + (y-(\frac{a_2+b_2}{2}) )^2 + (z-(\frac{a_3+b_3}{2}) )^2 + a_1b_1 + a_2b_2 + a_3b_3 - ((\frac{a_1+b_1}{2})^2 + (\frac{a_2+b_2}{2})^2 + (\frac{a_3+b_3}{2})^2 ) = 0\\[/tex]
Comparing it with the equation of sphere, we get:
Center = (h,k,l) = [tex]( \frac{a_1+b_1}{2} , \frac{a_2+b_2}{2},\frac{a_3+b_3}{2} )[/tex]
Radius
[tex]=\sqrt{-(a_1b_1 + a_2b_2 + a_3b_3) + ((\frac{a_1+b_1}{2})^2 + (\frac{a_2+b_2}{2})^2 + (\frac{a_3+b_3}{2})^2 ) \\}\\= \sqrt{(\frac{a_1+b_1}{2})^2 + (\frac{a_2+b_2}{2})^2 + (\frac{a_3+b_3}{2})^2 -(a_1b_1 + a_2b_2 + a_3b_3)} \\=\sqrt{(\frac{a_1-b_1}{2})^2 + (\frac{a_2-b_2}{2})^2 + (\frac{a_3-b_3}{2})^2}\\=\dfrac{\sqrt{(a_1 - b_1)^2 + (a_2 -b_2)^2 + (a_3 - b_3)^2}}{2}\\\\= Length( a - b)/2\\\\= \dfrac{|a-b|}{2}[/tex]
Thus,
Center of the considered sphere = [tex]( \frac{a_1+b_1}{2} , \frac{a_2+b_2}{2},\frac{a_3+b_3}{2} )[/tex]
Radius of the considered sphere [tex]\dfrac{|a-b|}{2}[/tex] units.
Learn more about equation of sphere here:
https://brainly.com/question/4536228?
