math dude solve this question
Answers
Answer:
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Step-by-step explanation:
Given:
If r is the radius of the circle given by ;
x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0; lx + my + nz = 0
To prove:
prove that -
(r2 + d2)(l2 + m2 + n2) = (mw − nv)2 + (nu − lw)2 + (lv − mu)2
Solution:
According to question;
The center of the sphere S = 0 is C(-u,-v,-w)C(−u,−v,−w) and radius
R =\sqrt{(u^{2} + v^{2} + w^{2} -d)}R=(u2+v2+w2−d)
Let O be the center of the circle determined by S = 0 and \pi=0π=0 .
Then CO is perpendicular to the plane \pi=0π=0 .
⇒ d. r of CO are the d. r of the normal to the plane \pi=0π=0
and they are l, m, n.
Therefore the equation of the line CO are
\dfrac{x+u}{l}=\dfrac{y+v}{m}=\dfrac{z+w}{n} .lx+u=my+v=nz+w.
The point O is (kl-u, km-v, kn-w)(kl−u,km−v,kn−w) for some k.
Since, O lies on the plane \piπ ,
\begin{gathered}l((kl-u)+m(km-v)+n(kn-w)=0\\\\implies,\\ k(l^{2}+m^{2}+n^{2})=lu+mv+nw\end{gathered}l((kl−u)+m(km−v)+n(kn−w)=0implies,k(l2+m2+n2)=lu+mv+nw
Therefore,
k=\dfrac{lu+mv+nw}{l^{2} +m^{2} +n^{2} }k=l2+m2+n2lu+mv+nw
Now,
\begin{gathered}r^{2}=R^{2}-OC^{2}\\ =(u^2+v^2+w^2-d)-[(kl)^2+(km)^2+(kn)^2]\\\\=(u^2+v^2+w^2-d)-k^2(l^2+m^2+n^2)\\\\=(u^2+v^2+w^2-d)-\dfrac{lu+mv+nw}{l^2+m^2+n^2}(l^2+m^2+n^2) \\implies,\\r^2(l^2+m^2+n^2)=(u^2+v^2+w^2-d)(l^2+m^2+n^2)-(lu+mv+nw)^2\\\\(r^2+d^2)(l^2+m^2+n^2)=(l^2+m^2+n^2)(u^2+v^2+w^2)-(lu+mv+nv)^2\\\\=(mw-nv)^2 +(nu-lw)^2 +(lv-mu)^2\end{gathered}r2=R2−OC2=(u2+v2+w2−d)−[(kl)2+(km)2+(kn)2]=(u2+v2+w2−d)−k2(l2+m2+n2)=(u2+v2+w2−d)−l2+m2+n2lu+mv+nw(l2+m2+n2)implies,r2(l2+m2+n2)=(u2+v2+w2−d)(l2+m2+n2)−(lu+mv+nw)2(r2+d2)(l2+m2+n2)=(l2+m2+n2)(u2+v2+w2)−(lu+mv+nv)2=(mw−nv)2+(nu−lw)2+(lv−mu)2
Hence, proved.
Answer:
Given:
If r is the radius of the circle given by ;
x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0; lx + my + nz = 0
To prove:
prove that -
(r2 + d2)(l2 + m2 + n2) = (mw − nv)2 + (nu − lw)2 + (lv − mu)2
Solution:
According to question;
The center of the sphere S = 0 is C(-u,-v,-w)C(−u,−v,−w) and radius
R =\sqrt{(u^{2} + v^{2} + w^{2} -d)}R=(u2+v2+w2−d)
Let O be the center of the circle determined by S = 0 and \pi=0π=0 .
Then CO is perpendicular to the plane \pi=0π=0 .
⇒ d. r of CO are the d. r of the normal to the plane \pi=0π=0
and they are l, m, n.
Therefore the equation of the line CO are
\dfrac{x+u}{l}=\dfrac{y+v}{m}=\dfrac{z+w}{n} .lx+u=my+v=nz+w.
The point O is (kl-u, km-v, kn-w)(kl−u,km−v,kn−w) for some k.
Since, O lies on the plane \piπ ,
\begin{gathered}l((kl-u)+m(km-v)+n(kn-w)=0\\\\implies,\\ k(l^{2}+m^{2}+n^{2})=lu+mv+nw\end{gathered}l((kl−u)+m(km−v)+n(kn−w)=0implies,k(l2+m2+n2)=lu+mv+nw
Therefore,
k=\dfrac{lu+mv+nw}{l^{2} +m^{2} +n^{2} }k=l2+m2+n2lu+mv+nw
Now,
\begin{gathered}r^{2}=R^{2}-OC^{2}\\ =(u^2+v^2+w^2-d)-[(kl)^2+(km)^2+(kn)^2]\\\\=(u^2+v^2+w^2-d)-k^2(l^2+m^2+n^2)\\\\=(u^2+v^2+w^2-d)-\dfrac{lu+mv+nw}{l^2+m^2+n^2}(l^2+m^2+n^2) \\implies,\\r^2(l^2+m^2+n^2)=(u^2+v^2+w^2-d)(l^2+m^2+n^2)-(lu+mv+nw)^2\\\\(r^2+d^2)(l^2+m^2+n^2)=(l^2+m^2+n^2)(u^2+v^2+w^2)-(lu+mv+nv)^2\\\\=(mw-nv)^2 +(nu-lw)^2 +(lv-mu)^2\end{gathered}r2=R2−OC2=(u2+v2+w2−d)−[(kl)2+(km)2+(kn)2]=(u2+v2+w2−d)−k2(l2+m2+n2)=(u2+v2+w2−d)−l2+m2+n2lu+mv+nw(l2+m2+n2)implies,r2(l2+m2+n2)=(u2+v2+w2−d)(l2+m2+n2)−(lu+mv+nw)2(r2+d2)(l2+m2+n2)=(l2+m2+n2)(u2+v2+w2)−(lu+mv+nv)2=(mw−nv)2+(nu−lw)2+(lv−mu)2
Hence, proved.
Step-by-step explanation:
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