The sphere ${x}^{2}+{y}^{2}+{z}^{2}-2x+6y+14z+3=0$ meets the line joining $A(2,-1,4),B(5,5,5)$ in the points $C$ and $D$. Prove that $AC:CB=-AD:DB=1:2$

Glenn Hopkins
2022-07-23
Answered

The sphere ${x}^{2}+{y}^{2}+{z}^{2}-2x+6y+14z+3=0$ meets the line joining $A(2,-1,4),B(5,5,5)$ in the points $C$ and $D$. Prove that $AC:CB=-AD:DB=1:2$

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Sandra Randall

Answered 2022-07-24
Author has **17** answers

Hint : The equation of the line through $A$ and $B$ is $(x,y,z)=(2+3t,-1+6t,4+t)$. Substitute this into the equation for the circle, solve the quadratic to find the points $C$ and $D$ ... etc ...

Livia Cardenas

Answered 2022-07-25
Author has **5** answers

The line through $A(2,-1,4),B(5,5,5)$ has a direction given by $\overrightarrow{u}=B-A=\{3,6,1\}$

so the line $AB$ has equation $A+s\overrightarrow{v}=\{2+3s,-1+6s,4+s\}$

plug in the equation of the sphere

$(s+4{)}^{2}+(3s+2{)}^{2}+(6s-1{)}^{2}-2(3s+2)+6(6s-1)+14(s+4)+3=0$

$2(23{s}^{2}+26s+35)=0$

Which has no real solutions

so the line $AB$ has equation $A+s\overrightarrow{v}=\{2+3s,-1+6s,4+s\}$

plug in the equation of the sphere

$(s+4{)}^{2}+(3s+2{)}^{2}+(6s-1{)}^{2}-2(3s+2)+6(6s-1)+14(s+4)+3=0$

$2(23{s}^{2}+26s+35)=0$

Which has no real solutions

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I don't find Postulate 13 of Spivak's Calculus trivial, nor can I understand why it's true.

Postulate 13: Every non-empty set of real numbers that is bounded above has a least upper bound (sup).

Why is this postulate true? Any proof/intuition behind it?

Edit: Let me pose a few questions.

1. Suppose I decide to devise a pathological function $\mathbb{R}$$\to $$\mathbb{R}$ which is bounded above but has no supremum. Why will I fail to find such $f$? If you simply take it as an axiom, there's no guarantee I won't be successful.

2. Suppose $S$ is an arbitrary non-empty set of real numbers that is bounded above. Does there exist an algorithm to determine $sup(S)$?

Postulate 13: Every non-empty set of real numbers that is bounded above has a least upper bound (sup).

Why is this postulate true? Any proof/intuition behind it?

Edit: Let me pose a few questions.

1. Suppose I decide to devise a pathological function $\mathbb{R}$$\to $$\mathbb{R}$ which is bounded above but has no supremum. Why will I fail to find such $f$? If you simply take it as an axiom, there's no guarantee I won't be successful.

2. Suppose $S$ is an arbitrary non-empty set of real numbers that is bounded above. Does there exist an algorithm to determine $sup(S)$?

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Consider the triangle $\mathrm{\Delta}ABC$ , which D is the midpoint of segment BC, and let the point G be defined such that $(GD)=\frac{1}{3}(AD)$ . Assuming that ${z}_{A},{z}_{B},{z}_{C}$ are the complex numbers representing the points (A, B, C):

a. Find the complex number ${z}_{G}$ that represents the point G

b. Show that $(CG)=\frac{2}{3}(CF)$ and that F is the midpoint of the segment (AB)

a. Find the complex number ${z}_{G}$ that represents the point G

b. Show that $(CG)=\frac{2}{3}(CF)$ and that F is the midpoint of the segment (AB)

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What does "external angle bisector" mean? How an angle can be bisected externally?

What does "external angle bisector" mean? How an angle can be bisected externally?

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I had a question about a problem that I was working on for my pre-calculus class.

Here's the problem statement:

The area of the parallelogram with vertices 0, $\u043c$, $w$, and $v+w$ is 34. Find the area of the parallelogram with vertices 0, $Av$, $Aw$, and $Av+Aw$, where

$A=\left(\begin{array}{cc}3& -5\\ -1& -3\end{array}\right).$

I got the answer by doing something very tedious. I set $v={\textstyle (}\genfrac{}{}{0ex}{}{17}{0}{\textstyle )}$ and $w={\textstyle (}\genfrac{}{}{0ex}{}{0}{2}{\textstyle )}$, and did some really crazy matrix multiplication and a lot of plotting points of GeoGebra to get the answer of: $\overline{){\displaystyle 476}}$.

Now, I'm 100% sure that was not the fastest way, can someone tell me the non-bash way to do the problem?

Here's the problem statement:

The area of the parallelogram with vertices 0, $\u043c$, $w$, and $v+w$ is 34. Find the area of the parallelogram with vertices 0, $Av$, $Aw$, and $Av+Aw$, where

$A=\left(\begin{array}{cc}3& -5\\ -1& -3\end{array}\right).$

I got the answer by doing something very tedious. I set $v={\textstyle (}\genfrac{}{}{0ex}{}{17}{0}{\textstyle )}$ and $w={\textstyle (}\genfrac{}{}{0ex}{}{0}{2}{\textstyle )}$, and did some really crazy matrix multiplication and a lot of plotting points of GeoGebra to get the answer of: $\overline{){\displaystyle 476}}$.

Now, I'm 100% sure that was not the fastest way, can someone tell me the non-bash way to do the problem?

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The internal bisectors of the angles of a triangle ABC meet the sides in D,E,and F.Show that area of the triangle DEF is equal to $\frac{2\mathrm{\Delta}\times abc}{(b+c)(c+a)(a+b)}$,here $\mathrm{\Delta}$ is area of triangle ABC

If I choose B as origin,C as(a,0),a is side length BC.what should i take coordinates of A,can i get answer with this approach or any better approach?

If I choose B as origin,C as(a,0),a is side length BC.what should i take coordinates of A,can i get answer with this approach or any better approach?