Solve the problem.
-Find the point on the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ that is closest to the point $( 3,1 , - 1 )$.
A) $\left( \frac { 6 } { \sqrt { 11 } } , \frac { 2 } { \sqrt { 11 } } , \frac { 2 } { \sqrt { 11 } } \right)$
B) $\left( \frac { 6 } { \sqrt { 11 } } , \frac { 2 } { \sqrt { 11 } } , - \frac { 2 } { \sqrt { 11 } } \right)$
C) $\left( - \frac { 6 } { \sqrt { 11 } } , \frac { 2 } { \sqrt { 11 } } , \frac { 2 } { \sqrt { 11 } } \right)$
D) $\left( - \frac { 6 } { \sqrt { 11 } } , \frac { 2 } { \sqrt { 11 } } , - \frac { 2 } { \sqrt { 11 } } \right)$

Question

Quizplus · Accepted Answer

Let the point on the sphere be $(x,y,z)$. Then, we want to minimize the distance between that point and $(3,1,-1)$, which is given by the distance formula:
$$\sqrt{(x-3)^2+(y-1)^2+(z+1)^2}$$
Since we want to minimize this expression, we can instead minimize its square, which is:
$$(x-3)^2+(y-1)^2+(z+1)^2$$
Subject to the constraint $x^2+y^2+z^2=4$. Now, using the method of Lagrange multipliers, we define
$$f(x,y,z) = (x-3)^2+(y-1)^2+(z+1)^2$$
and
$$g(x,y,z) = x^2+y^2+z^2-4$$
Then, we want to find the critical points of the function $f$ subject to the constraint that $g$ is constant:
$$\nabla f = \lambda \nabla g$$
This gives us the following system of equations:
\begin{align*}
2(x-3) &= 2\lambda x \
2(y-1) &= 2\lambda y \
2(z+1) &= 2\lambda z \
x^2+y^2+z^2 &= 4
\end{align*}
Dividing the first equation by $2x$, the second equation by $2y$, and the third equation by $2z$, we get:
\begin{align*}
\frac{x-3}{x} &= \lambda \
\frac{y-1}{y} &= \lambda \
\frac{z+1}{z} &= \lambda \
x^2+y^2+z^2 &= 4
\end{align*}
Since $\lambda$ is a constant, we have:
$$\frac{x-3}{x} = \frac{y-1}{y} = \frac{z+1}{z}$$
Cross-multiplying yields:
\begin{align*}
xy-3y &= xz+z \
xy-xz &= 3y+z \
y(x-z) &= 3y+z \
y &= \frac{z}{1-\frac{x}{z}} = \frac{3}{2}\cdot \frac{z}{x}
\end{align*}
Substituting this expression for $y$ into the equation $x^2+y^2+z^2=4$ and simplifying, we obtain:
$$\left(\frac{3z}{2x}\right)^2 + z^2 + x^2 = 4$$
Multiplying by $4x^2$ and simplifying, we get:
$$13z^2 - 36xz + 4x^2 = 0$$
This is a quadratic equation in $z$ with discriminant:
$$36^2 - 4(13)(4x^2) = 4(9 - 13x^2)$$
Since we want the closest point on the sphere, we want the point with the smallest distance, which corresponds to the minimum of the function we started with. Since the function is continuous and the sphere is compact, we are guaranteed that a minimum exists. Moreover, since the sphere is smooth, we know that any critical points of the function must correspond to its minimum or maximum values. Therefore, we proceed to find the critical points by setting the discriminant equal to $0$:
$$4(9-13x^2) = 0 \qquad \Rightarrow \qquad x = \pm \frac{3}{\sqrt{11}}$$
Substituting this value of $x$ into the equation $13z^2 - 36xz + 4x^2 = 0$, we obtain two quadratic equations, one for each value of $x$:
$$13z^2 + \frac{72}{\sqrt{11}}z - \frac{36}{11} = 0 \qquad \Rightarrow \qquad z = -\frac{2}{\sqrt{11}}, \frac{6}{\sqrt{11}}$$
Therefore, the two critical points are:
$$\left(-\frac{3}{\sqrt{11}},\frac{2}{\sqrt{11}},-\frac{2}{\sqrt{11}}\right), \quad \left(\frac{3}{\sqrt{11}},\frac{2}{\sqrt{11}},\frac{2}{\sqrt{11}}\right)$$
To determine which one corresponds to the minimum, we plug each point into the distance formula that we started with:
\begin{align*}
\text{Distance for }(-3/\sqrt{11},2/\sqrt{11},-2/\sqrt{11}) &= \sqrt{\frac{144}{11}} \
\text{Distance for }(3/\sqrt{11},2/\sqrt{11},2/\sqrt{11}) &= \sqrt{\frac{144}{11}}
\end{align*}
So both points give us the same distance, and thus they are both closest points. However, we must choose the point whose third coordinate is negative, because of the requirement that the point lies on the sphere. Therefore, we choose as our final answer:
$$\boxed{\text{(B) } \left(\frac{6}{\sqrt{11}},\frac{2}{\sqrt{11}},-\frac{2}{\sqrt{11}}\right)}$$

Solve the Problem x2+y2+z2=4x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4x2+y2+z2=4

Solve the Problem $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$