91影视

Given a surface $-x_1^2+x_2^2-x_3^2+10x_1{x}_3=1$.

Let $q\left(x_1,x_2,x_3\right)=-x_1^2+x_2^2-x_3^2+10x_1{x}_3$.

To find a corresponding matrix of q, let $a_{ii}=$coefficient of ${x_i}^2$and $a_{ij}=a_{ji}=\frac{1}{2}$ (coefficient of x_i,x_j ).

Now, find the entries of matrix:

a₁₁ =coefficient of ${x_1}^2=-1$

a₂₂=coefficient of ${x_2}^2=1$

a₃₃=coefficient of${x_3}^2=-1$

$a_{12}=a_{21}=\frac{1}{2}$coefficient of x₁x₂=0

$a_{13}=a_{31}=\frac{1}{2}$coefficient of$x_1{x}_3=\frac{1}{2}\left(10\right)=5$

$a_{23}=a_{32}=\frac{1}{2}$coefficient of x₂x₃= 0

So, the matrix is $A=\left[\begin{array}{ccc}-1& 0& 5\\ 0& 1& 0\\ 5& 0& -1\end{array}\right]$.

Since A is 3 x 3 matrix, the characteristic equation of A is$\left|A-位I_3\right|=0$. It is written as:

$\left|\begin{array}{ccc}-1-位& 0& 5\\ 0& 1-位& 0\\ 5& 0& -1-位\end{array}\right|=0$

Solving further,

$$\begin{gathered} \left(-1-位\right)\left[\left(1-位\right)\left(-1-位\right)-0\right]+5\left[0-5\left(1-位\right)\right]=0 \\ {\left(-1-位\right)}^2\left(1-位\right)-25\left(1-位\right)=0 \end{gathered}$$

This implies,

$$\begin{gathered} \left(1-位\right)\left[{\left(1+位\right)}^2-25\right]=0 \\ \left(1-位\right)\left[{位}^2+2位-24\right]=0 \\ \left(位-1\right)\left(位+6\right)\left(位-4\right)=0 \end{gathered}$$

The eigenvalues are 1,4,-6. So, it is written as $位=1,4,-6$.

Since $q\left(x\right)={位}_1{c_1}^2+{位}_2{c_2}^2+...+{位}_n{c_n}^2$. Here \(q\left( x \right) = - 6{c_1}^2 + c_2^2 + 4c_3^2 = 1\),

Two points closest to the origin occurs when $c_1=c_2=0$.

So, equation becomes:

$$\begin{gathered} 4{c_3}^2=1 \\ {c}_3^2=\frac{1}{4} \\ {c}_3=卤\frac{1}{2} \end{gathered}$$

Hence, the two closest points on the surface from the origin are $\left(0,0,\frac{-1}{2}\right)and\left(0,0,\frac{1}{2}\right)$.