91Ó°ÊÓ

10 Let's find a normal vector \(\boldsymbol{n}\) to the plane defined by the vectors \(\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right)\) and \(\left(\begin{array}{c} -1 \\ -1 \\ 1 \end{array}\right)\). To do this we will take the cross product of these two vectors:$$\boldsymbol{n}=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right)\times\left(\begin{array}{c} -1 \\ -1 \\ 1 \end{array}\right)$$After calculating the cross product, we get:$$\boldsymbol{n}=\left(\begin{array}{c} 2 \\ -2 \\ 0 \end{array}\right)$$Now we have a normal vector for the plane.

To find the equation of the plane, we can use the point-normal form:$$ax + by + cz = d$$where (a, b, c) are the components of the normal vector \(\boldsymbol{n}\), and d is the dot product of \(\boldsymbol{n}\) and any point on the plane. We can use the point \(\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right)\) for this:$$d=\boldsymbol{n}\cdot\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right)=2(1)-2(1)+0(1)=0$$So the equation of the plane is:$$2x-2y+0z=0$$or$$x-y=0$$

To find the orthogonal projection of the point \(\boldsymbol{v}=\left(\begin{array}{c}3 \\ 1 \\ -1\end{array}\right)\) onto the plane, we can use the formula: $$\boldsymbol{p}=\boldsymbol{v}-\frac{\boldsymbol{n}\cdot(\boldsymbol{v}-\boldsymbol{a})}{\boldsymbol{n}\cdot\boldsymbol{n}}\boldsymbol{n}$$where \(\boldsymbol{a}\) is any point on the plane, and \(\boldsymbol{n}\) is the normal vector to the plane (Step 1), and \(p\) is the projection point in the plane. Using the equation of the plane found in Step 2, and the point \(\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right)\) we can calculate the projection point:$$\boldsymbol{p}=\left(\begin{array}{c}3 \\ 1 \\ -1\end{array}\right)-\frac{\left(\begin{array}{c} 2 \\ -2 \\ 0 \end{array}\right)\cdot\left(\begin{array}{c} 2 \\ 0 \\ -2 \end{array}\right)}{\left(\begin{array}{c} 2 \\ -2 \\ 0 \end{array}\right)\cdot\left(\begin{array}{c} 2 \\ -2 \\ 0 \end{array}\right)}\left(\begin{array}{c} 2 \\ -2 \\ 0 \end{array}\right)=\left(\begin{array}{c} 3 \\ 1 \\ -1 \end{array}\right)-\frac{4}{8}\left(\begin{array}{c} 2 \\ -2 \\ 0 \end{array}\right) $$This simplifies to:$$\boldsymbol{p}=\left(\begin{array}{c} 3 \\ 1 \\ -1 \end{array}\right)-\left(\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right) = \left(\begin{array}{c} 2 \\ 2 \\ -1 \end{array}\right)$$

To find the minimum distance from the given point \(\boldsymbol{v}\) to the plane, we can calculate the distance between \(\boldsymbol{v}\) and its orthogonal projection \(\boldsymbol{p}\):$$d=\sqrt{(v_{1} - p_{1})^2 + (v_{2} - p_{2})^2 + (v_{3} - p_{3})^2}$$where \(v_{1},v_{2},v_{3}\) and \(p_{1},p_{2},p_{3}\) are the coordinates of \(\boldsymbol{v}\) and \(\boldsymbol{p}\) respectively.$$d=\sqrt{(3 - 2)^2 + (1 - 2)^2 + (-1 - (-1))^2}=\sqrt{1 + 1 + 0}=\sqrt{2}$$So, the minimal distance from point \(\boldsymbol{v}\) to the plane is \(\boxed{\sqrt{2}}\).