answer:The normalized form of the vector is: left( begin{array}{c} -frac{5}{sqrt{11}} -frac{7}{sqrt{11}} -frac{5}{sqrt{11}} end{array} right) Explanation: To normalize a vector, divide each component by the magnitude (or length) of the vector. The magnitude of the given vector is sqrt{left(frac{-5}{3}right)^2 + left(frac{-7}{3}right)^2 + left(frac{-5}{3}right)^2} = sqrt{11}. Thus, dividing each component by sqrt{11} yields the normalized vector.

answer:Nodal surfaces and nodal planes are regions in atomic orbitals where the probability of finding an electron is zero. Nodal Surfaces, also known as radial nodes, are generally spherical regions in which the electron density is zero. For instance, the hydrogen atom's #"1s"# orbital has no nodal surface, but higher-energy #"s"# orbitals like #"2s"# and #"3s"# have spherical nodal surfaces. These nodes represent areas where the wavefunction changes sign. On the other hand, Nodal Planes, or angular nodes, are flat regions of zero electron probability. These are unique to #"p"# orbitals, which have one nodal plane each, aligned along the #xy#, #xz#, or #yz# axes. Higher-energy #"p"# orbitals, like #"3p"#, may also have spherical nodal surfaces in addition to the planar ones. As we move to more complex orbitals, such as #"d"# and #"f"# orbitals, the nodal structures become more intricate, with multiple nodal planes and surfaces. These nodes contribute to the unique shapes and properties of these orbitals, allowing for a better understanding of electron distribution in atoms.

answer:To prove this, we'll use the definition of openness in a metric space. (Rightarrow) Assume U is open in (A,d). For each x in U, there exists an open ball B_x^A = {y in A | d(x,y) < r_x} in (A,d) centered at x with radius r_x such that B_x^A subset U. Now, define V = bigcup_{x in U} B_x^M, where B_x^M = {y in M | d(x,y) < r_x} is the open ball in (M,d) with the same center and radius. Since the union of open balls is an open set, V is open in (M,d). Moreover, U = A cap V since B_x^A = A cap B_x^M for each x in U. (Leftarrow) Conversely, suppose there exists an open set V in (M,d) such that U = A cap V. For any x in U, we have x in V, and since V is open, there exists r_x > 0 such that B(x, r_x) subset V. Thus, B_x^A = A cap B(x, r_x) is an open ball in (A,d) centered at x and contained in U. It follows that U can be written as the union of open balls in (A,d), making U an open set in (A,d). In conclusion, U subset A is open in (A,d) if and only if there exists an open set V in (M,d) with U = A cap V. The proof does not rely on connectedness properties of the sets.

answer:{eq}text{Consider the matrix} A=begin{bmatrix} 20 & 30 -12 & -16 end{bmatrix} text{The characteristic polynomial of matrix is as follows} left | A-lambda I right |=0 small small begin{vmatrix} 20 -lambda & 30 -12 & -16-lambda end{vmatrix}=0 lambda^2-4lambda +40=0 lambda =frac{-left(-4right)pm sqrt{left(-4right)^2-4cdot :1cdot :40}}{2cdot :1} Rightarrowlambda= 2pm 6i {color{Red} {text{Therefore,}quad a=2,enspace b=6}} text{The eigen vector corresponding to eigen value};: : lambda_1=2+6i left ( A-lambda_1I right )vec{V^1}=0 begin{bmatrix} 20 -2-6i& 30 -12 & -16-2-6i end{bmatrix}begin{bmatrix} v_1 v_2 end{bmatrix}=0 begin{bmatrix} 18-6i& 30 -12 & -18-6i end{bmatrix}sim begin{bmatrix}2&3+i 0&0end{bmatrix}begin{bmatrix} v_1 v_2 end{bmatrix}=0 2v_1+(3+i)v_2=0Rightarrow 2v_1=-(3+i)v_2 text{Therefore, the eigen vector corresponding to eigen value};: : lambda_1=2+6i vec{V^1}=begin{bmatrix} -3-i 2 end{bmatrix} text{The eigen vector corresponding to eigen value};: : lambda_2=2-6i left ( A-lambda_2I right )vec{V^2}=0 begin{bmatrix} 20 -2+6i& 30 -12 & -16-2+6i end{bmatrix}begin{bmatrix} v_1 v_2 end{bmatrix}=0 begin{bmatrix} 18+6i& 30 -12 & -18+6i end{bmatrix}sim begin{bmatrix}2&3-i 0&0end{bmatrix}begin{bmatrix} v_1 v_2 end{bmatrix}=0 2v_1+(3-i)v_2=0Rightarrow 2v_1=-(3-i)v_2 text{Therefore, the eigen vector corresponding to eigen value};: : lambda_2=2-6i vec{V^2}=begin{bmatrix} -3+i 2 end{bmatrix} {color{Red} {text{Therefore,}quad c=-3,enspace d=1}} {/eq}