91Ó°ÊÓ

Use vectors to prove the following theorems from geometry; The diagonals of a parallelogram bisect each other.

The Pauli spin matrices in quantum mechanics are $$ A=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) . \quad B=\left(\begin{array}{rr} 0 & -i \\ i & 0 \end{array}\right), \quad C=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) $$ Show that \(A^{2}=B^{2}=C^{2}=1\). (Note carefully that this 1 means the 2 by 2 unit matrix and not the number 1 ; this notation is customary in quantum mechanics.) Also show that any pair of these matrices anticommute, that is, \(A B=-B A\), etc. Show that the commutator of \(A\) and \(B\), that is, \(A B-B A\), is \(2 i C\), and similarly for other pairs in cyclic order.

The median of a trapezoid means the line joining the midpoints of the two nonparallel sides. Prove that the median bisects both diagonals; that the median is parallel to the two parallel bases and equal to half the sum of their lengths.

Solve the given set of equations by reducing the matrix to echelon form. Sa! geometrically what the solution is (one point, all points on a line or on a plane, or no solution). If the solution is a line, write its vector equation. \(\left\\{\begin{array}{l}5 x+7 y-4 z+w=2 \\ 2 x-y-2 z+3 w=3 \\ 4 x+3 y-5 z+2 w=5 \\ 3 x+2 y+z+5 w=1\end{array}\right.\)

Find the distance from the origin to the plane \(3 x-2 y-6 z=7\).

Find a condition for four points in space to lie in a plane. Your answer should be in the form of a determinant which must be equal to zero. Hint: The equation of a plane is of the form \(a x+b y+c z=d\), where \(a, b, c, d\) are constants. The four points \(\left(x_{1}, y_{1}, z_{1}\right)\) \(\left(x_{2}, y_{2}, z_{2}\right)\), etc., are all to satisfy this equation. When can you find \(a, b, c, d\) not all zero?