1. ## vectors

1. Prove that $\alpha^{2} + \beta^{2} + \gamma^{2} = 1$ (direction cosines).

So $\alpha = \frac{A_x}{|\bold{A}|}$ , $\beta = \frac{A_y}{|\bold{A}|}$, $\gamma = \frac{A_z}{|\bold{A}|}$ which implies $\alpha^{2} + \beta^{2} + \gamma^{2} = 1$.

2. Show that if $|\bold{A} - \bold{B}| = |\bold{A}+ \bold{B}|$ then $\bold{A}$ is perpendicular to $\bold{B}$.

So $\sqrt{(A_x-B_x)^{2} + (A_y-B_y)^{2} + (A_z - B_z)^{2}}$ $= \sqrt{(A_x+B_x)^{2} + (A_y+B_y)^{2} + (A_z + B_z)^{2}}$.

Then $(A_x - B_x)^{2} = (A_x+B_x)^{2}$ or $A_{x}^{2} - 2A_{x}B_{x} + B_{x}^{2} = A_{x}^{2} + 2A_{x}B_{x} + B_{x}^{2}$ so that $A_{x}B_{x} = A_{y}B_{y} = A_{z}B_{z} = 0$ which implies that $\bold{A} \cdot \bold{B} = 0$.

Are these correct?

2. In # 2, Square both sides.
$\begin{array}{rcl}
\left\| {A + B} \right\|^2 & = & \left( {A + B} \right) \cdot \left( {A + B} \right) \\ & = & A \cdot A + 2A \cdot B + B \cdot B \\ \end{array}$

$\begin{array}{rcl}
\left\| {A - B} \right\|^2 & = & \left( {A - B} \right) \cdot \left( {A - B} \right) \\
& = & A \cdot A - 2A \cdot B + B \cdot B \\ \end{array}$

This implies that $A \cdot B = 0$