Answer to Question #115197 in Linear Algebra for Bradley du Buy

Question

Let u = (4,2,-1), v = (3, 1, 1) and w = (0, 2, 1). Compute the following:
(i) 2v - 3w -u
(ii) u(w+v)
(iii) ||u . w ||
(iv) the orthogonal projection of u on w
(v) the vector component of u orthogonal to w

Answer 1

$\bold i) \\2v - 3w - u = (6, 2, 2) - (0, 6, 3) - (4, 2, -1) = (2, -6, 0)\\ \bold{ii)} \\ w + v = (3,3,2) \\u(w + v) = (4,2,-1)(3,3,2) =4*3 +3*2 +2*(-1) = 16 \\ \bold{iv)} \\ proj_w u=\frac{u•w} {|w| }*w\\ Find\ the\ dot\ product\ of u \ and \ w\\ u•w=4•0+2•2+1•(-1)=3\\ Find\ the \ length \ of \ w\\ |w|= \sqrt{0^2+2^2+1^2}=\sqrt{5}\\ proj_wu=(0,\frac{6\sqrt{5}} {5},\frac{3\sqrt {5}}{5})\\ \bold{v)}\\ comp_w u = u -\frac{u•w}{|w|}w=(4,2,-1) - (0,\frac{6\sqrt{5}}{5},\frac{3\sqrt{5}}{5})=\\ =(4, \frac{10-6\sqrt{5}}{5}, \frac{-5-3\sqrt{5}}{5})\\ \bold{iii)} \begin{Vmatrix} u*w \end{Vmatrix} = \begin{Vmatrix} i & j & k\\ 4 & 2 & -1\\ 0 & 2 & 1 \end{Vmatrix} = \begin{Vmatrix} \begin{vmatrix} 2 & -1 \\ 2 & 1\\ \end{vmatrix},& \begin{vmatrix} -1 & 4 \\ 1 & 0 \end{vmatrix}, \begin{vmatrix} 4 & 2 \\ 0 & 2 \end{vmatrix}\\ \end{Vmatrix} =\begin{Vmatrix} (4, -4 , 8) \end{Vmatrix} = \sqrt{4^2+(-4)^2+8^2} = 4\sqrt{6}$