Answer to Question #178531 in Physics for Francis cyril

Question

If vector A=I+j+2k, vector B=2i+j+k, vector=1-2j-2k fine the magnitude and direction cos theta of

(A+B+C)

(A-B+C)

Answer 1

a)

$\bold{A+B+C}=(1+2+1,1+1-2,2+1-2)\\=(4,0,1)$

$d=\sqrt{4^2+1^2}=\sqrt{15}$

$\alpha=\frac{4}{\sqrt{15}},\beta=0,\gamma=\frac{1}{\sqrt{15}}$

b)

$\bold{A-B+C}=(1-2+1,1-1-2,2-1-2)\\=(0,-2,-1)$

$d=\sqrt{2^2+1^2}=\sqrt{5}$

$\alpha=0,\beta=\frac{-2}{\sqrt{15}},\gamma=\frac{-1}{\sqrt{15}}$