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)

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

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

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