Question

If →a=4^i+7^j−5^ka→=4i^+7j^−5k^ and →b=3^i+4^j+^kb→=3i^+4j^+k^ find the direction cosines of →a−→b

Answer 1

$\vec a-\vec b=\vec i+3\vec j-6\vec k$

$(\vec a-\vec b)/|\vec a-\vec b|=(\vec i+3\vec j-6\vec k)/\sqrt{1^2+3^2+6^2}=$

$=0.1474\vec i+0.4423\vec j-0.8847\vec k$

$\cos\alpha=0.1474;\ \cos\beta=0.4423;\ \cos\gamma=-0.8847$