Question

# Find a unit vector in the direction of the resultant of the vectors $\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k},2\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}\mathrm{and}\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}.$

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Solution

## Given: $\stackrel{\to }{a}=\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k},\stackrel{\to }{b}=2\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}$ and $\stackrel{\to }{c}=\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}$ are the position vectors. Then, Resultant of the vectors = $\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}$ $=\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}+2\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}+\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=4\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}$ So, $\left|\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right|=\sqrt{{4}^{2}+{2}^{2}+{1}^{2}}=\sqrt{16+4+1}=\sqrt{21}$ ∴ Unit vector in the direction of the resultant vector = $\frac{\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}}{\left|\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right|}=\frac{1}{\sqrt{21}}\left(4\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)$

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