How do you find the unit vector having the same direction as vector u = i - 6 j + 5 k?

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Answer 1 · 2016-07-01T10:48:17

Thus $ˆ u = \frac{1}{\sqrt{61}} i - \frac{6}{\sqrt{61}} j + \frac{5}{\sqrt{61}} k$ $" "hat(u)=1/sqrt(61)i-6/sqrt(61)j+5/sqrt(61)k$

Or if you prefer

$ˆ u = \frac{\sqrt{61}}{61} i - \frac{6 \sqrt{61}}{61} j + \frac{5 \sqrt{61}}{61} k$ $" "hat(u)=sqrt(61)/61i-(6sqrt(61))/61j+(5sqrt(61))/61k$

Given that $\to u = i - 6 j + 5 k$ $vec(u)=i-6j+5k$

Then $| | u | | = \sqrt{1^{2} + {(- 6)}^{2} + 5^{2}} = \sqrt{61}$ $||u||=sqrt(1^2+(-6)^2+5^2)" "=" "sqrt(61)$

Not that 61 is a prime number

Thus $ˆ u = \frac{1}{\sqrt{61}} i - \frac{6}{\sqrt{61}} j + \frac{5}{\sqrt{61}} k$ $" "hat(u)=1/sqrt(61)i-6/sqrt(61)j+5/sqrt(61)k$

Or if you prefer

$ˆ u = \frac{\sqrt{61}}{61} i - \frac{6 \sqrt{61}}{61} j + \frac{5 \sqrt{61}}{61} k$ $" "hat(u)=sqrt(61)/61i-(6sqrt(61))/61j+(5sqrt(61))/61k$