How do you find the unit vector in the direction of the given vector of $u = < 0, - 2 >$ $u=<0,-2>$ ?

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Answer 1 · 2017-03-24T16:38:53

$ˆ u = ⟨ 0, - 1 ⟩$ $hat u = langle 0, -1 rangle$

The unit vector, $ˆ u$ $hat u$ , is:

$ˆ u = \frac{\to u}{∣ ∣ \to u ∣ ∣}$ $hat u = (vec u)/(abs ( vec u))$

And:

$∣ ∣ \to u ∣ ∣ = \sqrt{0^{2} + {(- 2)}^{2}} = 2$ $abs ( vec u) = sqrt(0^2 + (-2)^2) = 2$

(But we knew that anyway as it only has a $j$ $mathbf j$ component of magnitude 2.)

So:

$ˆ u = \frac{⟨ 0, - 2 ⟩}{2} = ⟨ 0, - 1 ⟩$ $hat u = (langle 0, -2 rangle)/(2) = langle 0, -1 rangle$

How do you find the unit vector in the direction of the given vector of u=<0,−2>u=<0,-2>?