How do you find a unit vector u in the same direction as the vector ⟨1,−2,−3⟩?

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How do you find a unit vector u in the same direction as the vector ⟨1,−2,−3⟩?

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Answer 1 · 2016-06-20T08:06:24

$\to v = \frac{{1, - 2, - 3}}{\sqrt{14}}$ $vec v = { {1,-2,-3}}/sqrt(14)$

Explanation:

A unit vector is a vector $\to u$ $vec u$ such that

$∥ ∥ \to u ∥ ∥ = 1$ $norm vec u = 1$

Given a vector $\to V = {1, - 2, - 3}$ $vec V ={1,-2,-3}$ the way to find its associated unit vector $\to v$ $vec v$ is normalizing it. Then

$\to v = \frac{\to V}{∥ ∥ ∥ \to V ∥ ∥ ∥} = \frac{{1, - 2, - 3}}{\sqrt{1^{2} + {(- 2)}^{2} + {(- 3)}^{2}}} = \frac{{1, - 2, - 3}}{\sqrt{14}}$ $vec v = vec V/(norm vec V) ={ {1,-2,-3}}/sqrt(1^2+(-2)^2+(-3)^2} = { {1,-2,-3}}/sqrt(14)$ .

We know that $⟨ \to V, \to V ⟩ = {∥ ∥ ∥ \to V ∥ ∥ ∥}^{2}$ $<< vecV,vec V >> = norm(vec V)^2$ then

$⟨ \to v, \to v ⟩ = ⟨ \frac{\to V}{∥ ∥ ∥ \to V ∥ ∥ ∥}, \frac{\to V}{∥ ∥ ∥ \to V ∥ ∥ ∥} ⟩ = \frac{⟨ \to V, \to V ⟩}{{∥ ∥ ∥ \to V ∥ ∥ ∥}^{2}} = \frac{{∥ ∥ ∥ \to V ∥ ∥ ∥}^{2}}{{∥ ∥ ∥ \to V ∥ ∥ ∥}^{2}} = 1$ $<< vec v, vec v >> = << vec V/(norm vec V) , vec V/(norm vec V) >> = << vec V, vec V>>/norm(vec V)^2 = norm(vecV)^2/norm vecV^2 = 1$

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How do you find a unit vector u in the same direction as the vector ⟨1,−2,−3⟩?

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