How do you normalize $(i - 2 j + 3 k)$ $(i -2j + 3k)$ ?

Question

1952 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-28T16:16:12

$\frac{1}{\sqrt{14}} (ˆ i - 2 ˆ j + 3 ˆ k)$ $1/sqrt14(hati-2hatj+3hatk)$

to normalize a vector means to change it to a unit vector, that is ensure that its magnitude is $= 1$ $=1$

$ˆ \to a = \frac{\to a}{∣ ∣ \to a ∣ ∣}$ $hatveca=(veca)/|veca|$

we ahve

$\to a = ˆ i - 2 ˆ j + 3 ˆ k$ $veca=hati-2hatj+3hatk$

$| a | 14$ $|a|14$

$= \sqrt{1^{2} + 2^{2} + 3^{2}} = \sqrt{14}$ $=sqrt(1^2+2^2+3^2)=sqrt14$

$ˆ \to a = \frac{1}{\sqrt{14}} (ˆ i - 2 ˆ j + 3 ˆ k)$ $hatveca=1/sqrt14(hati-2hatj+3hatk)$

How do you normalize (i−2j+3k) (i -2j + 3k) ?