How do you find a vector of length 1 in the same direction as (6,-8)?

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Answer 1 · 2016-08-29T13:29:44

The reqd. vector is $(6/10,-8/10)$ .

The Reqd. Vector, say $vecx$ is in the same direction as $(6,-8)$ , we must have,

$vecx=k(6,-8)=(6k,-8k),$ where, $k in RR^+$ .

As the second requirement for $vecx$ is $||vecx||=1$ , we have,

$36k^2+64k^2=1$ , i.e., $100k^2=1$ , or, $k=+-1/10$ , but, as, $k>0$ ,

$k=+1/10$ .

Therefore, the reqd. vector is $(6/10,-8/10)$ .

Enjoy Maths.!