Question

find the unit vector above the above​

Answer 1

Answer:

$\\\tt \dfrac{i + 4j - 2k}{\sqrt{21}}$

Logic:

A vector is given by:

vec(A) = |vec(A)|direction(A)

To get a unit vector in the direction of vector A use:

vec(A) / |vec(A)|

Using it here on $\\\tt i + 4j - 2k$ we have the unit vector in the same direction given by:

$\\\tt \dfrac{i + 4j - 2k}{\sqrt{1^2+ 4^2 + 2^2}}$

$\\\tt = \dfrac{i + 4j - 2k}{\sqrt{21}}$