Question

find a unit vector parallel to the a vector is given by A vector = 2l +4j+k

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Answer 1

Answer:

ok

Explanation:

Answer 2

$\mathrm{Answer:}\\ \\ \mathrm{\overrightarrow{A} = 2 \vec{i}\ +\ 4 \vec{j}\ +\ \vec{k} } \\ \\ \mathrm{|\overrightarrow{A}| = \sqrt{(2)^2+(4)^2+(1)^2} = \sqrt{21}}\\ \\ \mathrm{\hat{A} = \dfrac{\overrightarrow{A}}{|\mathrm{\overrightarrow{A}}|}= \dfrac{2 \vec{i}\ +\ 4 \vec{j}\ +\ \vec{k} }{\sqrt{21}} } \\ \\ \\ \\ \boxed{\mathrm{Therefore\ unit\ vector\ parallel\ to\ A\ is: \hat{A}=\dfrac{2 \vec{i}\ +\ 4 \vec{j}\ +\ \vec{k} }{\sqrt{21}} }} \\ \\ \\ \mathrm{Hope\ this\ helps}$