Notes: `u = 20i + 25j` Use the dot product to find the magnitude of u.

Monday, 8 September 2014

The magnitude of a vector $\displaystyle{u}$ is the square root of its dot product by itself, because

$\displaystyle{u}\cdot{u}={\left|{\left|{u}\right|}\right|}\cdot{\left|{\left|{u}\right|}\right|}\cdot{\cos{{\left(\theta\right)}}},$

and $\displaystyle\theta={0},$ $\displaystyle{\cos{{\left(\theta\right)}}}={1}.$

I suppose that $\displaystyle{i}$ and $\displaystyle{j}$ are orthonormal. Therefore

$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{u}\cdot{u}}}=\sqrt{{{\left({20}{i}+{25}{j}\right)}\cdot{\left({20}{i}+{25}{j}\right)}}}=$

$\displaystyle=\sqrt{{{20}\cdot{20}+{25}\cdot{25}}}={5}\sqrt{{{4}\cdot{4}+{5}\cdot{5}}}={5}\sqrt{{{41}}}\approx{32}.$

This is the answer.

The magnitude of a vector $\displaystyle{u}$ is the square root of its dot product by itself, because

$\displaystyle{u}\cdot{u}={\left|{\left|{u}\right|}\right|}\cdot{\left|{\left|{u}\right|}\right|}\cdot{\cos{{\left(\theta\right)}}},$

and $\displaystyle\theta={0},$ $\displaystyle{\cos{{\left(\theta\right)}}}={1}.$

I suppose that $\displaystyle{i}$ and $\displaystyle{j}$ are orthonormal. Therefore

$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{u}\cdot{u}}}=\sqrt{{{\left({20}{i}+{25}{j}\right)}\cdot{\left({20}{i}+{25}{j}\right)}}}=$

$\displaystyle=\sqrt{{{20}\cdot{20}+{25}\cdot{25}}}={5}\sqrt{{{4}\cdot{4}+{5}\cdot{5}}}={5}\sqrt{{{41}}}\approx{32}.$

This is the answer.

Notes