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Tuesday, September 29, 2015

Length of vector

Let $\vec{a}, \vec{b}$ be vectors such that:

$$|\vec{a}|=11, \;\; |\vec{b}|=23, \;\; |\vec{a}-\vec{b}|=30$$

Find the length of the vector $\vec{a}+\vec{b}$.

Solution

We are really insterested in evaluating $|\vec{a}+\vec{b}|$.So,


$$\begin{align*}
\left | \vec{a}-\vec{b} \right |^2 =900 &\Leftrightarrow \left | \vec{a} \right |^2 -2\vec{a}\vec{b}+ |\vec{b}|^2 =900 \\
 &\Leftrightarrow 121 -2\vec{a}\vec{b}+ 529 =900 \\
 &\Leftrightarrow  \vec{a}\vec{b}= -125
\end{align*}$$

Now, in order to evaluate the wanted length we also take its square, so:

$$\begin{align*}
\left | \vec{a}+\vec{b} \right |^2 &=\left | \vec{a} \right |^2 +2\vec{a}\vec{b} +|\vec{b}|^2  \\
 &=121 - 250 + 529 \\
 &=400
\end{align*}$$

Hence:

$$|\vec{a}+\vec{b}|=20$$

and the exercise is complete.

1 comment:

  1. Hello Tolaso.

    Also, according to the rule of parallelogram, we get :

    \(\displaystyle{||a+b||^2+||a-b||^2=2\,||a||^2+2||b||^2}\), so :

    \(\displaystyle{||a+b||^2=2\,||a||^2+2||b||^2-||a-b||^2}\)

    or

    \(\displaystyle{||a+b||^2=2\cdot 121+2\cdot 529-900=400\iff ||a+b||=20}\) .

    ReplyDelete