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.
$$|\vec{a}|=11, \;\; |\vec{b}|=23, \;\; |\vec{a}-\vec{b}|=30$$
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.
Hello Tolaso.
ReplyDeleteAlso, 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
\(\displaystyle{||a+b||^2=2\cdot 121+2\cdot 529-900=400\iff ||a+b||=20}\) .