Question

If the vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitude, prove that $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$ is equally inclined with vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$.

Answer 1

Hint: We first assume the magnitude of the three vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$. They are three mutually perpendicular vectors of equal magnitude. We, based on the given condition find the mathematical form of $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$. Then we find the magnitude of the vector form $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$. We also assume the angle between them. We find the angles and prove the requirement.

Complete step by step solution:
It’s given that $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitude.
We assume that $\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|=\lambda $.
Since the vectors are mutually perpendicular, we can say $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$.
We have to show that $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$ is equally inclined with vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$.
We take the magnitude of $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$.
So, ${{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}^{2}}=\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}+2\overrightarrow{a}.\overrightarrow{b}+2\overrightarrow{b}.\overrightarrow{c}+2\overrightarrow{c}.\overrightarrow{a}$.
We put the values to get
$\begin{align}
  & {{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}^{2}} \\
 & =\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}+2\overrightarrow{a}.\overrightarrow{b}+2\overrightarrow{b}.\overrightarrow{c}+2\overrightarrow{c}.\overrightarrow{a} \\
 & ={{\lambda }^{2}}+{{\lambda }^{2}}+{{\lambda }^{2}}+0 \\
 & =3{{\lambda }^{2}} \\
\end{align}$
Taking square root, we get $\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|=\sqrt{3}\lambda $.
Now we assume the angle these vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ make with $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$ are ${{\theta }_{1}},{{\theta }_{2}},{{\theta }_{3}}$ respectively.
Therefore, $\cos {{\theta }_{1}}=\dfrac{\overrightarrow{a}\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)}{\left| \overrightarrow{a} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}\overrightarrow{c}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{{{\left| \overrightarrow{a} \right|}^{2}}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\left| \overrightarrow{a} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}$.
Putting the values, we get \[\cos {{\theta }_{1}}=\dfrac{\left| \overrightarrow{a} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\lambda }{\sqrt{3}\lambda }=\dfrac{1}{\sqrt{3}}\].
This gives \[{{\theta }_{1}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
similarly, $\cos {{\theta }_{2}}=\dfrac{\overrightarrow{b}\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)}{\left| \overrightarrow{b} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{b}\overrightarrow{c}}{\left| \overrightarrow{b} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{{{\left| \overrightarrow{b} \right|}^{2}}}{\left| \overrightarrow{b} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\left| \overrightarrow{b} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}$.
Putting the values, we get \[\cos {{\theta }_{2}}=\dfrac{\left| \overrightarrow{b} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\lambda }{\sqrt{3}\lambda }=\dfrac{1}{\sqrt{3}}\].
This gives \[{{\theta }_{2}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
Therefore, $\cos {{\theta }_{3}}=\dfrac{\overrightarrow{c}\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)}{\left| \overrightarrow{c} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\overrightarrow{c}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{b}+\overrightarrow{c}\overrightarrow{c}}{\left| \overrightarrow{c} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{{{\left| \overrightarrow{c} \right|}^{2}}}{\left| \overrightarrow{c} \right|\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\left| \overrightarrow{c} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}$.
Putting the values, we get \[\cos {{\theta }_{3}}=\dfrac{\left| \overrightarrow{c} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}=\dfrac{\lambda }{\sqrt{3}\lambda }=\dfrac{1}{\sqrt{3}}\].
This gives \[{{\theta }_{3}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
Therefore, ${{\theta }_{1}}={{\theta }_{2}}={{\theta }_{3}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$. Hence, $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$ is equally inclined with vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$.

Note: We need not to find the exact angle for the vector $\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right)$. The equality of the angles is enough to find the proof. As the three vectors are perpendicular, we can say $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0=\overrightarrow{b}.\overrightarrow{a}=\overrightarrow{c}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}$.