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Question: If the vectors \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) are three mutually perpe...

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

Explanation

Solution

We first assume the magnitude of the three vectors a→,b→,c→\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 a→.b→=b→.c→=c→.a→=0\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0. Then we find the magnitude of the vector form (a→+b→+c→)\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 a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} are three mutually perpendicular vectors of equal magnitude.
We assume that ∣a→∣=∣b→∣=∣c→∣=λ\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|=\lambda .
Since the vectors are mutually perpendicular, we can say a→.b→=b→.c→=c→.a→=0\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0.
We have to show that (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) is equally inclined with vectors a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}.
We take the magnitude of (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right).
So, ∣a→+b→+c→∣2=a→.a→+b→.b→+c→.c→+2a→.b→+2b→.c→+2c→.a→{{\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
∣a→+b→+c→∣2 =a→.a→+b→.b→+c→.c→+2a→.b→+2b→.c→+2c→.a→ =λ2+λ2+λ2+0 =3λ2 \begin{aligned} & {{\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{aligned}
Taking square root, we get ∣a→+b→+c→∣=3λ\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|=\sqrt{3}\lambda .
Now we assume the angle these vectors a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} make with (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) are θ1,θ2,θ3{{\theta }_{1}},{{\theta }_{2}},{{\theta }_{3}} respectively.
Therefore, cos⁡θ1=a→(a→+b→+c→)∣a→∣∣a→+b→+c→∣=a→.a→+a→.b→+a→c→∣a→∣∣a→+b→+c→∣=∣a→∣2∣a→∣∣a→+b→+c→∣=∣a→∣∣a→+b→+c→∣\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⁡θ1=∣a→∣∣a→+b→+c→∣=λ3λ=13\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 θ1=cos⁡−1(13){{\theta }_{1}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right).
similarly, cos⁡θ2=b→(a→+b→+c→)∣b→∣∣a→+b→+c→∣=b→.a→+b→.b→+b→c→∣b→∣∣a→+b→+c→∣=∣b→∣2∣b→∣∣a→+b→+c→∣=∣b→∣∣a→+b→+c→∣\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⁡θ2=∣b→∣∣a→+b→+c→∣=λ3λ=13\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 θ2=cos⁡−1(13){{\theta }_{2}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right).
Therefore, cos⁡θ3=c→(a→+b→+c→)∣c→∣∣a→+b→+c→∣=c→.a→+c→.b→+c→c→∣c→∣∣a→+b→+c→∣=∣c→∣2∣c→∣∣a→+b→+c→∣=∣c→∣∣a→+b→+c→∣\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⁡θ3=∣c→∣∣a→+b→+c→∣=λ3λ=13\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 θ3=cos⁡−1(13){{\theta }_{3}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right).
Therefore, θ1=θ2=θ3=cos⁡−1(13){{\theta }_{1}}={{\theta }_{2}}={{\theta }_{3}}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right). Hence, (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) is equally inclined with vectors a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}.

Note: We need not to find the exact angle for the vector (a→+b→+c→)\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 a→.b→=b→.c→=c→.a→=0=b→.a→=c→.b→=a→.c→\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}.