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Question: If \[\overrightarrow{a}\overrightarrow{,b},\overrightarrow{c}\] are mutually perpendicular vectors o...

If a→,b→,c→\overrightarrow{a}\overrightarrow{,b},\overrightarrow{c} are mutually perpendicular vectors of equal magnitude, then a→+b→+c→\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} is equally inclined to
(a). a→\overrightarrow{a} only
(b). a→,c→\overrightarrow{a},\overrightarrow{c} only
(c). All a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}
(d). None of these

Explanation

Solution

First of all we have to determine the relation between the vectors a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} as in the mutually perpendicular case. Then we have to assume that ∣a→∣=∣b→∣=∣c→∣\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right| which is equal to its constant. Then we have to determine the sum of the square for a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}. After that you have to use the formula for finding the angle between the vectors a→\overrightarrow{a} and (a→+b→+c→),b→\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right),\overrightarrow{b} and (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) and c(a→+b→+c→)c\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right). Now use can easily find out the vector which is equally inclined.

Complete step by step solution:
In the given question we have given that vectors are equal in magnitudes a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} are of equal magnitudes
So, ∣a→∣=∣b→∣=∣c→∣\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|
Also as per given in the question a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} are mutually perpendicular to each other.
So, a→.b→=b→.c→=c→.a→=0\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0
Now we know that
(a→+b→+c→).a→=∣a→+b→+c→∣∣a→∣cos⁡α\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right).\overrightarrow{a}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{a} \right|\cos \alpha Where α=angle\alpha =angle between (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) and a→\overrightarrow{a}.
Now we will open the bracket of the light hand side and the resulted equation will be,
a→,a→+b→.a→+c→.a→=∣a→+b→+c→∣∣a→∣cos⁡α.............(i)\overrightarrow{a},\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{a}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{a} \right|\cos \alpha .............\left( i \right)
Now we will use the property of vectors
Property: a→.b→=b→.a→\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a} and a→,a→=∣a→∣2\overrightarrow{a},\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}} apply the properties on equation (i)
⇒∣a→∣2+0+0=∣a→+b→+c→∣∣a→∣cos⁡α\Rightarrow {{\left| \overrightarrow{a} \right|}^{2}}+0+0=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{a} \right|\cos \alpha
Now we will find the value of cos⁡α\cos \alpha from the above equation
⇒cos⁡α=∣a→∣∣a→+b→+c→∣\Rightarrow \cos \alpha =\dfrac{\left| \overrightarrow{a} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}
Now we will take
(a→+b→+c→),b→=∣a→+b→+c→∣∣b→∣cos⁡β\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right),\overrightarrow{b}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \beta Where β=angle\beta =angle between (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) and b→\overrightarrow{b}.
Now we will open the bracket of the left-handed side and the resulted equation will be,
⇒a→.b→+b→.b→+c→.b→=∣a→+b→+c→∣∣b→∣cos⁡β........(ii)\Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{b}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \beta ........(ii)
Now we will again use the property of vectors
Property: a→,b→=b→.a→\overrightarrow{a},\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a} and a→.a→=∣a→∣2\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}} apply these properties on equation (ii)
⇒0+∣b→2∣+0=∣a→+b→+c→∣∣b→∣cos⁡β\Rightarrow 0+\left| {{\overrightarrow{b}}^{2}} \right|+0=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \beta
Now we will find the value of cos⁡β\cos \beta from the above equation
⇒cos⁡β=∣b→∣∣a→+b→+c→∣\Rightarrow \cos \beta =\dfrac{\left| \overrightarrow{b} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}
Now we will take
(a→+b→+b→).c→=∣a→+b→+c→∣∣b→∣cos⁡γ\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{b} \right).\overrightarrow{c}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos \gamma Where γ=\gamma = angle between (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) and c→\overrightarrow{c}.
Now we will open the bracket of the left hand side and the resulted equation will be,
⇒a→.c→+b→.c→=c→.c→=∣a→+b→+c→∣∣c→∣cos⁡γ.........(iii)\Rightarrow \overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{c}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{c} \right|\cos \gamma .........(iii)
Now we will again use the property of vectors
Property: a→.b→=b→.a→\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a} and a→.a→=∣a→∣2\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}} apply the perpendicular on equation (iii)
⇒0+0+∣c→∣2=∣a→+b→+c→∣∣c→∣cos⁡γ\Rightarrow 0+0+{{\left| \overrightarrow{c} \right|}^{2}}=\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|\left| \overrightarrow{c} \right|\cos \gamma
Now we will find the value of cos⁡γ\cos \gamma from the above equation
⇒cos⁡γ=∣c→∣∣a→+b→+c→∣\Rightarrow \cos \gamma =\dfrac{\left| \overrightarrow{c} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}
As calculated above in the question
⇒cos⁡α=∣a→∣∣a→+b→+c→∣,cos⁡β=∣b→∣∣a→+b→+c→∣,cos⁡γ=∣c→∣∣a→+b→+c→∣\Rightarrow \cos \alpha =\dfrac{\left| \overrightarrow{a} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|},\cos \beta =\dfrac{\left| \overrightarrow{b} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|},\cos \gamma =\dfrac{\left| \overrightarrow{c} \right|}{\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|}
According to the information given in the question we know the
⇒∣a→∣=∣b→∣=∣c→∣\Rightarrow \left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|
Thus we can say that cos⁡α=cos⁡β=cos⁡γ\cos \alpha =\cos \beta =\cos \gamma and according to the above equation we can say that
⇒α=β=γ\Rightarrow \alpha =\beta =\gamma
Therefore, (a→+b→+c→)\left( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right) is equation inclined to a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}.

So, the correct answer is “Option c”.

Note: We know that the dot product in between 22 unit vectors is very simple to compute. If the vector is length type then we have 33 vector of dot product i.e. i→.j→,k→\overrightarrow{i}.\overrightarrow{j},\overrightarrow{k}. It becomes, i→,i→=j→.j→=k→.k→=1\overrightarrow{i},\overrightarrow{i}=\overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1 if vector are of length a→\overrightarrow{a} then, we get i→,i→=j→.j→=k→.k→=a\overrightarrow{i},\overrightarrow{i}=\overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=a.