Question

If \[\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,,\overset{\to }{\mathop{c}}\,\]are three mutually perpendicular vectors, then the vector which is equally inclined to three vectors are
(a) \[\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\,\]
(b) \[\dfrac{\overset{\to }{\mathop{a}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|}+\dfrac{\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}+\dfrac{\overset{\to }{\mathop{c}}\,}{\left| \overset{\to }{\mathop{c}}\, \right|}\]
(c) \[\dfrac{\overset{\to }{\mathop{a}}\,}{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}}+\dfrac{\overset{\to }{\mathop{b}}\,}{{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}}+\dfrac{\overset{\to }{\mathop{c}}\,}{{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}}\]
(d) \[\left| \overset{\to }{\mathop{a}}\, \right|\overset{\to }{\mathop{a}}\,-\left| \overset{\to }{\mathop{b}}\, \right|\overset{\to }{\mathop{b}}\,+\left| \overset{\to }{\mathop{c}}\, \right|\overset{\to }{\mathop{c}}\,\]

Answer 1

Hint: Find the relation between vectors \[\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,\]and \[\overset{\to }{\mathop{c}}\,\] in case of being mutually perpendicular. Assume \[\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|\] is equal to a constant. Find the sum of square of \[\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,\]and\[\overset{\to }{\mathop{c}}\,\]. Use the formula to find the angle between vector\[\overset{\to }{\mathop{a}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\], \[\overset{\to }{\mathop{b}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]and\[\overset{\to }{\mathop{c}}\,\]and\[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]. Find the vector which is equally inclined.

Complete step by step answer:
Given to us are the three vectors \[\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,\]and \[\overset{\to }{\mathop{c}}\,\]which are mutually perpendicular to each other.
Two vectors \[\overset{\to }{\mathop{a}}\,\]and \[\overset{\to }{\mathop{b}}\,\]whose dot product \[\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=0\]are said to be orthogonal. Therefore, the vectors \[\overset{\to }{\mathop{a}}\,\]and \[\overset{\to }{\mathop{b}}\,\]are mutually perpendicular. Similarly, dot product of vector \[\overset{\to }{\mathop{b}}\,\]and \[\overset{\to }{\mathop{c}}\,\], is 0, and so, they are mutually perpendicular.
\[\begin{align}
  & \Rightarrow \overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{.b}}\,=0,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{.c}}\,=0,\overset{\to }{\mathop{c}}\,\overset{\to }{\mathop{.a}}\,=0 \\
 & \Rightarrow \overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{.b}}\,=\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{.c}}\,=\overset{\to }{\mathop{c}}\,\overset{\to }{\mathop{.a}}\,=0-(1) \\
\end{align}\]
Now, let us consider that\[\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=\lambda \], where \[\lambda \]is a constant.
We know the formula, \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)\].
\[\therefore \]The value of \[{{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}\]will be,
\[{{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\, \right)\]
We know from equation (1) that\[\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\,=0\].
\[\begin{align}
  & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( 0+0+0 \right) \\
 & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\times 0 \\
 & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}} \\
\end{align}\]
Let us substitute \[\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=\lambda \]on the RHS of the expression.
\[\begin{align}
  & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\lambda }^{2}}+{{\lambda }^{2}}+{{\lambda }^{2}} \\
 & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}=3{{\lambda }^{2}} \\
\end{align}\]
Taking square root on both sides,
\[\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|=\sqrt{3}\lambda \].
Now, let us find the angle between \[\overset{\to }{\mathop{a}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\].
The formula for find the angle between the two vectors is,
\[\cos \alpha =\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|.\left| \overset{\to }{\mathop{b}}\, \right|}\].
The cosine of the angle between 2 vectors is equal to the dot product of this vector divided by the product of vector magnitude.
Here, let us take the two vectors as \[\overset{\to }{\mathop{a}}\,\]and\[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\].
\[\therefore \cos \alpha =\dfrac{\overset{\to }{\mathop{a}}\,.\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)}{\left| \overset{\to }{\mathop{a}}\, \right|.\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}\]
Let us take dot product in the numerator. Substitute the value of \[\overset{\to }{\mathop{a}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]in the denominator.
\[\therefore \cos \alpha =\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{c}}\,}{\lambda \times \sqrt{3}\lambda }\]
We know, \[\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{c}}\,=0\]and \[{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}={{\lambda }^{2}}\].

\[\begin{align}
  & \therefore \cos \alpha =\dfrac{{{\lambda }^{2}}}{\sqrt{3}{{\lambda }^{2}}}=\dfrac{1}{\sqrt{3}} \\
 & \therefore \alpha ={{\cos }^{-1}}\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Similarly, we can find the angle between \[\overset{\to }{\mathop{b}}\,\]and\[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\].\[\begin{align}
  & \therefore \cos \alpha =\dfrac{\overset{\to }{\mathop{b}}\,.\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)}{\left| \overset{\to }{\mathop{b}}\, \right|.\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}=\dfrac{\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,}{\lambda \times \sqrt{3}\lambda } \\
 & =\dfrac{{{\lambda }^{2}}}{\sqrt{3}{{\lambda }^{2}}}=\dfrac{1}{\sqrt{3}} \\
\end{align}\]
We know, \[\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{a}}\,=0,\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,=0\], so above equation can be written as,
\[\therefore \cos \alpha =\dfrac{{{\lambda }^{2}}}{\sqrt{3}{{\lambda }^{2}}}=\dfrac{1}{\sqrt{3}}\]
\[\therefore \alpha ={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\]
Similarly, the angle between \[\overset{\to }{\mathop{c}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]is\[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
Hence, angle between \[\overset{\to }{\mathop{a}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]=\[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
Angle between \[\overset{\to }{\mathop{b}}\,\]and\[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]=\[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
Angle between \[\overset{\to }{\mathop{c}}\,\]and \[\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)\]=\[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\].
\[\therefore \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\,\]is equally inclined with \[\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,,\overset{\to }{\mathop{c}}\,\].

So, the correct answer is “Option A”.

Note: The dot product between 2 unit vectors and itself is simple to compute. If the vectors are of length one, then the dot product of three vectors \[\overset{\to }{\mathop{i}}\,,\overset{\to }{\mathop{j}}\,\]and \[\overset{\to }{\mathop{k}}\,\]becomes,
\[\overset{\to }{\mathop{i}}\,.\overset{\to }{\mathop{i}}\,=\overset{\to }{\mathop{j}}\,.\overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{k}}\,.\overset{\to }{\mathop{k}}\,=1\].
If the vectors are of length a, then the dot product becomes,
\[\overset{\to }{\mathop{i}}\,.\overset{\to }{\mathop{i}}\,=\overset{\to }{\mathop{j}}\,.\overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{k}}\,.\overset{\to }{\mathop{k}}\,=a\]