Question

The unit vector perpendicular to both $\overset{\hat{\ }}{\mathop{i}}\,+\overset{\hat{\ }}{\mathop{j}}\,$and $\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$is
(a) $\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$
(b) $\overset{\hat{\ }}{\mathop{i}}\,+\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$
(c) $\dfrac{\overset{\hat{\ }}{\mathop{i}}\,+\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{3}}$
(d) $\dfrac{\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{3}}$

Answer 1

Hint:Vector is a quantity that has both magnitudes as well as direction. It is generally represented by an arrow. Whose direction is as same as that of the quantity but the length of it is proportional to the quantity's magnitude. By the way, a vector has magnitude and direction, it doesn't have any position. That is, it can be as long as its length is not changed, a vector can't be changed if it is displaced parallel to itself. Also, a Unit vector is one whose length is unity as well as,$\overset{\hat{\ }}{\mathop{i}}\,$,$\overset{\hat{\ }}{\mathop{j}}\,$, $\overset{\hat{\ }}{\mathop{k}}\,$ are the unit vectors in the x, y, z-direction

Formula Used: For finding unit vector, \[\overset{\hat{\ }}{\mathop{c}}\,=\dfrac{\left( \overset{\to }{\mathop{g}}\,\text{ }x\text{ }\overset{\to }{\mathop{h}}\, \right)}{\left| \overset{\to }{\mathop{g}}\,\text{ }x\text{ }\overset{\to }{\mathop{h}}\, \right|}\]



Complete step-by-step solution:
Let the both vectors perpendicular to a$\overset{\hat{\ }}{\mathop{i}}\,$+ b$\overset{\hat{\ }}{\mathop{j}}\,$+ c$\overset{\hat{\ }}{\mathop{k}}\,$.
That’s why, vector$\overset{\to }{\mathop{g}}\,$= $\overset{\hat{\ }}{\mathop{i}}\,$+ $\overset{\hat{\ }}{\mathop{j}}\,$+ 0$\overset{\hat{\ }}{\mathop{k}}\,$
                    vector$\overset{\to }{\mathop{h}}\,$= 0$\overset{\hat{\ }}{\mathop{i}}\,$+ $\overset{\hat{\ }}{\mathop{j}}\,$+ $\overset{\hat{\ }}{\mathop{k}}\,$
So, unit vector \[\overset{\hat{\ }}{\mathop{c}}\,=\dfrac{\left( \overset{\to }{\mathop{g}}\,\text{ }x\text{ }\overset{\to }{\mathop{h}}\, \right)}{\left| \overset{\to }{\mathop{g}}\,\text{ }x\text{ }\overset{\to }{\mathop{h}}\, \right|}\]
     \[\left( \overset{\to }{\mathop{g}}\,\text{ }x\text{ }\overset{\to }{\mathop{h}}\, \right)=\left| \begin{matrix}
   \overset{\hat{\ }}{\mathop{i}}\, \\
   1 \\
   0 \\
\end{matrix}\begin{matrix}
   \overset{\hat{\ }}{\mathop{j}}\, \\
   1 \\
   1 \\
\end{matrix}\begin{matrix}
   \overset{\hat{\ }}{\mathop{k}}\, \\
   0 \\
   1 \\
\end{matrix} \right|\]
                               $=\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$
Now,
\[\left| \overset{\to }{\mathop{g}}\,\text{ }x\text{ }\overset{\to }{\mathop{h}}\, \right|=\sqrt{\left( 1+1+1 \right)}=\sqrt{3}\]
Therefore,
Unit Vector \[\overset{\hat{\ }}{\mathop{c}}\,\text{ }=\dfrac{\left( i-j+k \right)}{\sqrt{3}}\]
Hence, the unit vector which is perpendicular to both the vector $\overset{\hat{\ }}{\mathop{i}}\,+\overset{\hat{\ }}{\mathop{j}}\,$and $\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$is $\dfrac{\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{3}}$
So, the correct option for this question is option (d).

Additional Information: Since all quantities are not a vector. Only some of them are vectors, those having a direction with magnitude. Examples of these quantities are:- Displacement, Velocity, Acceleration, Force, etc. While some of them have magnitude but not direction. So, these quantities are called scalar quantities. Examples of these quantities are:- volume, density, speed, energy, mass, etc.

Note: Don't get confused with the vector symbol as an arrow ($\to $) is a symbol of the normal vector, as well as a cap ( ^ ), which is a symbol of the unit vector. Also, don't get confused about the formula for finding vectors. If you do step-wise, then you won't get into any trouble.