Question

The unit vector perpendicular to vectors \[\widehat{i}-\widehat{j}\] and \[\widehat{i}+\widehat{j}\] forming a right-handed system is:
\[\left( \text{a} \right)\text{ }\widehat{k}\]
\[\left( \text{b} \right)\text{ }-\widehat{k}\]
\[\left( \text{c} \right)\text{ }\dfrac{1}{\sqrt{2}}\left( \widehat{i}-\widehat{j} \right)\]
\[\left( \text{d} \right)\text{ }\dfrac{1}{\sqrt{2}}\left( \widehat{i}+\widehat{j} \right)\]

Answer 1

Hint: To solve the given question, we will first see the definition of vectors. Then, after doing this, we will take the cross – product of vectors given in the question to obtain the perpendicular vector. Then, to find the unit vector, we will divide the vector obtained by its magnitude.

Complete step by step solution:
Before solving this question, we must know what vectors are. A vector is an object or an entity that has both a magnitude and a direction. In other words, a vector is a directed line segment whose length is the magnitude of the vector and with an arrow indicating the direction. Now, we assume that the first vector is denoted by \[\overrightarrow{a}\] and the second vector is denoted by \[\overrightarrow{b}.\] Thus, we have,
\[\overrightarrow{a}=\widehat{i}-\widehat{j}\]
\[\overrightarrow{b}=\widehat{i}+\widehat{j}\]
Now, the vector which is perpendicular to both the vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] is obtained by the cross – product of these vectors. The product of two vectors: \[\overrightarrow{A}=p\widehat{i}+q\widehat{j}+r\widehat{k}\] and \[\overrightarrow{B}=x\widehat{i}+y\widehat{j}+z\widehat{k}\] is given by
\[\overrightarrow{A}\times \overrightarrow{B}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   p & q & r \\
   x & y & z \\
\end{matrix} \right|\]
Thus, the cross product of \[\overrightarrow{a}\] and \[\overrightarrow{b}\] will be given by
\[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   1 & -1 & 0 \\
   1 & 1 & 0 \\
\end{matrix} \right|\]
\[\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\widehat{i}\left[ \left( -1 \right)\left( 0 \right)-\left( 0 \right)\left( 1 \right) \right]-\widehat{j}\left[ \left( 1 \right)\left( 0 \right)-\left( 0 \right)\left( 1 \right) \right]+\widehat{k}\left[ \left( 1 \right)\left( 1 \right)-\left( -1 \right)\left( 1 \right) \right]\]
\[\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=0\widehat{i}+0\widehat{j}+2\widehat{k}\]
\[\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=2\widehat{k}\]
Now, we denote the cross product obtained above as
\[\Rightarrow \overrightarrow{c}=2\widehat{k}\]
But we are given the question that we have to find the perpendicular vector which is also a unit vector. The unit vector of a vector can be obtained by dividing the vector with its magnitude. Thus, the unit vector of the vector \[\overrightarrow{A}=p\widehat{i}+q\widehat{j}+r\widehat{k}\] is represented by \[\widehat{A}\] and is given by
\[\widehat{A}=\dfrac{p\widehat{i}+q\widehat{j}+r\widehat{k}}{\sqrt{{{p}^{2}}+{{q}^{2}}+{{r}^{2}}}}\]
In our case, p = 0, q = 0 and r = 2. So, we have,
\[\widehat{c}=\dfrac{2\widehat{k}}{\sqrt{{{0}^{2}}+{{0}^{2}}+{{2}^{2}}}}\]
\[\Rightarrow \widehat{c}=\dfrac{2\widehat{k}}{\sqrt{{{2}^{2}}}}\]
\[\Rightarrow \widehat{c}=\dfrac{2\widehat{k}}{2}\]
\[\Rightarrow \widehat{c}=\widehat{k}\]
Hence, option (a) is the right answer.

Note: The alternate method of solving the question is shown below. As we know that the vectors \[\overrightarrow{a}=\widehat{i}-\widehat{j}\] and \[\overrightarrow{b}=\widehat{i}+\widehat{j}\] are the vectors in the x – y plane. So, if a vector is perpendicular to x – y plane, it should be in the direction of \[\widehat{k}.\] Thus, the perpendicular vector will be
\[\overrightarrow{c}=n\widehat{k}\]
Thus, the perpendicular unit vector is \[\overrightarrow{c}=\widehat{k}.\]