Question

A unit vector perpendicular to both the vectors \[i+j\] and \[j+k\] is
1) \[\dfrac{(-i-j+k)}{\sqrt{3}}\]
2) \[\dfrac{(i+j-k)}{3}\]
3) \[\dfrac{(i+j+k)}{\sqrt{3}}\]
4) \[\dfrac{(i-j+k)}{\sqrt{3}}\]

Answer 1

Hint: Cross product is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both vectors. And then to find the unit vector fro that vector we will use the formula for unit vector that is \[\widehat{v}=\dfrac{\overrightarrow{v}}{|\overrightarrow{v}|}\] .

Complete step by step answer:
First we should know the basic definition of vector and that is below:
The quantity which has both magnitude and direction is known as a vector. It is represented with the help of a straight line with an arrow head. This length of the straight line denotes the magnitude of the vector and the arrowhead shows the direction of the vector. Vectors are represented as \[\overrightarrow{v}\] .
Now coming to the question two vectors are given as:
Let’s say vector \[\overrightarrow{a}=\widehat{i}+\widehat{j}\]
And let’s say vector \[\overrightarrow{b}=\widehat{j}+\widehat{k}\]
Now as we must observe the definition of cross product of two vectors and that is :
Cross product is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both vectors. The Vector product of two vectors, \[\overrightarrow{a}\] and \[\overrightarrow{b}\] , is denoted by \[\overrightarrow{a}\times \overrightarrow{b}\] . Its resultant vector is perpendicular to \[\overrightarrow{a}\] and \[\overrightarrow{b}\] .
Therefore, let’s say vector \[\overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\]
By using Determinant method to find cross-product:
\[\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   1 & 1 & 0 \\
   0 & 1 & 1 \\
\end{matrix} \right|\]
\[\Rightarrow \overrightarrow{c}=\widehat{i}(1-0)-\widehat{j}(1-0)+\widehat{k}(1-0)\]
On simplifying further,
\[\Rightarrow \overrightarrow{c}=\widehat{i}-\widehat{j}+\widehat{k}\]
Here we got the vector that is perpendicular to both the vectors.
But in the question it is asked to find a unit vector that is perpendicular to both vectors
So, to find that we must know first what is a unit vector and how to find it for any vector.
Unit Vector is represented by the symbol \[\widehat{{}}\] , which is called a cap or hat, such as: \[\widehat{v}\] . It is given by \[\widehat{v}=\dfrac{\overrightarrow{v}}{|\overrightarrow{v}|}\] .
Where \[|\overrightarrow{v}|\] is for the norm or magnitude of the vector \[\overrightarrow{v}\] .
So, now we will find the unit vector by using the above definition and formulas we discussed.
Therefore,
We need to find, \[\widehat{c}=\dfrac{\overrightarrow{c}}{\left| \overrightarrow{c} \right|}\]
So first we need to find magnitude of vector \[\overrightarrow{c}\] and that can be calculated as:
\[\left| \overrightarrow{c} \right|=\sqrt{{{1}^{2}}+{{(-1)}^{2}}+{{1}^{2}}}\]
\[\Rightarrow \left| \overrightarrow{c} \right|=\sqrt{3}\]
Now we will put this value in the above formula of the unit vector.
Therefore we get,
\[\widehat{c}=\dfrac{1}{\sqrt{3}}\left( \widehat{i}-\widehat{j}+\widehat{k} \right)\]

So, the correct answer is “Option 4”.

Note: Most commonly in physics, vectors are used to represent displacement, velocity, and acceleration. Vectors are a combination of magnitude and direction, and are drawn as arrows. The length represents the magnitude and the direction of that quantity is the direction in which the vector is pointing.