Question

How do you find the unit vector perpendicular to the vector (\[4\widehat{i}-3\widehat{j}\])?

Answer 1

Hint: We need to understand the relation between the vectors when they are perpendicular to each other and the one of the required vectors is a unit vector. We can use this relation in order to find the perpendicular vector to solve this problem.

Complete answer:
We know that the two vectors are said to be perpendicular if the scalar product or the dot product of the two of them becomes zero. The dot product is defined as the product of two quantities when they are in the same direction, i.e., there exists a parallel element of both of the quantities. In case of two quantities which are in perpendicular directions, as a result of the definition, the magnitude turns out to be zero.

Using this let us consider a vector \[\overrightarrow{A}=a\widehat{i}+b\widehat{j}\] which is a unit vector and is along a perpendicular direction to the given vector. From the definition of the dot products, we can write the product of these two being equal to zero. This can be given as –
\[\begin{align}
  & \overrightarrow{A}.(4\widehat{i}-3\widehat{j})=(a\widehat{i}+b\widehat{j}).(4\widehat{i}-3\widehat{j}) \\
 & \Rightarrow 0=4a-3b \\
 & \therefore a=\dfrac{3b}{4} \\
\end{align}\]

Now, we know that the vector A is a unit vector, i.e., the magnitude of this vector will be unity, which can be found as –
\[\begin{align}
  & \left| \overrightarrow{A} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
 & \Rightarrow \left| \overrightarrow{A} \right|=\sqrt{{{\left( \dfrac{3b}{4} \right)}^{2}}+{{b}^{2}}} \\
 & \Rightarrow \left| \overrightarrow{A} \right|=\sqrt{\dfrac{9{{b}^{2}}+16{{b}^{2}}}{16}} \\
 & \Rightarrow \pm 1=\dfrac{5b}{4} \\
 & \therefore b=\pm \dfrac{4}{5} \\
\end{align}\]

Also, we can find ‘a’ as –
 \[\begin{align}
  & a=\dfrac{3b}{4} \\
 & \Rightarrow a=\dfrac{3}{4}\times \pm \dfrac{4}{5} \\
 & \therefore a=\pm \dfrac{3}{5} \\
\end{align}\]

So, the required vector A can be given as –
\[\overrightarrow{A}=\pm \left( \dfrac{3}{5}\widehat{i}+\dfrac{4}{5}\widehat{j} \right)\]

These are the vectors which are perpendicular to the given vector and are unit vectors as required by the problem.
This is the required solution.

Note:
We know that we have obtained two vectors which satisfy the condition of being unit vectors and perpendicular to the given vector. This is possible mathematically as for every vector there can be an anti-vector which is in the opposite direction.