A vector perpendicular to the vector \[(\widehat{i}+2\widehat{j})\] and having the magnitude \[3\sqrt{5}\]units is –
\[\begin{align}
  & \text{A) }3\widehat{i}+6\widehat{j} \\
 & \text{B) 6}\widehat{i}-3\widehat{j} \\
 & \text{C) }4\widehat{i}-2\widehat{j} \\
 & \text{D) }\widehat{i}-2\widehat{j} \\
\end{align}\]

Hint: Two vectors can be perpendicular to each other when they meet a certain condition. We can find the magnitude and direction of a vector which is perpendicular to another very easily. The vector products play an important role in finding the solution.

Complete answer:
Two vectors are said to be perpendicular, if and only if the dot product or the scalar product of the two vectors become zero.
We know that there are two types of vector multiplication – the scalar and the cross products.
The scalar product is given by –
\[A.B=\left| A \right|\left| B \right|\cos \theta \]
The vector product is given by –
\[A\times B=\left| A \right|\left| B \right|\sin \theta \widehat{n}\]
From the equations we can understand that when,
\[\theta ={{90}^{0}}\]
The dot product becomes zero.
So, let us compute for the unknown vector using this concept.
Let,
\[\widehat{A}=x\widehat{i}+y\widehat{j}\]
be the unknown vector and
\[\widehat{B}=\widehat{i}+2\widehat{j}\]
Now, we can just verify this as -
\[\begin{align}
  & \left| A \right|=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
 & \text{which is given as 3}\sqrt{5} \\
 & \Rightarrow \text{ }\sqrt{{{x}^{2}}+{{y}^{2}}}=3\sqrt{5}\text{ } \\
 & \text{or} \\
 & \Rightarrow \text{ }{{x}^{2}}\text{+}{{\text{y}}^{2}}\text{=45 --(1)} \\
 & \\
\end{align}\]
Now let us do the dot product of the two vectors A and B.
\[\begin{align}
  & A.B=(x\widehat{i}+y\widehat{j}).(\widehat{i}+2\widehat{j}) \\
 & \text{We know that } \\
 & \widehat{i}.\widehat{i}=1, \\
 & \widehat{j}.\widehat{j}=1\text{ } \\
 & \text{and } \\
 & \widehat{i}.\widehat{j}=0 \\
 & A.B=x+2y \\
 & \text{but,} \\
 & A.B=0 \\
 & \Rightarrow \text{ }x+2y=0 \\
\end{align}\]
Now we have two equations in x and y. We can substitute one of them as –
\[x=-2y\]
Substitute in (1),
\[\begin{align}
  & x=-2y \\
 & {{(-2y)}^{2}}+{{y}^{2}}=45 \\
 & \Rightarrow \text{ }4{{y}^{2}}+{{y}^{2}}=45 \\
 & \Rightarrow \text{ }5{{y}^{2}}=45 \\
 & \Rightarrow \text{ }y=\pm 3 \\
 & \text{Also, } \\
 & x=-2y \\
 & \Rightarrow \text{ }x=\mp 6 \\
\end{align}\]
From the above we can conclude that the unknown vector is given by –
\[\widehat{A}=\pm 6\widehat{i}\mp 3\widehat{j}\]
Both the equations of A are valid and are perpendicular to the given vector B. We have to choose the option which has the correct answer.

The option which has the right answer is B.

Note:
Any vector can have an image vector which exists as an anti-parallel pair if they are lines. In the given case, the two position vectors which we solved live in diagonally opposite quadrants making both of them equally perpendicular to the given vector
The two vectors which are found to be perpendicular to the given vector are in opposite directions.