Question

Let \({\rm{v}} = \left( {\begin{aligned}a\\b\end{aligned}} \right)\). Describe the set \(H\) of vectors \(\left( {\begin{aligned}x\\y\end{aligned}} \right)\) that are orthogonal to \({\bf{v}}\). (Hint: Consider \({\rm{v}} = 0{\rm{ and v}} \ne 0\))

Short answer

The required set is \(H = \left\{ {\left( {\begin{aligned}{*{20}{c}}b\\{ - a}\end{aligned}} \right)} \right\}\).

Step-by-step solution

Definition of Orthogonal sets

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} = 0\end{aligned}\)

 Computing the required vector

As per the question, we have:

\({\bf{v}} = \left( {\begin{aligned}{*{20}{c}}a\\b\end{aligned}} \right)\)

Since \(0\)is orthogonal to all possible vectors.

Therefore, Let \({\bf{v}} \ne 0\),

Then, the span of \({\bf{v}} \ne 0\) is in \({\mathbb{R}^2}\). And, the line in \(H\) will be perpendicular to this vector.

So, let us have a vector of opposite entriesof \({\bf{v}}\) as:

\({\bf{u}} = \left( {\begin{aligned}{*{20}{c}}b\\{ - a}\end{aligned}} \right)\)

Find the product \({\bf{u}} \cdot {\bf{v}}\).

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{aligned}{*{20}{c}}a\\b\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}b\\{ - a}\end{aligned}} \right)\\ = ab - ab\\ = 0\end{aligned}\)

Now,

We can say:

\(H = \left\{ {\left( {\begin{aligned}{*{20}{c}}b\\{ - a}\end{aligned}} \right)} \right\}\)

Hence, this is the required answer.