Q. 63 - Calculus

Question

Let \(u\) be a nonzero vector and let \(v\) be any vector. Show that the decomposition \(v=v_{\parallel }+v_{\perp }\), where \(v_{\parallel }\) is parallel to \(u\) and \(v_{\perp }\) is orthogonal to \(u\), is unique.

Step-by-Step Solution

Answer

It is shown that the decomposition \(v=v_{\parallel }+v_{\perp }\), where \(v_{\parallel }\) is parallel to \(u\) and \(v_{\perp }\) is orthogonal to \(u\), is unique. 

Step 1. Given Information.

It is given that \(u\) is a nonzero vector and \(v\) is any vector.  We have to show that the decomposition \(v=v_{\parallel }+v_{\perp }\), where \(v_{\parallel }\) is parallel to \(u\) and \(v_{\perp }\) is orthogonal to \(u\), is unique.

Step 2. Show

To show that the decomposition \(v=v_{\parallel }+v_{\perp }\) is unique.

Let the decomposition of vector \(v\) is not unique and there are two decompositions of it.

The two decompositions of \(v\) are \(v=v_{\parallel }+v_{\perp }\) and \(v=v^{'}_{\parallel }+v^{'}_{\perp }\).

Now, in \(v\) are \(v=v_{\parallel }+v_{\perp }\) let \(v_{\parallel }\) is parallel to \(u\) and \(v_{\perp }\) is orthogonal to \(u\).

As both decompositions are of \(v\), thus, both are equal.

\(v_{\parallel }+v_{\perp }=v^{'}_{\parallel }+v^{'}_{\perp }\) ......(i)

Step 3. Show

We can write equation (i) as

\(v_{\parallel }-v^{'}_{\parallel }=v^{'}_{\perp }-v_{\perp }\) ...(ii)

Since \(v_{\parallel }\) and \(v^{'}_{\parallel }\) are both parallel to \(u\), thus, \(v_{\parallel }-v^{'}_{\parallel }\) is parallel to \(u\).

And \(v_{\perp }\) and  \(v^{'}_{\perp }\) are orthogonal to \(u\), thus, \(v^{'}_{\perp }-v_{\perp }\) is orthogonal to \(u\).  

This can be possible if both sides of equation (ii) can be equal to \(0\).

Thus, \(v_{\parallel }-v^{'}_{\parallel }=0\), so, \(v_{\parallel }=v^{'}_{\parallel }\).

Similarly,  \(v^{'}_{\perp }=v_{\perp }\).

Thus, the decomposition of \(v\) is unique.