Question

Let vand wdenote two nonzero vectors. Show that the vector$\mathit{v}-\alpha \mathit{w}$is orthogonal to wif$\alpha =\frac{\mathit{v}\mathbf{·}\mathit{w}}{{\left|\left|\mathit{w}\right|\right|}^2}$

Short answer

The given vectors are orthogonal.

Step-by-step solution

Step 1. Given Information 

There are two nonzero vectorsvandw.

We have to show that the vectors$\mathbf{v}-\mathrm{\alpha }\mathbf{w}$is orthogonal towif$\alpha =\frac{\mathbf{v}·\mathbf{w}}{{\left|\left|\mathbf{w}\right|\right|}^2}.$

Step 2. Showing the vectors are orthogonal 

As we know if any two vectors are orthogonal then their dot product is equal to zero.

Let's find the dot product of $\left(v\mathrm{-}\mathrm{\alpha }w\right)\mathrm{and}\left(w\right)$

So,

$$\begin{gathered} \mathbf{(}\mathbf{v}\mathbf{-}\mathbf{\alpha }\mathbf{w}\mathbf{)}·\mathbf{\left(}\mathbf{w}\mathbf{\right)} \\ =\mathbf{v}·\mathbf{w}\mathbf{}-\mathrm{\alpha }{\left|\left|\mathbf{w}\right|\right|}^2\left[\mathbf{a}·\mathbf{a}={\left|\left|\mathbf{a}\right|\right|}^2\right] \\ =\mathbf{v}·\mathbf{w}-\left(\frac{\mathbf{v}·\mathbf{w}}{{\left|\left|\mathbf{w}\right|\right|}^2}\right){\left|\left|\mathbf{w}\right|\right|}^2 \\ =\mathbf{v}·\mathbf{w}\mathbf{}-\mathbf{v}·\mathbf{w} \\ =0 \end{gathered}$$

Therefore, given vectors are orthogonal.