Question

Show that the projection of vonto iis $\left(\mathbf{v}\mathbf{·}\mathbf{i}\right)\mathbf{i}\mathbf{.}$ Then show that we can always write a vector vas

$\mathbf{v}=\left(v·i\right)\mathbf{i}\mathbf{}+\mathbf{}\left(v·j\right)\mathbf{j}$

Short answer

Yes, the projection ofvontoiis$\left(\mathbf{v}·\mathbf{i}\right)\mathbf{i}$and the vectorvis$\left(\mathbf{v}·\mathbf{i}\right)\mathbf{i}\mathbf{}+\left(\mathbf{v}·\mathbf{j}\right)\mathbf{j}\mathbf{.}$

Step-by-step solution

Step 1. Given Information

We have to show that the projection of vonto iis$\left(\mathbf{v}·\mathbf{i}\right)\mathbf{i}$and vectorvis$\left(\mathbf{v}·\mathbf{i}\right)\mathbf{i}\mathbf{}+\left(\mathbf{v}·\mathbf{j}\right)\mathbf{j}\mathbf{.}$

Step 2. Showing the projection of v onto i

As we know If vand ware two nonzero vectors, the vector projection of vonto w is

${\mathbf{v}}_{\mathbf{1}}=\frac{\mathbf{v}\mathbf{·}\mathbf{w}}{{\left|\left|\mathbf{w}\right|\right|}^2}\mathbf{w}$

iis the unit vector, so the projection of vonto iis

role="math" localid="1646824336209" $$\begin{gathered} {\mathbf{v}}_{\mathbf{1}}=\frac{\mathbf{v}\mathbf{·}\mathbf{i}}{{\left|\left|\mathbf{i}\right|\right|}^2}\mathbf{i} \\ {\mathbf{v}}_{\mathbf{1}}=\left(\mathbf{v}\mathbf{·}\mathbf{i}\right)\mathbf{i}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[{\left|\left|\mathbf{i}\right|\right|}^2=1\right]..........\left(a\right) \end{gathered}$$

Hence proved that the projection ofvontoiis$\left(\mathbf{v}\mathbf{·}\mathbf{i}\right)\mathbf{i}.$

Step 3. Showing the vector v

We have found the projection of vonto i,now let's find the projection of vonto j.

jis the unit vector so,

$$\begin{gathered} {\mathbf{v}}_2=\frac{\mathbf{v}\mathbf{·}\mathbf{j}}{{\left|\left|\mathbf{j}\right|\right|}^2}\mathbf{j} \\ {\mathbf{v}}_2=\left(\mathbf{v}\mathbf{·}\mathbf{j}\right)\mathbf{j}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[{\left|\left|\mathbf{j}\right|\right|}^2=1\right].........\left(b\right) \end{gathered}$$

We can write vector vas

$$\begin{gathered} \mathbf{v}={\mathbf{v}}_{\mathbf{1}}+{\mathbf{v}}_{\mathbf{2}} \\ \mathbf{v}=\left(\mathbf{v}\mathbf{·}\mathbf{i}\right)\mathbf{i}\mathbf{}+\left(\mathbf{v}\mathbf{·}\mathbf{j}\right)\mathbf{j} \end{gathered}$$

Hence proved.