Question

Projection

Show that ${proj}_v\mathbf{u}$ and $\mathbf{u}-{proj}_{\mathbf{v}}\mathbf{u}$ are orthogonal.

Short answer

This is proved that ${proj}_v\text{u}$and $\mathbf{u}-{\mathbf{proj}}_v\mathbf{u}$ are orthogonal.

Step-by-step solution

Step 1. Given information

A force is given as $\mathbf{F}=4\mathbf{i}-7\mathbf{j}$and displacement of body is $4$

Step 2. Concept used

Projection of u onto v is given by:

${proj}_{v}u=\left(\frac{u·v}{|v{|}^2}\right)v$

Vector scaler product or Dot product:

$a·b=|a\Vert b|\cos \theta$

The dot product of two orthogonal vectors is zero.

Step 3. Calculation

The dot product of two orthogonal vectors is zero,

$\begin{array}{l}{\mathrm{proj}}_{v}u\cdot \left(u-{\mathrm{proj}}_{v}u\right)=\left({\mathrm{proj}}_{v}u\right)\cdot u-\left({\mathrm{proj}}_{v}u\right)\cdot \left({\mathrm{proj}}_{v}u\right)\\ {\mathrm{proj}}_{v}u\cdot \left(u-{\mathrm{proj}}_{v}u\right)=\left(\left(\frac{u\cdot v}{|v{|}^2}\right)v\right)\cdot u-{\left({\mathrm{proj}}_{v}u\right)}^2\\ {\mathrm{proj}}_{v}u\cdot \left(u-{\mathrm{proj}}_{v}u\right)=\left(\left(\frac{u\cdot v}{|v{|}^2}\right)v\right)\cdot u-{\left(\left(\frac{u\cdot v}{|v{|}^2}\right)v\right)}^2\\ {\mathrm{proj}}_{v}u\cdot \left(u-{\mathrm{proj}}_{v}u\right)=\frac{{(u\cdot v)}^2}{|v{|}^2}-\frac{{(u\cdot v)}^2|v{|}^2}{|v{|}^4}\\ =\frac{{\left(uv\right)}^2}{|v{|}^2}-\frac{{\left(uv\right)}^2}{|v{|}^2}\\ =0\end{array}$