Question

Vector Projection of $v$ onto $v$

(a) Calculate ${proj}_{\mathbf{v}}\mathbf{u}$

(b) Resolve $u$ into $u1$ and $u2$, where $u1$ is parallel to $v$ and $v2$ is orthogonal to $v$

$\mathbf{u}=⟨11,3⟩,\mathbf{v}=⟨-3,-2⟩$

Short answer

1. The projection is ${proj}_{v}u=⟨9,6⟩$
2. By resolving u into u1 and u2 we get,$u_1=⟨9,6⟩,u_2=⟨2,-3⟩$

Step-by-step solution

a.Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨11,3⟩,\mathbf{v}=⟨-3,-2⟩$

Step 2. Concept used

If the vector u is resolved into u1 and u2 , where u1 is parallel to v and u2 is orthogonal to v , then

$\begin{array}{l}{u}_1={\mathrm{Proj}}_{v}u\\ u_2=u-{\mathrm{Proj}}_{v}u\end{array}$

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:

$a·b=|a\Vert b|\cos \theta$

Use the formula of projection of vector.

Step 3. Calculation

Projection of u on $v$,

$\begin{array}{l}{\mathrm{proj}}_{v}u=\left\{\frac{u\cdot v}{|v{|}^2}\right\}v\\ =\left(\frac{⟨11,3⟩\cdot ⟨-3,-2⟩}{\left({(-3)}^2+{(-2)}^2\right)}\right)⟨-3,-2⟩\\ =\left(\frac{-33-6}{(9+4)}\right)⟨-3,-2⟩\\ =\left(\frac{-39}{13}\right)⟨-3,-2⟩\\ =-3⟨-3,-2⟩\\ =⟨9,6⟩\\ =〈\frac{-1}{2},\frac{3}{2}〉\end{array}$

b.Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨11,3⟩,\mathbf{v}=⟨-3,-2⟩$ &

${proj}_{v}u=⟨9,6⟩$

Step 2. Concept used

If the vector u is resolved into u1 and u2 , where u1 is parallel to v and u2 is orthogonal to v , then

$\begin{array}{l}{u}_1={\mathrm{Proj}}_{v}u\\ u_2=u-{\mathrm{Proj}}_{v}u\end{array}$

Projection of u onto v is given by:

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

Resolve by using the given definition.

Step 3. Calculation

$\begin{array}{l}pro{j}_{v}u=⟨9,6⟩\\ u_1=pro{j}_{v}u\\ u_1=⟨9,6⟩\\ u_2=u-Pro{j}_{v}u\\ u_2=⟨11,3⟩-⟨9,6⟩\\ u_2=⟨11-9,3-6⟩\\ u_2=⟨2,-3⟩\end{array}$