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 $v$

$\mathbf{u}=⟨-2,4⟩,\mathbf{v}=⟨1,1⟩$

Short answer

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

Step-by-step solution

a.Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨-2,4⟩,\mathbf{v}=⟨1,1⟩$

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:

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

Use the formula of projection of vector.

Step 3. Calculation

Projection of u on $v$,

$\begin{array}{l}{\mathrm{proj}}_{v}u=\left\{\frac{u\cdot v}{|v{|}^2}\right\}v\\ =\left\{\frac{(-2)\left(1\right)+\left(4\right)\left(1\right)}{{\left\{\sqrt{1^2+1^2}\right\}}^2}\right\}⟨1,1⟩\\ =\left\{\frac{-2+4}{2}\right\}⟨1,1⟩\\ =1⟨1,1⟩\\ =⟨1,1⟩\end{array}$

b.Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨-2,4⟩,\mathbf{v}=⟨1,1⟩$ &

${proj}_{v}u=⟨1,1⟩$

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

$\begin{array}{l}pro{j}_{v}u=⟨1,1⟩\\ u_1=pro{j}_{v}u\\ u_1=⟨1,1⟩\\ u_2=u-{\mathrm{Proj}}_{v}u\\ u_2=⟨-2,4⟩-⟨1,1⟩\\ u_2=⟨-2-1,4-1⟩\\ u_2=⟨-3,3⟩\end{array}$