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 $v$

$\mathbf{u}=⟨2,9⟩,\mathbf{v}=⟨-3,4⟩$

Short answer

1. The projection is
2. By resolving u into u1 and u2 we get,

Step-by-step solution

a.Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨2,9⟩,\mathbf{v}=⟨-3,4⟩$

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 $v$,

$\begin{array}{l}{\mathrm{proj}}_{v}u=\left\{\frac{u\cdot v}{|v{|}^2}\right\}v\\ =\left(\frac{⟨2,9⟩\cdot ⟨-3,4⟩}{\left({(-3)}^2+{\left(4\right)}^2\right)}\right)(-3,4⟩\\ =\left(\frac{-6+36}{(9+16)}\right)⟨-3,4⟩\\ =\left(\frac{30}{25}\right)⟨-3,4⟩\\ =\frac{6}{5}⟨-3,4⟩\\ =〈\frac{-18}{5},\frac{24}{5}〉\end{array}$

b.Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨2,9⟩,\mathbf{v}=⟨-3,4⟩$ &

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

$\begin{array}{l}{u}_1={\mathrm{proj}}_{v}u\\ u_1=〈\frac{-18}{5},\frac{24}{5}〉\\ u_2=u-{\mathrm{Proj}}_{v}u\\ u_2=⟨2,9⟩-〈\frac{-18}{5},\frac{24}{5}〉\\ u_2=〈2-\left(\frac{-18}{5}\right),9-\frac{24}{5}〉\\ u_2=〈2+\frac{18}{5},9-\frac{24}{5}〉\\ u_2=〈\frac{10+18}{5},\frac{45-24}{5}〉\\ u_2=〈\frac{28}{5},\frac{21}{5}〉\end{array}$