Let $\mathbf{a}$ and $\mathbf{b}$ be orthogonal vectors.  If $\operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix},$ then find $\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 3 \\ -3 \end{pmatrix}.$
Since $\begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix}$ is the projection of $\begin{pmatrix} 3 \\ -3 \end{pmatrix}$ onto $\mathbf{a},$
\[\begin{pmatrix} 3 \\ -3 \end{pmatrix} - \begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix} = \begin{pmatrix} \frac{18}{5} \\ -\frac{9}{5} \end{pmatrix}\]is orthogonal to $\mathbf{a}.$  But since $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, $\begin{pmatrix} \frac{18}{5} \\ -\frac{9}{5} \end{pmatrix}$ is a scalar multiple of $\mathbf{b}.$

[asy]
usepackage("amsmath");

unitsize(1 cm);

pair A, B, O, P, Q, V;

A = (1,2);
B = (2,-1);
O = (0,0);
V = (3,-3);
P = (V + reflect(O,A)*(V))/2;

draw(O--V,Arrow(6));
draw(O--P,Arrow(6));
draw(P--V,Arrow(6));
draw((-1,0)--(4,0));
draw((0,-4)--(0,1));

label("$\begin{pmatrix} 3 \\ -3 \end{pmatrix}$", V, SE);
label("$\begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix}$", P, W);
[/asy]

Furthermore,
\[\begin{pmatrix} 3 \\ -3 \end{pmatrix} - \begin{pmatrix} \frac{18}{5} \\ -\frac{9}{5} \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix}\]is a scalar multiple of $\mathbf{a},$ and therefore orthogonal to $\mathbf{b}.$  Hence, $\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \boxed{\begin{pmatrix} \frac{18}{5} \\ -\frac{9}{5} \end{pmatrix}}.$
\begin{pmatrix}\frac{18}{5}\-\frac{9}{5}\end{pmatrix}
Level 4
Precalculus
Problem: Let $\mathbf{a}$ and $\mathbf{b}$ be orthogonal vectors.  If $\operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix},$ then find $\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 3 \\ -3 \end{pmatrix}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the projection of the vector \begin{pmatrix} 3 \\ -3 \end{pmatrix} onto the vector \mathbf{b}, we first need to understand the relationship between the projections onto orthogonal vectors. Given that \mathbf{a} and \mathbf{b} are orthogonal, we can use the property that the sum of the projections onto orthogonal vectors equals the original vector.

We know that the projection of a vector \mathbf{v} onto a vector \mathbf{u} is given by the formula:

\operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}.

From the problem, we have:

\operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix}.

This means that the vector \begin{pmatrix} 3 \\ -3 \end{pmatrix} can be expressed as the sum of its projections onto \mathbf{a} and \mathbf{b}:

\begin{pmatrix} 3 \\ -3 \end{pmatrix} = \operatorname{proj}_{\mathbf{a}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} + \operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 3 \\ -3 \end{pmatrix}.

Let \operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}. Then we can substitute the known projection into the equation:

\begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} \\ -\frac{6}{5} \end{pmatrix} + \begin{pmatrix} x \\ y \end{pmatrix}.

This leads to the system of equations:

3 = -\frac{3}{5} + x,
-3 = -\frac{6}{5} + y.

Solving for x in the first equation:

x = 3 + \frac{3}{5} = \frac{15}{5} + \frac{3}{5} = \frac{18}{5}.

Now solving for y in the second equation:

y = -3 + \frac{6}{5} = -\frac{15}{5} + \frac{6}{5} = -\frac{9}{5}.

Thus, we find that:

\operatorname{proj}_{\mathbf{b}} \begin{pmatrix} 3 \\ -3 \end{pmatrix} = \begin{pmatrix} \frac{18}{5} \\ -\frac{9}{5} \end{pmatrix}.<|END_OF_LATENT|> The final answer is $\begin{pmatrix}\frac{18}{5}\-\frac{9}{5}\end{pmatrix}$.