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Difference between revisions of "2013 Mock AIME I Problems/Problem 8"

(Created page with "== Problem == Let <math>\textbf{u}=4\textbf{i}+3\textbf{j}</math> and <math>\textbf{v}</math> be two perpendicular vectors in the <math>x-y</math> plane. If there are <math>n<...")
 
(→‎Solution)
 
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== Solution ==
 
== Solution ==
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<math>\boxed{020}</math>.
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Let <math>\theta</math> be the angle that <math>u</math> makes with the positive <math>x</math>-axis. Now, rotate the plane clockwise by <math>\theta</math> such that <math>u</math> is on the new <math>x'</math>-axis, perpendicular to the <math>y'</math>-axis, on which <math>v</math> now lies, as seen in the diagram below. Rotations do not affect the magnitude of the vectors <math>r_i</math>, so our answer will be the same.
 +
 
 +
<asy>
 +
 
 +
import geometry;
 +
 
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// x' and y' Axes
 +
draw((-8,0)--(8,0), arrow=Arrows);
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label("$x'$", (9,0));
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draw((0,-8)--(0,8), arrow=Arrows);
 +
label("$y'$", (0,9));
 +
 
 +
// Vectors
 +
draw((0,0)--(5,0), red+linewidth(1.5), arrow=Arrow(TeXHead));
 +
label("$u$", midpoint((0,0)--(5,0)), S);
 +
draw((0,0)--(0,6), blue+linewidth(1.5), arrow=Arrow(TeXHead));
 +
label("$v$", midpoint((0,0)--(0,6)), W);
 +
 
 +
// Right Angle Mark
 +
markscalefactor = 0.12;
 +
draw(rightanglemark((1,0),(0,0),(0,1)));
 +
 
 +
</asy>
 +
 
 +
Now, from the information given by the problem, it is clear that the rotated vectors <math>r_i'</math> are those with coordinates <math>(\pm1, \pm2)</math>. Thus, there are <math>4</math> of these vectors, each with a squared magnitude of <math>(\pm 1)^2+(\pm 2)^2 = 1+4 = 5</math>. Thus, our answer is <math>4 \cdot 5 = \boxed{020}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 14:20, 30 July 2024

Problem

Let $\textbf{u}=4\textbf{i}+3\textbf{j}$ and $\textbf{v}$ be two perpendicular vectors in the $x-y$ plane. If there are $n$ vectors $\textbf{r}_i$ for $i=1, 2, \ldots, n$ in the same plane having projections of $1$ and $2$ along $\textbf{u}$ and $\textbf{v}$ respectively, then find \[\sum_{i=1}^{n}\|\textbf{r}_i\|^2.\] (Note: $\textbf{i}$ and $\textbf{j}$ are unit vectors such that $\textbf{i}=(1,0)$ and $\textbf{j}=(0,1)$, and the projection of a vector $\textbf{a}$ onto $\textbf{b}$ is the length of the vector that is formed by the origin and the foot of the perpendicular of $\textbf{a}$ onto $\textbf{b}$.)

Solution

Let $\theta$ be the angle that $u$ makes with the positive $x$-axis. Now, rotate the plane clockwise by $\theta$ such that $u$ is on the new $x'$-axis, perpendicular to the $y'$-axis, on which $v$ now lies, as seen in the diagram below. Rotations do not affect the magnitude of the vectors $r_i$, so our answer will be the same.

[asy] import geometry; // x' and y' Axes draw((-8,0)--(8,0), arrow=Arrows); label("$x'$", (9,0)); draw((0,-8)--(0,8), arrow=Arrows); label("$y'$", (0,9)); // Vectors draw((0,0)--(5,0), red+linewidth(1.5), arrow=Arrow(TeXHead)); label("$u$", midpoint((0,0)--(5,0)), S); draw((0,0)--(0,6), blue+linewidth(1.5), arrow=Arrow(TeXHead)); label("$v$", midpoint((0,0)--(0,6)), W); // Right Angle Mark markscalefactor = 0.12; draw(rightanglemark((1,0),(0,0),(0,1))); [/asy]

Now, from the information given by the problem, it is clear that the rotated vectors $r_i'$ are those with coordinates $(\pm1, \pm2)$. Thus, there are $4$ of these vectors, each with a squared magnitude of $(\pm 1)^2+(\pm 2)^2 = 1+4 = 5$. Thus, our answer is $4 \cdot 5 = \boxed{020}$.

See also

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