In the plane xOy find the vector vec r perpendicular to vec…

In the plane $xOy$ find the vector $\vec r$ perpendicular to $\vec a\{5,-3,4\}$ with $|\vec r|=|\vec a|$. Can $\vec r$ be determined for an arbitrary vector $\vec a$?

Explanation of Method

A vector r in the xOy plane, perpendicular to a and of the same length.

Step 1: Unknowns and conditions

\[ \vec r=\{x,y,0\},\qquad |\vec a|=\sqrt{25+9+16}=5\sqrt2. \]

\[ \vec r\cdot\vec a=5x-3y=0\ \Longrightarrow\ y=\tfrac53x,\qquad x^2+y^2=50. \]

Step 2: Solution

\[ x^2\left(1+\tfrac{25}{9}\right)=50\ \Longrightarrow\ x^2=\frac{225}{17},\qquad x=\pm\frac{15}{\sqrt{17}},\quad y=\pm\frac{25}{\sqrt{17}}. \]

\[\vec r_{1,2}=\pm\frac{\sqrt{17}}{17}\{15,\ 25,\ 0\}\approx\pm\{3.64,\ 6.06,\ 0\}\]

Check: $5\cdot15-3\cdot25=0$ $\checkmark$, and $\dfrac{225+625}{17}=50=|\vec a|^2$ $\checkmark$

$O$ $\vec a$ $\vec r_1$ $\vec r_2$

$\vec a=\{5,-3,4\}$
$\vec r_{1,2}=\pm\tfrac{1}{\sqrt{17}}\{15,25,0\}$

Both solutions lie in the shaded plane $xOy$, are perpendicular to $\vec a$ and have length $5\sqrt2=|\vec a|$; they are opposite vectors.

Step 3: Arbitrary $\vec a=\{a_1,a_2,a_3\}$

The conditions become $a_1x+a_2y=0$ and $x^2+y^2=|\vec a|^2$.

If $a_1^2+a_2^2\neq0$ ($\vec a$ not parallel to $Oz$), the direction in $xOy$ is fixed. There are exactly two solutions,
\[ \vec r=\pm\frac{|\vec a|}{\sqrt{a_1^2+a_2^2}}\{a_2,\,-a_1,\,0\}. \]
If $a_1=a_2=0$ ($\vec a\parallel Oz$, $\vec a\neq\vec 0$), every vector of $xOy$ is perpendicular to $\vec a$. Any vector of the plane of length $|a_3|$ works (infinitely many), so $\vec r$ is not determined.

For the given $\vec a$ the formula gives $\pm\tfrac{5\sqrt2}{\sqrt{34}}\{-3,-5,0\}$, the same pair as above.

Result

\[\boxed{\,\vec r_{1,2}=\pm\frac{\sqrt{17}}{17}\{15,\ 25,\ 0\}\approx\pm\{3.64,\ 6.06,\ 0\}\,}\]