Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 29/Solution 2

(Alternative solution) First, we find a vector ${\textbf {u}}$ orthogonal to ${\textbf {v}}$ with the same magnitude with ${\mathbf {v}}$

As a solution to the equation ${\textbf {u}}\cdot {\textbf {v}}=0$, we can find an orthogonal vector ${\mathbf {u}}=c[0,5,-1]$ to ${\mathbf {v}}$. To make the same magnitude with ${\mathbf {v}}$, we choose $c$ which solves

$3^{2}+1^{2}+5^{2}=|{\mathbf {v}}|^{2}=|{\mathbf {u}}|^{2}=c^{2}(5^{2}+(-1)^{2})\iff 35=26c^{2}\iff c=\pm {\sqrt {\frac {35}{26}}}$.

Now, we find the desired vector ${\mathbf {u}}={\sqrt {\frac {35}{26}}}\ [0,5,-1]$.

Note that the addition of two orthogonal vectors with the same magnitude makes an angle of ${\frac {\pi }{4}}$ radians with both vectors. Therefore, ${\textbf {w}}={\textbf {u}}+{\textbf {v}}=\color {blue}\left[3,1+5{\sqrt {\frac {35}{26}}},5-{\sqrt {\frac {35}{26}}}\right]$ is a vector that we are looking for.