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

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

As a solution to the equation $\text{u}\cdot \text{v}=0$, 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

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

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, $\text{w}=\text{u}+\text{v}=[3,1+5\sqrt{\frac{35}{26}},5-\sqrt{\frac{35}{26}}]$ is a vector that we are looking for.