# How do you find a unit vector orthogonal to both (2,0,1,-4) and (2,3,0,1)?

Jul 12, 2016

See below

#### Explanation:

Calling $\vec{u} = \left\{2 , 0 , 1 , - 4\right\}$ and $\vec{v} = \left\{2 , 3 , 0 , 1\right\}$

we need a vector

$\vec{x} = \left\{a , b , c , d\right\}$

such that

$\left\langle\vec{u} , \vec{x}\right\rangle = 0$
$\left\langle\vec{v} , \vec{x}\right\rangle = 0$
$\left\lVert \vec{x} \right\rVert = 1$

Solving

{ (2 a + c - 4 d = 0), (2 a + 3 b + d = 0), (a^2 + b^2 + c^2 + d^2 = 1) :}

for $a , b , c$ we obtain

((a = 1/7 (10 d - 3 sqrt[1 - 6 d^2])),( b = 1/7 (-9 d + 2 sqrt[1 - 6 d^2])), (c = 2/7 (4 d + 3 sqrt[1 - 6 d^2])))

or

((a = 1/7 (10 d + 3 sqrt[1 - 6 d^2])),( b = 1/7 (-9 d - 2 sqrt[1 - 6 d^2])), (c = 2/7 (4 d - 3 sqrt[1 - 6 d^2])))

then if we choose $1 - 6 {d}^{2} \ge 0$ or $- \frac{1}{\sqrt{6}} \le d \le \frac{1}{\sqrt{6}}$ we will have solutions to this problem