Question

Find a unit vector that is orthogonal to both u = i − 4j + k and v = 2i + 3j.

 

Answer 1

Given vectors are:

\(u=i-4 j+k\) and \(v=2 i+3 j\)

Or

\(u=i-4 j+k\) and \(v=2 i+3 j+0 k\)

The unit vector of \(u\) and \(v\), which is orthogonal to both the vectors, is given by:

\(\frac{\vec{u} \times \vec{v}}{|\vec{u} \times \vec{v}|}\)

Where \(\vec{u} \times \vec{v}\) is a cross product of vectors \(u\) and \(v\), and \(|\vec{u} \times \vec{v}|\) is the absolute value of \(\vec{u} \times \vec{v}\).

 

For the given vectors \(u\) and \(v\), the \(\vec{u} \times \vec{v}\) is given by:

$$ \begin{aligned} \vec{u} \times \vec{v} &=\left|\begin{array}{ccc} i & j & k \\ 1 & -4 & 1 \\ 2 & 3 & 0 \end{array}\right| \\ &=i(-4 \times 0-3 \times 1)-j(1 \times 0-2 \times 1)+k(1 \times 3-2 \times(-4)) \\ &=i(-3)-j(-2)+k(3+8) \\ &=-3 i+2 j+11 k \end{aligned} $$

 

Absolute value of \(\vec{u} \times \vec{v}\) is:

$$ \begin{aligned} |\vec{u} \times \vec{v}| &=|-3 i+2 j+11 k| \\ &=\sqrt{(-3)^{2}+2^{2}+11^{2}} \\ &=\sqrt{9+4+121} \\ &=\sqrt{134} \end{aligned} $$

Now, substitute the obtained values of \(\vec{u} \times \vec{v}\) and \(|\vec{u} \times \vec{v}|\) in \(\frac{\vec{u} \times \vec{v}}{|\vec{u} \times \vec{v}|}\).

$$ \frac{\vec{u} \times \vec{v}}{|\vec{u} \times \vec{v}|}=\frac{-3 i+2 j+11 k}{\sqrt{134}} $$

A unit vector that is orthogonal to both \(u=i-4 j+k\) and \(v=2 i+3 j\) is \(\frac{-3 i+2 j+11 k}{\sqrt{134}}\).