Answer to Question #196381 in Differential Geometry | Topology for Fatowore Samson

Question

Suppose that \\( \\alpha=2i-3j+k\\) and \\(\\beta=7i-5j+k\\) , find a unit vector perpendicular to \\(\\alpha\\) and \\(\\beta) respectively.

Answer 1

$\alpha = 2i-3j+k \\ \beta = 7i-5j+k$

Let $d = \alpha \times \beta$

Consider the computation

$d= \begin{vmatrix} i&j&k\\ 2&-3&1\\7&-5&1 \end{vmatrix} = (-3+5)i -(2-7)j +(-10+21)k = 2i+5j+11k$

Let c be the unit vector of d.$c= \dfrac{2i+5j+11k}{\sqrt{(2)^2+(5)^2+(11)^2}} = \dfrac{2i+5j+11k}{\sqrt{150}} = \dfrac{2i+5j+11k}{5\sqrt{6}}$

We claim that c is the desired unit vector perpendicular to $\alpha$ and $\beta$ . To see this consider

$c \cdot \alpha = \frac{1}{5\sqrt6}(2i+5j+11k) \cdot (2i-3j+k) = \frac{1}{5\sqrt6} (4-15+11) = 0$

Also, consider

$c \cdot \beta = \frac{1}{5\sqrt6}(2i+5j+11k) \cdot (7i-5j+k) = \frac{1}{5\sqrt6} (14-25+11) = 0$