Question #196381

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




1
Expert's answer
2021-05-23T19:36:02-0400

α=2i3j+kβ=7i5j+k\alpha = 2i-3j+k \\ \beta = 7i-5j+k

Let d=α×βd = \alpha \times \beta

Consider the computation

d=ijk231751=(3+5)i(27)j+(10+21)k=2i+5j+11kd= \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=2i+5j+11k(2)2+(5)2+(11)2=2i+5j+11k150=2i+5j+11k56c= \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 α\alpha and β\beta . To see this consider

cα=156(2i+5j+11k)(2i3j+k)=156(415+11)=0c \cdot \alpha = \frac{1}{5\sqrt6}(2i+5j+11k) \cdot (2i-3j+k) = \frac{1}{5\sqrt6} (4-15+11) = 0

Also, consider

cβ=156(2i+5j+11k)(7i5j+k)=156(1425+11)=0c \cdot \beta = \frac{1}{5\sqrt6}(2i+5j+11k) \cdot (7i-5j+k) = \frac{1}{5\sqrt6} (14-25+11) = 0





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