Question

4.Suppose that α=2i−3j+k and \eta=7i−5j+k , find a unit vector perpendicular to α
and \(\eta) respectively.
\\(\\frac{-2i+5j+11k }{5\\sqrt(6)}\\)
\\(\\frac{2i-5j+11k }{5\\sqrt(6)}\\)
\\(\\frac{2i+5j-11k }{5\\sqrt(6)}\\)
\\(\\frac{2i+5j+11k }{5\\sqrt(6)}\\)

Answer 1

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

The cross product $\alpha \times \beta$ is a vector that is perpendicular to both $\alpha$ and  $\beta$ .

$\gamma =\alpha \times \beta =\begin{vmatrix} i&j&k \\ 2&-3&1\\7&-5&1 \end{vmatrix}=(-3\times 1-(-5)\times 1)i-(2\times 1-7\times 1)j+(2\times (-5)-7(-3))k=2i+5j+11k$

$|\gamma|=\sqrt{2^2+5^2+11^2}=\sqrt{150}=5\sqrt6$

Unit vector is $\frac{\gamma}{|\gamma|}$ .

Answer: $\frac{2i+5j+11k}{5\sqrt6}$  a unit vector perpendicular to $\alpha$ and $\beta$ respectively.