Question

(a) Find the unit vector which is parallel to the vector $\overrightarrow{A}=2\hat{i}+2\hat{j}-2\hat{k}$

(b) Find the unit vector which is perpendicular to both of the

vector $\overrightarrow{A}=2\hat{i}and\overrightarrow{B}=3\hat{i}+4\hat{j}+12\hat{k}$

Answer 1

$a)\left|\overline{A}\right|=\sqrt{2^2+2^2+{(-2)}^2}=\sqrt{12}$

$\stackrel{\wedge }{A}=\frac{\overline{A}}{\left|\overline{A}\right|}=\frac{2\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}-2\stackrel{\wedge }{k}}{\sqrt{12}}$

$=\frac{2(\stackrel{\wedge }{i}+\stackrel{\wedge }{j}-\stackrel{\wedge }{k})}{2\sqrt{3}}$

$=\frac{\stackrel{\wedge }{i}+\stackrel{\wedge }{j}-\stackrel{\wedge }{k}}{\sqrt{3}}$

$b)Vectorperpendicularto\overline{A}and\overline{B}.$

$\overline{A}\times \overline{B}=\left|\begin{array}{ccc}\stackrel{\wedge }{i}& \stackrel{\wedge }{j}& \stackrel{\wedge }{k}\\ 2& 0& 0\\ 3& 4& 12\end{array}\right|$

$⟹\stackrel{\wedge }{i}\left(0\right)-\stackrel{\wedge }{j}\left(24\right)+\stackrel{\wedge }{k}\left(8\right)$

$⟹-24\stackrel{\wedge }{j}+8\stackrel{\wedge }{k}$

$⟹8(-3\stackrel{\wedge }{j}+\stackrel{\wedge }{k})$

$unitvector:\frac{\overline{A}\times \overline{B}}{|\overline{A}\times \overline{B}|}$

$⟹\frac{8(-3\stackrel{\wedge }{j}+\stackrel{\wedge }{k})}{8\sqrt{9+1}}$

$⟹\frac{-3\stackrel{\wedge }{j}+\stackrel{\wedge }{k}}{\sqrt{10}}$