Question

If vector A=2i+2j+3k and vector B=3i-2j-4k, find cross product and angle b/w A and B?​

Answer 1

$\textbf{Given:}$

$\mathrm{\overrightarrow{a}=2\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k}}$

$\mathrm{\overrightarrow{b}=3\overrightarrow{i}-2\overrightarrow{j}-4\overrightarrow{k}}$

$\textbf{To find:}$

$\mathrm{\overrightarrow{a}{\times}\overrightarrow{b}}$

$\text{and angle between the given vectors}$

$\textbf{Solution:}$

$\text{The cross product of given two vectors is,}$

$\mathrm{\overrightarrow{a}{\times}\overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\2&2&3\\3&-2&-4\end{array}\right|}$

$\mathrm{=(-8+6)\overrightarrow{i}-(-8-9)\overrightarrow{j}+(-4-6)\overrightarrow{k}}$

$\mathrm{=-2\overrightarrow{i}+17\overrightarrow{j}-10\overrightarrow{k}}$

$\implies\mathrm{\bf\overrightarrow{a}{\times}\overrightarrow{b}=-2\overrightarrow{i}+17\overrightarrow{j}-10\overrightarrow{k}}$

$\text{Let \theta be the angle between given two vectors}$

$\mathrm{cos\,\theta=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|\;|\overrightarrow{b}|}}$

$\mathrm{cos\,\theta=\dfrac{2(3)+2(-2)+3(-4)}{\sqrt{4+4+9\;}\sqrt{9+4+16}}}$

$\mathrm{cos\,\theta=\dfrac{6-4-12}{\sqrt{17}\;\sqrt{29}}}$

$\mathrm{cos\,\theta=\dfrac{-10}{\sqrt{17}\;\sqrt{29}}}$

$\implies\boxed{\mathrm{\bf\theta=cos^{-1}(\dfrac{-10}{\sqrt{17}\;\sqrt{29}})}}$

Answer 2

$\textbf{Given:}$

$\mathrm{\overrightarrow{a}=2\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k}}$

$\mathrm{\overrightarrow{b}=3\overrightarrow{i}-2\overrightarrow{j}-4\overrightarrow{k}}$

$\textbf{To find:}$

$\mathrm{\overrightarrow{a}{\times}\overrightarrow{b}}$

$\text{and angle between the given vectors}$

$\textbf{Solution:}$

$\text{The cross product of given two vectors is,}$

$\mathrm{\overrightarrow{a}{\times}\overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\2&2&3\\3&-2&-4\end{array}\right|}$

$\mathrm{=(-8+6)\overrightarrow{i}-(-8-9)\overrightarrow{j}+(-4-6)\overrightarrow{k}}$

$\mathrm{=-2\overrightarrow{i}+17\overrightarrow{j}-10\overrightarrow{k}}$

$\implies\mathrm{\bf\overrightarrow{a}{\times}\overrightarrow{b}=-2\overrightarrow{i}+17\overrightarrow{j}-10\overrightarrow{k}}$

$\text{Let \theta be the angle between given two vectors}$

$\mathrm{cos\,\theta=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|\;|\overrightarrow{b}|}}$

$\mathrm{cos\,\theta=\dfrac{2(3)+2(-2)+3(-4)}{\sqrt{4+4+9\;}\sqrt{9+4+16}}}$

$\mathrm{cos\,\theta=\dfrac{6-4-12}{\sqrt{17}\;\sqrt{29}}}$

$\mathrm{cos\,\theta=\dfrac{-10}{\sqrt{17}\;\sqrt{29}}}$

$\implies\boxed{\mathrm{\bf\theta=cos^{-1}(\dfrac{-10}{\sqrt{17}\;\sqrt{29}})}}$