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  • Find a unit vector perpendicular to each of the vectors where <img alt="\overrig

Find a unit vector perpendicular to each of the vectors \overrightarrow{a}and \overrightarrow{b}where \overrightarrow{a}=5\widehat{i}+6\widehat{j}-2\widehat{k} and \overrightarrow{b}=7\widehat{i}+6\widehat{j}-2\widehat{k}

 

 
 
 
 
 

Answers (1)

Given:

\overrightarrow{a}=5\widehat{i}+6\widehat{j}-2\widehat{k}

 \overrightarrow{b}=7\widehat{i}+6\widehat{j}-2\widehat{k}

The vector perpendicular to each of the vectors \overrightarrow{a} \ and \overrightarrow{b} is given by \\ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}

 

\\ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 5 & 6 & -2 \\ 7 & 6 & 2 \end{array}\right| \\\\= \hat{\mathrm{i}}(12+12)- \hat{\mathrm{j}}(10+14)+ \hat{\mathrm{k}}(30-42)\\\\ = 24 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}- 12 \hat{\mathrm{k}}

\\\text { Unit vector perpendicular to both } \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} = \frac{24 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}- 12 \hat{\mathrm{k}}}{\sqrt{24^2+(-24)^2+12^2}} \\\\ = \frac{24 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}- 12 \hat{\mathrm{k}}}{\sqrt{24^2+(-24)^2+12^2}} \\\\ = \frac{24 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}- 12 \hat{\mathrm{k}}}{36} \\\\= \frac{2}{3} \hat{\mathrm{i}}-\frac{2}{3} \hat{\mathrm{j}}-\frac{1}{3} \hat{\mathrm{k}}

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