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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra4 Marks
Question
Find a unit vector perpendicular to each of the vector $\vec a + \vec b\;$ and $\vec a - \vec b$, where $\vec a = 3\hat i + 2\hat j + 2\hat k\;$ and $\;\vec b = \hat i + 2\hat j - 2\hat k$.
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Answer

It is given that:
$\vec a = 3\hat i+2 \hat j+2 \hat k$ and $\vec b=\hat i +2 \hat j-2\hat k$
$\therefore \vec a +\vec b=4\hat i+4\hat j$ and $\vec a -\vec b=2\hat i +4 \hat k$
$\therefore (\vec a + \vec b) \times (\vec a - \vec b) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 4&4&0 \\ 2&0&4 \end{array}} \right| = 16\hat i - 16\hat j - 8\hat k$
$\therefore| (\vec a + \vec b) \times (\vec a - \vec b) |= \sqrt{576}=24$
Therefore, the unit vector perpendicular to both the vectors $(\overrightarrow{a}+\overrightarrow{b})$
and
$(\overrightarrow{a}-\overrightarrow{b})$ is given by:
$=\pm \frac{(16\widehat{i}-16\widehat{j}-8\widehat{k})}{24}=\pm \frac{1}{3}(2\widehat{i}-2\widehat{j}-\widehat{k}) .$

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