Question

Find a unit vector perpendicular to each of the vector \(\vec{a}+\vec{b} \space and\space \vec{a}-\vec{b}\),where \(\vec{a}=3\hat{i}+2\vec{j}+2\vec{k}\space and \space \vec{b}=\hat{i}+2\hat{j}-2\hat{k}.\)

Answer 1

We have,
\(\vec{a}=3\hat{i}+2\vec{j}+2\vec{k}\space and \space \vec{b}=\hat{i}+2\hat{j}-2\hat{k}.\)
\(∴\) \(\vec{a}+\vec{b} =\)\(4\hat{i}+4\hat{j},\vec{a}-\vec{b}=2\hat{i}+4\hat{k}\)
\((\vec{a}+\vec{b})\times (\vec{a}-\vec{b})=\)\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & 4 & 0 \\2&0&4\end{vmatrix}=\hat{i}(16)-\hat{j}(16)+\hat{k}(-8)=16\hat{i}-16\hat{j}-8\hat{k}\)
\(∴|(\vec{a}+\vec{b})|×(\vec{a}-\vec{b})|\)=\(\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}\)
\(=\sqrt{2^{2}×8^{2}+2^{2}×8^{2}+8^{2}}\)
\(=8\sqrt{2^{2}+2^{2}+1^{2}}=8\sqrt{9}=8×3=24\)
Hence,the unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b} \space and\space \vec{a}-\vec{b}\) is given by the relation,
\(=\pm\frac{(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}→)×(a→-b→)|}=±16i^-16j^-8k^/24\)
\(=\pm\frac{2\hat{i}-2\hat{j}-\hat{k}}{3}=\pm\frac{2}{3}\hat{i}\mp\frac{2}{3}\hat{j}\mp\frac{1}{3}\hat{k}\)