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Mathematics Question on Vector Algebra

Find a unit vector perpendicular to each of the vector a⃗+b⃗  and  a⃗−b⃗\vec{a}+\vec{b} \space and\space \vec{a}-\vec{b},where a⃗=3i^+2j⃗+2k⃗  and  b⃗=i^+2j^−2k^.\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

We have,
a⃗=3i^+2j⃗+2k⃗  and  b⃗=i^+2j^−2k^.\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}
∴∣(a⃗+b⃗)∣×(a⃗−b⃗)∣∴|(\vec{a}+\vec{b})|×(\vec{a}-\vec{b})|=162+(−16)2+(−8)2\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}
=22×82+22×82+82=\sqrt{2^{2}×8^{2}+2^{2}×8^{2}+8^{2}}
=822+22+12=89=8×3=24=8\sqrt{2^{2}+2^{2}+1^{2}}=8\sqrt{9}=8×3=24
Hence,the unit vector perpendicular to each of the vectors a⃗+b⃗  and  a⃗−b⃗\vec{a}+\vec{b} \space and\space \vec{a}-\vec{b} is given by the relation,
=±(a⃗+b⃗)×(a⃗−b⃗)∣(a⃗+b⃗→)×(a→−b→)∣=±16i−16j−8k/24=\pm\frac{(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}→)×(a→-b→)|}=±16i^-16j^-8k^/24
=±2i^−2j^−k^3=±23i^∓23j^∓13k^=\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}