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5.3 · Q39

Q.Find two unit vectors each of which is perpendicular to both uˉ\bar u and vˉ\bar v, where uˉ=2i^+j^−2k^\bar u=2\hat i+\hat j-2\hat k, vˉ=i^+2j^−2k^\bar v=\hat i+2\hat j-2\hat k.

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✓ Free question

uˉ=2i^+j^−2k^, vˉ=i^+2j^−2k^\bar u=2\hat i+\hat j-2\hat k,\ \bar v=\hat i+2\hat j-2\hat k. Compute

uˉ×vˉ=∣i^j^k^21−212−2∣=i^(1⋅(−2)−(−2)⋅2)−j^(2⋅(−2)−(−2)⋅1)+k^(2⋅2−1⋅1)=i^(−2+4)−j^(−4+2)+k^(4−1)=2i^+2j^+3k^.\bar u\times\bar v=\begin{vmatrix}\hat i&\hat j&\hat k\\ 2&1&-2\\ 1&2&-2\end{vmatrix} =\hat i(1\cdot(-2)-(-2)\cdot2)-\hat j(2\cdot(-2)-(-2)\cdot1)+\hat k(2\cdot2-1\cdot1) =\hat i(-2+4)-\hat j(-4+2)+\hat k(4-1)=2\hat i+2\hat j+3\hat k.

Its magnitude is 22+22+32=4+4+9=17\sqrt{2^2+2^2+3^2}=\sqrt{4+4+9}=\sqrt{17}. The two unit vectors perpendicular to both νˉ\bar\nu and vˉ\bar v are ±uˉ×vˉ∣uˉ×vˉ∣=±2i^+2j^+3k^17\pm\dfrac{\bar u\times\bar v}{|\bar u\times\bar v|}=\pm\dfrac{2\hat i+2\hat j+3\hat k}{\sqrt{17}}.

✓Final answer

±2i^+2j^+3k^17\pm\dfrac{2\hat i+2\hat j+3\hat k}{\sqrt{17}}

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