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MiscII · Q123

Q.Two sides of a parallelogram are 3i^+4j^−5k^3\hat i+4\hat j-5\hat k and −2j^+7k^-2\hat j+7\hat k. Find the unit vectors parallel to the diagonals.

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Sides uˉ=3i^+4j^−5k^, vˉ=−2j^+7k^\bar u=3\hat i+4\hat j-5\hat k,\ \bar v=-2\hat j+7\hat k. The two diagonals (from the common

vertex, and connecting the two "far" vertices) are uˉ+vˉ\bar u+\bar v and uˉ−vˉ\bar u-\bar v:

uˉ+vˉ=3i^+(4−2)j^+(−5+7)k^=3i^+2j^+2k^,∣uˉ+vˉ∣=9+4+4=17.\bar u+\bar v=3\hat i+(4-2)\hat j+(-5+7)\hat k=3\hat i+2\hat j+2\hat k,\qquad |\bar u+\bar v|=\sqrt{9+4+4}=\sqrt{17}.

uˉ−vˉ=3i^+(4+2)j^+(−5−7)k^=3i^+6j^−12k^,∣uˉ−vˉ∣=9+36+144=189=321.\bar u-\bar v=3\hat i+(4+2)\hat j+(-5-7)\hat k=3\hat i+6\hat j-12\hat k,\qquad|\bar u-\bar v|=\sqrt{9+36+144}=\sqrt{189}=3\sqrt{21}.

Unit vector along uˉ+vˉ\bar u+\bar v: 3i^+2j^+2k^17\dfrac{3\hat i+2\hat j+2\hat k}{\sqrt{17}}. …

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