If the diagonals of a parallelogram are given by 3i+j-2k and i-3j+4k, then the lengths of its sides are

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Please find below the solution to the asked query:

We haved1=3i^+j^-2k^d2=i^-3j^+4k^We know thatLength of sides are d1+d22 and d1-d22 d1+d22=3i^+j^-2k^+i^-3j^+4k^2=4i^-2j^+2k^2=16+4+42=242=262=6 d1-d22=3i^+j^-2k^-i^+3j^-4k^2=2i^+4j^-6k^2=4+16+362=562=2142=14

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