Answer on Question #56113 – Math – Vector Calculus
If A=i−2j−3k, B=2i+3j+k and C=i+3j−2k, compute ∣(A×B)×C∣
2(√2)
5(√26)
3(√21)
4(√11)
Solution
A×B=⎣⎡i12j−23k−31⎦⎤=i(−2∗1−(−3)∗3)−j(1∗1−(−3)∗2)+k(1∗3−(−2)∗2)==i(−2+9)−j(1+6)+k(3+4)=7i−7j+7k.(A×B)×C=⎣⎡i71j−73k7−2⎦⎤=i(−7∗(−2)−7∗3)−j(7∗(−2)−7∗1)+k(7∗3−(−7)1)==i(14−21)−j(−14−7)+k(21+7)=−7i+21j+28k.∣(A×B)×C∣=(−7)2+212+282=49+441+784=1274=7∗7∗26=726.
Answer: 726.
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