If `bar"a" = hat"i" - 2hat"j"`, `bar"b" = hat"i" + 2hat"j" , bar"c" = 2hat"i" + hat"j" - 2hat"k"`, then find (i) `bar"a" xx (bar"b" xx bar"c")` (ii) `(bar"a" xx bar"b") xx bar"c"` Are the results same? Justify.
Solution
`bar"a" xx (bar"b" xx bar"c")`
`bar"b" xx bar"c" = |(hat"i",hat"j",hat"k"),(1,2,0),(2,1,-2)|`
`= (- 4 - 0)hat"i" - (- 2 - 0)hat"j" + (1 - 4)hat"k"`
`= - 4hat"i" + 2hat"j" - 3hat"k"`
∴ `bar"a" xx (bar"b" xx bar"c") = |(hat"i",hat"j",hat"k"),(1,-2,0),(- 4, 2, -3)|`
`= (6 - 0)hat"i" - ( - 3 - 0)hat"j" + (2 - 8)hat"k"`
`= 6hat"i" + 3hat"j" - 6hat"k"`
`(bar"a" xx bar"b") xx bar"c"`
`bar"a" xx bar"b" = |(hat"i",hat"j",hat"k"),(1,- 2,0),(1,2,0)|`
`= (0 - 0)hat"i" - (0 - 0)hat"j" + (2 - (- 2))hat"k"`
`= 4hat"k"`
∴ `(bar"a" xx bar"b") xx bar"c" = |(hat"i",hat"j",hat"k"),(0,0,4),(2, 1, -2)|`
`= (0 - 4)hat"i" - (0 - 8)hat"j" + (0 - 0)hat"k"`
`= - 4hat"i" + 8hat"j"`
`bar"a" xx (bar"b" xx bar"c") ≠ (bar"a" xx bar"b") xx bar"c"`
