Find the volume of the parallelepiped with a vertex at A(1,2,-1) and adjacent vertices at B(3,1,2), C(-1,2,4) and D(3,2,5).

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the vectors `bar(AB), bar(AC), bar(AD),` such that:

`bar(AB) = (x_B - x_A)bar i + (y_B - y_A)bar j + (z_B - z_A) bar k`

`bar(AB) = (3 -1)bar i + (1 - 2)bar j + (2 + 1) bar k`

`bar(AB) = 2bar i - bar j + 3 bar k`

`bar(AC) = (x_C - x_A)bar i + (y_C - y_A)bar j + (z_C - z_A) bar k`

`bar(AC) = (-1 -1)bar i + (2 - 2)bar j + (4 + 1) bar k`

`bar(AC) = -2bar i + 5bar k`

`bar(AD) = (x_D - x_A)bar i + (y_D - y_A)bar j + (z_D - z_A) bar k`

`bar(AD) = (3 -1)bar i + (2 - 2)bar j + (5 + 1) bar k`

`bar(AD) = 2bar i + 6bar k`

You need to evaluate the volume of parallelipiped, such that:

`V = bar(AD)*(bar(AC)Xbar(AB))`

You need to evaluate the cross product `(bar(AC)Xbar(AB))` such that:

`(bar(AC)Xbar(AB)) = [(bar i, bar j, bar k),(-2,0,5),(2,-1,3)]`

`(bar(AC)Xbar(AB)) = 2 bar k + 10 bar j + 5 bar i + 6 bar j`

`(bar(AC)Xbar(AB)) = 5 bar i + 16 bar j + 2 bar k `

You need to evaluate the dot product `bar(AD)*(bar(AC)Xbar(AB))` , such that:

`bar(AD)*(bar(AC)Xbar(AB)) = <2,0,6>*<5,16,2>`

`bar(AD)*(bar(AC)Xbar(AB)) = 2*5 + 0*16 + 6*2`

`bar(AD)*(bar(AC)Xbar(AB)) = 22`

Hence, evaluating the volume of parallelipiped, under the given conditions, yields `V = 22.`

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