Consider the vectors vector u = < 1,-1, 3 >, vetor v = < 1, k, k >, vector w =< k, 1, k^2 >Compute vector u dot product((vector v)cross product (vector w)).

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the cross product of vectors `bar v` and `bar w` , such that:

`bar v x bar w = [(bar i, bar j, bar k),(1,k,k),(k,1,k^2)]`

`bar v x bar w = k^3 bar i + bar k + k^2 bar j - k^2 bar k - k bar i - k^2 bar j`

`bar v x bar w = (k^3 - k) bar i + (k^2 - k^2) bar j + (1 - k^2) bar k`

`bar v x bar w = (k^3 - k) bar i + (1 - k^2) bar k`

You need to evaluate the cross product `bar u*(bar v x bar w)` , such that:

`bar u*(bar v x bar w) = (bar i - bar j + 3 bar k)((k^3 - k) bar i + (1 - k^2) bar k)`

`bar u*(bar v x bar w) = 1*(k^3 - k) - 1*0 + 3*(1 - k^2)`

`bar u*(bar v x bar w) = k^3 - k - 3(k - 1)(k + 1)`

`bar u*(bar v x bar w) = k(k - 1)(k + 1) - 3(k - 1)(k + 1)`

`bar u*(bar v x bar w) = (k - 1)(k + 1)(k - 3)`

`bar u*(bar v x bar w) = (k^2 - 1)(k - 3)`

Hence, performing the indicated operations yields `bar u*(bar v x bar w) = (k^2 - 1)(k - 3).`

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

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`vxxw =(k^3-k,0,1-k^2)`

`` `u*(vxxw)=1 *(k^3-k)+(-1)*0+3*(1-k^2)`

`u*(vxxw)=k^3-k+3-3k^2=k^3-3k^2- k+3`

let l,m, and n are muttually perpendicular directions.

`vxxw=` det( `[[l,m,n],[1,k,k],[k,1,k^2]]` )

=`l(k^3-k)-m(k^2-k^2)+n(1-k^2)`

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