Let `u=(u_1,u_2,u_3), v=(v_1,v_2,v_3), w=(w_1,w_2,w_3)`

then `u.(vxxw)=det[[u_1,u_2,u_3],[v_1,v_2,v_3],[w_1,w_2,w_3]]` .

Now we have to find `v.(2uxx4w)=8{v.(uxxw)}` (as we can take scalars out)

`=8det[[v_1,v_2,v_3],[u_1,u_2,u_3],[w_1,w_2,w_3]]`

`=8(-1)det[[u_1,u_2,u_3],[v_1,v_2,v_3],[w_1,w_2,w_3]]`

`=8(-1)20=-160.`

As by the rule if we interchange any two rows, the result is multiplied by (-1).

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Let `u=(u_1,u_2,u_3), v=(v_1,v_2,v_3), w=(w_1,w_2,w_3)`

then `u.(vxxw)=det[[u_1,u_2,u_3],[v_1,v_2,v_3],[w_1,w_2,w_3]]` .

Now we have to find `v.(2uxx4w)=8{v.(uxxw)}` (as we can take scalars out)

`=8det[[v_1,v_2,v_3],[u_1,u_2,u_3],[w_1,w_2,w_3]]`

`=8(-1)det[[u_1,u_2,u_3],[v_1,v_2,v_3],[w_1,w_2,w_3]]`

`=8(-1)20=-160.`

As by the rule if we interchange any two rows, the result is multiplied by (-1).