# Find a vector v for which the parallelogram formed by v and w = <2, 2, -1> has area equal to 6.

You can calculate this by using cross product.

`A=||vxxw||`

Let's first calculate cross product

`|(i,j,k),(v_1,v_2,v_3),(2,2,-1)|=i|(v_2,v_3),(2,-1)|-j|(v_1,v_3),(2,-1)|+k|(v_1,v_2),(2,2)|=`

`(-v_2-2v_3)i+(v_1+2v_3)j+(2v_1-2v_2)k`

Now we calculate norm of that vactor which needs to be equal to 6.

`6=sqrt((-v_2-2v_3)^2+(v_1+2v_3)^2+(2v_1-2v_2)^2)`

`6=sqrt(v_2^2+4v_2v_3+4v_3^2+v_1^2+4v_1v_3+4v_3^2+4v_1^2-8v_1v_2+4v_2^2)`

`6=sqrt(5v_1^2+5v_2^2+8v_3^2-8v_1v_2+4v_1v_3+4v_2v_3)`

So any vector `v=(v_1,v_2,v_3)` which satisfies the above equation will form parallelogram of area 6 with vector `w.`  Obviously...

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You can calculate this by using cross product.

`A=||vxxw||`

Let's first calculate cross product

`|(i,j,k),(v_1,v_2,v_3),(2,2,-1)|=i|(v_2,v_3),(2,-1)|-j|(v_1,v_3),(2,-1)|+k|(v_1,v_2),(2,2)|=`

`(-v_2-2v_3)i+(v_1+2v_3)j+(2v_1-2v_2)k`

Now we calculate norm of that vactor which needs to be equal to 6.

`6=sqrt((-v_2-2v_3)^2+(v_1+2v_3)^2+(2v_1-2v_2)^2)`

`6=sqrt(v_2^2+4v_2v_3+4v_3^2+v_1^2+4v_1v_3+4v_3^2+4v_1^2-8v_1v_2+4v_2^2)`

`6=sqrt(5v_1^2+5v_2^2+8v_3^2-8v_1v_2+4v_1v_3+4v_2v_3)`

So any vector `v=(v_1,v_2,v_3)` which satisfies the above equation will form parallelogram of area 6 with vector `w.`  Obviously there are infinitely many solutions. One is e.g. if you put `v_1=v_2=0` , your equation becomes

`6=sqrt(8v_3^2)=>v_3=pm3/sqrt2`

Hence two solutions (out of infinitely many) are `(0,0,3/sqrt2)` and `(0,0,-3/sqrt2).`

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