Determine the value(s) of x such that the area of the parallelogram formed by the vectors a=(x+1,1,−2) and b=(x,3,0) is sqrt(41)

comAttitRize8

comAttitRize8

Answered question

2022-07-20

How to determine the value of x such that the parallelogram has a given area?
Determine the value(s) of x such that the area of the parallelogram formed by the vectors a = ( x + 1 , 1 , 2 ) and b = ( x , 3 , 0 ) is 41
My work (using cross product)
| a × b | = ( ( 1 ) ( 0 ) ( 2 ) ( 3 ) , ( 2 ) ( x ) ( x + 1 ) ( 0 ) , ( x + 1 ) ( 3 ) ( 1 ) ( x ) ) | a × b | = ( 6 , 2 x , 3 x + 3 x ) | a × b | = ( 6 , 2 x , 2 x + 3 ) 41 = ( 6 , 2 x , 2 x + 3 )
I don't know what to do next

Answer & Explanation

Killaninl2

Killaninl2

Beginner2022-07-21Added 20 answers

You have to be careful with your equality signs. What you already showed is that
a × b = ( 6 , 2 x , 2 x + 3 ) .
In order to get the area to be 41 , we need
a × b 2 = 6 2 + ( 2 x ) 2 + ( 2 x + 3 ) 2 = 41.
This is a quadratic equation, which you can solve for x.

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