# Determine the value(s) of k such that the area of the parallelogram formed by vectors `veca` = (k+1, 1,-2) and `vecb` =(k,3,0) is `sqrt(41)`

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to use the following formula to evaluate area of parallelogram such that:

`A = bar a X bar b = |bara|*|barb|*sin alpha `

`alpha`  is the angle between `bar a`  and `bar b`

You need to evaluate the cross product between `bar a`  and `bar b ` using determinant of matrix formed from unit vectors `bar i, bar j, bar k`  and vectors `bar a`  and `bar b`  such that:

`bar a X bar b = [[bar i, bar j, bar k] , [k+1 , 1 , -2] , [k , 3, 0]]`

Substituting `sqrt 41`  for `bar a X bar b`  yields:

`sqrt 41 = bar i* [[1 , -2] , [3 , 0]] + bar j* [[(k+1) , -2] , [k , 0]] + bar k* [[(k+1) , 1] , [k , 3]]`

`sqrt 41 = 6 bar i + 2k*bar j + (3k+ 3 - 3k)*bar k`

`sqrt 41 = 6 bar i + 2k*bar j + 3 bar k`

You need to evaluate the magnitude of vector `bar a X bar b`  such that:

`sqrt 41 = sqrt (6^2+ (2k)^2 + 3^2)`

Raising to square both sides yields:

`41 = 36 + 4k^2 + 9`

`41 - 45 = 4k^2 =gt -4 = 4k^2 =gt k^2 = -1 =gt k_(1,2) = +-i ` (complex number theory)

Hence, evaluating k under given conditions yields `k_(1,2) = +-i .`