Question

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

Answer 1

recall that a×b has magnitude equal to the parallelogram formed by them.

a×b = |a| |b| sinθ
a•b = |a| |b| cosθ

a•b = (k+1)(k) + (1)(3) + (-2)(0) = k^2+k+3
|a| = √(k^2+2k+6)
|b| = √(k^2+9)

cosθ = (k^2+k+3)/√(k^2+9)(k^2+2k+6)
sinθ = √((8k^2 + 12k + 45)/((k^2+9)(k^2+2k+6)))

so we want

√41 = √((k^2+2k+6)(k^2+9))√((8k^2 + 12k + 45)/((k^2+9)(k^2+2k+6)))
or,
41 = ((k^2+2k+6)(k^2+9))((8k^2 + 12k + 45)/((k^2+9)(k^2+2k+6)))
41 = 8k^2 + 12k + 45
8k^2 + 12k + 4 = 0
2k^2 + 3k + 1 = 0
(2k+1)(k+1) = 0
k = -1/2 or -1

k=-1/2
(1/2,1,-2)x(-1/2,3,0) = (6,1,2) and has length √(36+1+4) = √41

k=-1
(0,1,-2)x(-1,3,0) = (6,2,1) and has length √41