+956 
Okay so: I did the cross product and I got:
(-7, 2k,4k+3) I thought of writing as i, j, and k but that didn't sound right so instead I put that in the formula, and from there I'm stuck.
\(A = \sqrt{(-7)^2+(2k)^2+(4k+3)^2} \)
I also thought of plugging in √(41) into A, but I'm not completely sure. Any help?
+129852 Let's see....I'm going to use "n" instead of "k" to prevent confusion
Cross product
i j k i j
n + 1 1 -2 n + 1 1
n 3 0 n 3
[ 0i + (-2n) j + (3n + 3) k ] - [ n k - 6i + 0j ] =
6i + ( - 2n) j + (2n + 3)k
So.....we have ( 6 , -2n , 3 + 2n )
So...we wnat to solve this
√ [ 6^2 + (-2n)^2 + (3 + 2n)^2 ] = √41 square both sides
6^2 + (-2n)^2 + (3 + 2n)^2 = 41 simplify
36 + 4n^2 + 4n^2 + 12n + 9 = 41
8n^2 + 12n + 45 = 41
8n^2 + 12n + 4 = 0
2n^2 + 3n + 1 = 0
(2n + 1) ( n + 1) = 0
Setting each factor to 0 and solving for n we get that n = -1/2 or n = -1
So...converting back to k....then k = -1/2 or k = -1
