+0  
 
+1
53
1
avatar+946 

 

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?

Julius  May 15, 2018
 #1
avatar+87334 
+2

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

 

 

 

cool cool cool

CPhill  May 15, 2018

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