We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
218
1
avatar+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?

 May 15, 2018
 #1
avatar+102905 
+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

 May 15, 2018

8 Online Users

avatar