Q.

Let a=4i^j^+k^, b=11i^j^+k^ and c be a vector that (a+b)×c=c×(2a+3b). If (2a+3b)·c=1670, then |c|2 is equal to :          [2024]

1 1600  
2 1618  
3 1627  
4 1609  

Ans.

(2)

We have, a=4i^j^+k^ and b=11i^j^+k^

a·b=44+1+1=46, |a|2=18, |b|2=123

(a+b)×c=c×(2a+3b)          ... (i)          [Given]

and (2a+3b)·c=1670          ... (ii)

  (a+b)×c=(2a3b)×c          [From (i)]

 (a+4b)×c=0  c=λ(4ba)

 (2a+3b)·λ(4ba)=1670          [From (ii)]

 λ(5a·b2|a|2+12|b|2)=1670

 λ=16705×462×18+12×123=1

Now, c=4ba=4(11i^j^+k^)(4i^j^+k^)

        =40i^3j^+3k^

  |c|2=1600+9+9=1618.