This two question are from an exam. They are not difficult but I think maybe the answer is not in the choices given in the test. I want to know if I am right or not
- Let vectors $a = (2,3,1) , b=(-2,0,3)$ be two sides of a Parallelogram. What is the length of its longest diagonal?
1.$\sqrt{26}$ ||| 2.$4$ ||| 3.$5$ ||| 4. $\sqrt{21}$
My solution is we know that the diameters are a +b and a-b so $a+b = (0 , 3,4)$ and $a-b= (4 , 3 , -2)$ So length of a diagonal is 5 and the other is $\sqrt{29}$ which is obviously greater than 5 so the answer is $\sqrt{29}$ which is not in the answers. Is my solution right?
The second question is about cauchy-schwartz inequality in vectors $|a||b|\geq |a.b|$
2.Let $6x-y+4z =12$ and $9x^2 + y^2 + 4z^2$ is minimum. Find $9x +3y +3z$ .
- $4$ ||| 2.$6$ ||| 3.$8$ ||| 4.$10$
My solution is $|(3x,y,2z).(2,-1,2)|=6x-y+4z=12 \leq |(3x,y,2z)||(2,-1,2)|$ so because of it says "$9x^2 + y^2 + 4z^2$ is minimum" so $|(3x,y,2z)|$ = 4. I also assumed that I must multiply this answer by $|(3,3,3/2)| = 9/2$ so I could get $9x +3y +2z$ and it is 18 which is not in the answers. Is my solution right? if not, what is the right solution?