Submitted by maths123 on Thu, 05/28/2015 - 10:56
Forums:
I reached the stage where you have to prove the expression for $ \lambda_1 $
I realised the use of $ x.x = 1 $ and I came to the below expression
$ \lambda^2 (3-2\gamma -2\beta + \alpha) + \lambda(2\alpha - \beta-\gamma) + \alpha + \beta + \gamma - 3 = 0 $
From looking at answers on TSR, the above has $(\lambda + 1) $ as a factor - how would you spot this? - If you can spot this the result follows easily from there...
In addition when I used the quadratic formula for the expression, why did I have to use the negative sign instead of the positive (in the $\pm$ part of formula) to get the answer, although we know $\lambda$ is positive - how would you know which sign to use?
Try sketching it
If you think about it, what you are doing is finding the intersection of a line with a circle. One point at which this happens is $Y_{1}$ but there is another obvious point where the line in question intersects the circle. Carry on with this logic and you should be able to get why $\lambda = -1$ is going to be a solution, and why we would use the other answer of the quadratic instead of $\lambda = -1$. Also, sketching it helps.
Alternatively, if you have gotten to $\lambda = -1$ some other way and are just wondering why we don't take it as our answer, just try plugging it back in and seeing what vector you get back out.