Consider the following three points in $\mathbb R^3$: $$P(1, 3, 7),~ Q(−1, 5, 8), ~R(2, 6, 4)$$ and let $a = \overrightarrow {PQ} \ \ \ b= \overrightarrow {PR} \ \ \ c = \overrightarrow {QR}$
Find the values of $~t ~\in \ \mathbb R$ for which $~b + tc~$ is perpendicular to $~a~$.
I have solved it but I got just one value so I am not sure if it's correct.
I calculated a = $\overrightarrow {PQ} = (\vec Q - \vec P) = (-2,2,1),\ \ $ $b = \overrightarrow {PR} = (\vec R - \vec P) = (1, 3, -3)\ \ $ and $c = \overrightarrow {QR} = (\vec R - \vec Q) = (3, 1, -4)$
Then I calculated $b+tc$ as $(1,3,-3) + t(3,1,-4) = (1+3t, 3+t, -3+4t)$
I set $(b+tc).a = 0$ because it's a perpendicular. $$(1+3t, 3+t, -3-4t) \ .\ (-2,2,1) = 0$$ $$-2(1+3t) + 2(3+t) + 1(-3-t)=0$$ $$t=\frac 18$$
I am not sure where I went wrong. I am usually clumsy with my arithmetic operations especially things like $\sqrt {x}$ having two solutions. Any help would be greatly appreciated.
your last equation is ok. the result wrong. Adding all t you find -5t adding the numbers you get 1 so you have -5t+1=0 just the last step too fast?