# Find the possible values of the constant alpha, given that the vectors a=<alpha,8,3 alpha+1> and b=<alpha+1,alpha-1,-2> are perpendicular to each other

Find the possible values of the constant $\alpha$, given that the vectors $a=<\alpha ,8,3\alpha +1>$ and $b=<\alpha +1,\alpha -1,-2>$ are perpendicular to each other
For this question do I have to assume a.b=0?
how can I find the value of constants?
my understanding is: $<\alpha ,8,3\alpha +1><\alpha +1,\alpha -1,-2>=0$
how do I proceed from here to find the values of constant 𝛼?
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postotnojeyf
Given a fixed number $z$, you can take each $q\in \mathbb{Q}$ and find the unique ${y}_{q}\in \mathbb{R}$ such that $z+{y}_{q}=q$. Since each ${y}_{q}$ is unique, clearly there are only countably many numbers with that can be found.
So out of the entire pool of $|\mathbb{R}|$ things that can go into the y of "$z+y$", only countably many will yield a rational number, and the rest (uncountably many) will yield an irrational number.
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Brenda Jordan
Yes if the vectors are perpendicular, their dot product is 0. Then use a⋅b $=\sum _{i=1}^{n}{a}_{i}{b}_{i}$. Then $\alpha \left(\alpha +1\right)+8\left(\alpha -1\right)-2\left(3\alpha +1\right)=0$