Textbook question on vector equations of planes

[ausmathgenius420]

Hi,
My textbook asks the question:

Find a Cartesian equation of the plane that is at right angles to the line given by x = 4 + t, y = 1 − 2t, z = 8t and goes through the point P(3, 2, 1).

For simplicity I will convert to vector form...
1. Convert to vector form
$$r=(4+t)i +(1-2t)j+8tk$$$$4i+j+t(i-2j+8k)$$
Therefore the line travels in a direction of $i-2j+8k$
Looking at the textbooks solution, they then conclude that the normal vector is also $i-2j+8k$

Can someone explain why this is true?

 

[blamocur]

This is one of the ways of defining a plane, i.e., given point $\mathbf p_0$ and a vector $\mathbf n$ the corresponding plane is defined as a set of points $\mathbf p$ such that $\mathbf p - \mathbf p_0$ is orthogonal to $\mathbf n$.
To put it differently: since you are asked to find the plane which is at the right angle with the line, the normal vector for the plane is the direction of that line.