Find an equation for the line that is parallel to the plane $2x - 3y + 5z - 10 = 0$ and passes through the point (-1, 7, 4) "Find an equation for the line that is parallel to the plane $2x - 3y + 5z - 10 = 0$ and passes through the point (-1, 7, 4)"
I'm just learning this and am pretty confused on how to do this problem. 
In class we went over distance between vectors, orthogonal projection, but I still don't know how to proceed with this question. How does this deal with vectors? 
 A: The normal vector for the supplied plane is (2, -3, 5). Any vector orthogonal to this will satisfy the constraint of being parallel to the supplied plane, and can be constructed by satisfying the equation: (2, -3, 5) x (x,y,z) = 0. (This is of course just the familiar inner or dot product.)
Then construct a line from the supplied point (-1,7,4) and the direction vector chosen above. There will be infinitely many of course, the entirety of which will define a new plane parallel to the original.
A: First of all, there is an infinite number of such lines. To see this, imagine two parallel planes and choose a random point on the first plane. Then draw a line through that point (while staying within the plane), and you can see that this line can be rotated 360 degrees while still being parallel to the other plane.
Now you are given the plane $2x−3y+5z−10=0$, and to find a parallel one you can simply choose a different value of $c$ in $2x - 3y + 5z - c = 0$ (to see that these planes are parallel, just note that any two non-parallel planes will have solutions in common, and by elimination we can see that these equations will have none).
Since $2x - 3y + 5z - c = 0$ will be parallel, we can sub in $(-1, 7, 4)$ to get a $c$ value of $c = 2(-1) - 3(7) + 5(4) = -3$, leaving us with the plane:
$$2x - 3y + 5z + 3 = 0$$
Now if we want to find the equation of a line which passes through $(-1, 7, 4)$, we need to create a vector equation, since a regular equation with 3 variables will always represent a plane. This is because holding a variable constant in the above equation (say $x$) would still allow you to modify both $y$ and $z$ freely, and since the rest of the equation would effectively be reduced to a single constant, the equation would reduce to that of a line with variables $y$ and $z$. But since $x$ is not held constant, but can vary over an infinite range of values, there is an infinite number of lines that is represented by the equation.
To create your vector equation, you simply need a vector that is in the plane. If you imagine a vector as an arrow pointing in some direction, a vector which is in the plane will point along the plane and not out of it. This means that for some distance moved in the $x$ direction, the vector moves a corresponding distance in both $y$ and $z$ so as to remain the plane. In other words, a vector is in the plane if adding it to a point produces another point in the same plane. Since a vector in the plane is the difference between two points, using any two points we can find such a vector.
What points do we have? We know that $(-1, 7, 4)$ is in the plane, and the other point we can select at random from any of the solutions to $2x - 3y + 5z + 3 = 0$. Since it's easy to obtain, I'll use the solution where both $x$ and $y$ are 0, namely $(0,0,{-3 \over 5})$. By subtracting these two points, we can construct the vector ${\vec m} = [0 - (-1), 0 - 7, {-3 \over 5} - 4] = [1, -7, {-23 \over 5}]$
To complete our vector equation that passes through $(-1, 7, 4)$, we simply write:
$$(x, y, z) = t[1, -7, {{-23 \over 5}}] + (-1, 7, 4)$$
For some value $t$.
