Find the line that passes through the point $(2, 5, 3)$ and is perpendicular to the plane $2x - 3y + 4z + 7 = 0$
My only real problem with this is how to shift the line
My first step is to find the norm of the plane which is $\vec{n}^{\ } = (2,-3,4)$
Is the answer as simple as $(x, y, z) = (2, 5, 3) + t(2, -3, 4)$
The question counts as 5 marks in the exam and this answer seems very simplistic.
That answer looks correct to me. I think I'd write something like
$L = \{ (2, 5, 3) + t(2, -3, 4) : t \in \mathbb R \}$
to be clear that you were giving a description of the line as a set. I'd call this a "parametric description", since there's a parameter, t, in the answer. If you want an "implicit" description like $ax + by + cz + d = 0$ (i.e., a condition that each point $(x,y,z)$ satisfies if and only if it's on the line, you've got a problem: you'll really need to have TWO equations, since each defines a plane, and therefore the two of them define the line of intersection of the planes (assuming they intersect nicely). But I'll bet that's not what you're supposed to do. Why? (1) It's hard to grade. There are a million correct answers, and it's hard to check that a student got the right one, (2) it's pretty useless in practice, and (3) it's a question that almost no one ever asks, in part because of answer #2. :)
That's it! That's all you need to do. It readily passes through the desired point when $t=0,$ and is perpendicular to the plane since it is parallel to a normal vector of the plane.
Perhaps the only thing left to do is rewrite the equation in (an) appropriate form. Perhaps $$(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$$ is an acceptable form. Perhaps $$(x,y,z)=(x_0+at,y_0+bt,z_0+ct)$$ is preferred. Without seeing the problem, itself, I really couldn't say. (It really depends how picky your instructor is, and what the instructions say.)
