Find the coordinates of the point a line meets a plane

Question

The line through the point (1, 2, 6) orthogonal to the plane $x + 3y + z = 4$ meets the plane at the point X. Find the coordinates of X.

I've attempted to illustrate what I've interpreted as what's going on in the following graphic:

The normal line (1, 3, 1) indicates where in space the plane is. The point (1, 2, 6) is a point on that normal line. Using the vector equation of a line, we can construct a way to express the coordinates of r, in the equation:

$$ r = a + t(n)$$

In which $a$ is the vector with (1, 2, 6) and $n$ is the vector with (1, 3, 1).

With this in mind, we can find the coordinates using the Cartesian equation of a line

$r_x = a_x + t(n_x)$

$r_y = a_y + t(n_y)$

$r_z = a_z + t(n_z)$

We obviously need to find the scalar here. According to lecture note solutions, we need to apply $r_x$, $r_y$, and $r_z$ coordinates into the plane equation. My main queries are as follows:

• What purpose does the vector equation of a plane serve? In terms of its derivation, it's based on the statement $n·(r - r_1)$ = 0, but what better does this do to describe the plane than simply the coordinates of the normal line? I know that the normal line isn't a function, so that isn't very useful for us to use, what information does this tell us?
• Are any of my assumptions, or anything about my graphic wrong? If so, why? Finally, why do we plug the values into the plane equation? Please take me through the logical forethought of that.

Answer 1

Simply, lines will be: $$\frac{x-1}{1}=\frac{y-2}{3}=\frac{z-6}{1}$$

Therefore take $x=k+1,y=3k+2,z=k+6$ and plug it into plane to get $k$ and thus your answer.

I took the Direction ratios of the line similar to that of normal because in the question, it's given that line is orthogonal to the plane

What purpose does the vector equation of a plane serve? In terms of its derivation, it's based on the statement $n⋅(r−r_1)n·(r−r_1) = 0$, but what better does this do to describe the plane than simply the coordinates of the normal line? I know that the normal line isn't a function, so that isn't very useful for us to use, what information does this tell us?

It gives you the information about the normal vector of the plane and an expression for the points lying on it.

Are any of my assumptions, or anything about my graphic wrong? If so, why? Finally, why do we plug the values into the plane equation? Please take me through the logical forethought of that.

Yes, What makes you say that we hit the plane at $(1,3,1)$ . They are direction ratios of the normal vector to the plane. Not the coordinates on it. To find the coordinates, follow the method i suggested up there.

Answer 2

A little bit more explicit:

Normal to the plane: $n = (1,3,1)$.

Line through $a = (1,2,6)$ :

$r(t) = (1,2,6) + t (1,3,1)$.

$x(t) = 1 + t; $ $y((t) = 2 + 3t;$

$z(t) = 6 + t$;

Plane and line intersect:

$x(t) + 3y(t) + z(t) = 4$, an equation for $t$.

Solve for $t$ and use this $t = t_0$ value in

$r = (1,2,3) + t_0 ( 1,3,1)$

to find the coordinates of the point of intersection.