How to find the equation of a plane given the equation of a line?

Question

Given the line $(x, y, z) = (1, -3, 2) + t(-2, 4, 7)$ , find planes to satisfy the following condition:

• A plane which is not intersected by the line

I'm just starting to learn this and find it a little confusing.

Answer 1

Go through each paragraph of below until you are sure of the answer yourself:

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First of all, in $\mathbb{R}^3$, what is required for a line and a plane to not intersect? Clearly, if the line and the plane is not parallel, then they must intersect at a point. So they better be parallel. So given they are parallel, we want one more condition: the line must not lie on the plane (obviously). Note that, these conditions are satisfied by infinitely many planes.

Now your line has direction vector $(-2,4,7)$. I assume you know how to find the equation of a plane using a vector normal to it. So how do we find a vector normal to $(-2,4,7)$?

Observe: $(-2)\times2+4\times1+7\times0=0$. Hence we see that the vector $(2,1,0)$ is normal to our direction vector (since their scalar product is 0). So we have found a normal vector to our plane.

Finally, we want any plane normal to this vector not passing through $(1,-3,2)$, or indeed any point on the line as the line must not lie on the plane. How do we make sure a point is not on the line? Consider a point $(2,-3,2)$. Notice that there is no $t$ satisfying

$ (1,-3,2)+t(-2,4,7)=(2,-3,2)$

(Notice also, that I just added 1 to x component of $(1,-3,2)$ to produce this vector)

So we are done: putting all together, our plane is normal to $(2,1,0)$ and must pass through $(2,-3,2)$, so we have our equation of plane:

$[\vec{r}-(2,-3,2)]\cdot (2,1,0)=0 \implies \vec{r}\cdot(2,1,0)=1 \implies 2x+y=1$

where $\vec{r}=(x,y,z)$

Answer 2

The equation for a plane can be written as

$$Ax+By+Cz+D=0$$

for some constants $A$, $B$, $C$, and $D$.

If the plane intersects the line, then for some $t$, we have

$$A(1-2t)+B(-3+4t)+C(2+7t)+D=0$$

Rearranging the above, we get

$$A-3B+2C+D+t(-2A+4B+7C)=0$$

Solving this for $t$, we get $$t=\frac{-A+3B-2C-D}{-2A+4B+7C}$$

Hence we have a solution except possibly when the denominator is equal to $0$, that is $-2A+4B+7C=0$. Additionally, when $-2A+4B+7C=0$, we find solutions in our equation if and only if $A-3B+2C+D=0$ as well (note that in this last case, the equation is $0$ for every $t$, so these are planes in which the line lies).

In conclusion, any plane in the form $Ax+By+Cz+D=0$ does not intersect the line if and only if $-2A+4B+7C=0$ and $A-3B+2C+D\ne 0$