If the line $r$ has direction vector $(0,2,0)$, how can I verify if it is parallel to the following plane $\pi : x+y+z-2=0$ with orthogonal direction vector $(1,1,1)$?
[Math] How to verify that a line is parallel to a plane
geometry
Related Solutions
After "cross multiplying" and canceling the factor of $\sqrt{2}$ on each side, we have $$ \sqrt{a^2 + b^2 + c^2} = |a - b|. $$ Now, square both sides and remove $a^2$ and $b^2$ terms that appear on both sides, yielding $$ c^2 = -2ab. $$
From the original equation, we know that $a + b + c = 0$, so $$ b = -a - c. $$
At this point, we can make the simplifying assumption that $a = 1$. Why? (Hover to reveal the answer after you've tried to answer this yourself.)
As long as $a \ne 0$ and we're looking for a direction vector, we may choose the scale of one coordinate. How do we know that $a \ne 0$? If it were true that $a = 0$, then the quadratic equation would give $c^2 = 0$ and so $c = 0$. From there, the linear equation would give $b = 0$. Now, $(a, b, c) = (0, 0, 0)$. No good.
Substitute $b = -1 - c$ into $c^2 = -2b$, producing $c^2 = 2 + 2c$ or $(c - 1)^2 = 3$. Therefore, $$ c = 1 \pm \sqrt{3}. $$
Now, $$ b = -1 - (1 \pm \sqrt{3}) = -2 \mp \sqrt{3}. $$
Putting this all together, their are two possible direction vectors: $$ \vec V_r = \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \mp \sqrt{3} \\ 1 \pm \sqrt{3} \end{bmatrix}. $$
Another condition "for a line to intersect another line, in 3D" is that there is a point that is on both lines. That sounds simplistic but that really is the condition.
Here is one way to solve your problem. The line that you want is in a plane that is parallel to the plane $3x-y+2z-15=0$. Any such plane has the equation
$$3x-y+2z+D=0$$
for some constant $D$. That plane must also go through the point $M(1,0,7)$, so we can substitute to find the value of $D$:
$$3\cdot 1-0+2\cdot 7+D=0$$
So $D=-17$ and the plane is $3x-y+3z-17=0$.
Combining that with your equations $\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}$ you can find the single point that is the intersection of that plane and your given line. The line you want goes through those two points.
You should be able to finish from here.
There are multiple ways to find the intersection of the plane and of the given line. You showed one way, by giving a parameterization of the line and solving for the parameter. You did make a computation error: the equation you get is
$$3(4t+1)-(2t+3)+2(t)-17=0$$
and the solution is
$$t=\frac{17}{12}$$
which makes $x=\frac{20}3,\ y=\frac{35}6,\ z=\frac{17}{12}$. Substituting this solution into all the equations checks, so this is the right answer. Thus the intersection point is
$$Q\left(\frac{20}3,\ \frac{35}6,\ \frac{17}{12}\right)$$
I got the intersection point by different means, by setting up and solving these simultaneous linear equations:
$$\begin{align} \color{white}{1}x \color{white}{+0y}-4z&=1 \\ y-2z&=3 \\ 3x-y+2z&=17 \end{align}$$
I got the same intersection point.
Best Answer
A line is parallel to a plane if the direction vector of the line is orthogonal to the normal vector of the plane.
To check whether two vectors are orthogonal, you can find their dot product, because two vectors are orthogonal if and only if their dot product is zero.
So in your example you need to check: $(0,2,0) \cdot (1,1,1)\overset{?}{=}0$