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 $(1,1,1)$?
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$\begingroup$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: $(0,2,0) \cdot (1,1,1)\overset{?}{=}0$
$\endgroup$ 6 $\begingroup$The angle say $\alpha$ between any two vectors say $\vec{a}$ & $\vec b$ is given as $$\color{red}{\cos \alpha=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}}$$ Hence, the angle say $\alpha$ between the direction vector of the line $r$: $(0\hat i+2\hat j+0\hat k)$ & the normal vector of plane $\pi$: $(\hat i+\hat j+\hat k)$ is given as $$\cos \alpha=\frac{(0\hat i+2\hat j+0\hat k)\cdot (\hat i+\hat j+\hat k)}{|0\hat i+2\hat j+0\hat k||\hat i+\hat j+\hat k|}$$ $$\implies \cos \alpha=\frac{0+2+0}{\sqrt{0^2+2^2+0^2}\sqrt{1^2+1^2+1^2}}=\frac{2}{2\sqrt{3}}$$ $$\implies \color{blue}{\alpha=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)}$$
But for the line $r$ to be normal to the plane $\pi$, the angle $\alpha$ (between the direction vector of line & the normal vector of the plane) must be $90^\circ$
Thus, the given line $r$ is not parallel to the plane $\pi$: $x+y+z-2=0$
$\endgroup$ $\begingroup$Take an arbitrary point in the plane, let $(1,1,0)$. Then moving in the direction of the line, $(1,1,0)+t(0,2,0)$ you must stay in the plane.
Does $$(1+0t)+(1+2t)+(0+0t)-2=0\ ??$$
Actually it is simpler to translate the plane to the origin, $x+y+z=0$ and start from it. $$0t+2t+0t=0 ?$$
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