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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 $(1,1,1)$?

MathLearner
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3 Answers3

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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: $(0,2,0) \cdot (1,1,1)\overset{?}{=}0$

coldnumber
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    How is a line parallel to a plane if the direction vector of the line orthogonal to the normal vector of the plane? I mean, why have I to check if they are orthogonal to verify if they are parallel? – MathLearner Jul 21 '15 at 07:32
  • The normal vector of a plane is orthogonal to every vector in the plane, so the direction vector of the line is parallel to a vector in the plane if and only if it is orthogonal to the normal vector of the plane. Does that make sense? – coldnumber Jul 21 '15 at 07:33
  • Yes it does, but why if direction vector of the line is parallel to every vector in the plane then it is also parallel to the plane? – MathLearner Jul 21 '15 at 07:35
  • What definition do you have for when a line is parallel to a plane? The definition I was using is that a line is parallel to a plane if they do not intersect; for this the line only has to be parallel to one vector in the plane. See here: http://mathworld.wolfram.com/ParallelLineandPlane.html – coldnumber Jul 21 '15 at 07:38
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    This makes sense now. I was missing the definition and was visualizing it into my mind, that's why I didn't fully understand what you were saying. Thank you for your time and help. – MathLearner Jul 21 '15 at 07:40
  • @coldnumber What is a "direction vector"? Do you mean a vector's direction or direction angle or angle (theta)? – KeyC0de Oct 11 '19 at 17:15
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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 $\pi$: $x+y+z-2=0$

Harish Chandra Rajpoot
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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. $$0t+2t+0t=0 ?$$