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I am struggling with the interpretation of this question:

Vectors: $u = \left(1,\ 0,\ \sqrt3 \right)$ and $v = (1,\ \sqrt3,\ 0)$ in standard position.

Find an equation of the plane containing $u$ and $v$.

Am I correct in interpreting this question that the plane is parallel to both vectors? Does this mean that I can form the cross product of $u$ and $v$ to find the normal and use one of the vectors as a point on the plane?

Thank you in advance

kimtahe6
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Will777
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  • Thank you. I see that the related problem has a point on the plane supplied as well. I think my main issue here is, how do I determine a point on the plane? Can I use the origin (0,0,0) as a point on the plane, or can I simply use one of the vectors as a point on the plane? – Will777 Nov 11 '13 at 07:08
  • Your problem is different from the one I referred you to. – Mhenni Benghorbal Nov 11 '13 at 07:25
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    @Will777: That's exactly what you do. Take the cross product and zero displacement. It's not wrong either if you pick one of the vectors instead. The non-parametric form of the equation of your plane shows why this does not matter... just give it a try ;-) – The Vee Nov 14 '13 at 16:14
  • @Vašek - thank you. That is the correct answer. After struggling a bit, I realised this is no different than being given three points on a plane, since you also then calculate 2 vectors on the plane using the points, then use the cross product to get the normal. Please post it as an answer so I can accept it and thank you very much – Will777 Nov 17 '13 at 06:09
  • Correct answer is -3x + sqrt(3)y + sqrt(3)z = 0 – Will777 Nov 17 '13 at 06:11

4 Answers4

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If you're using the cross product, this gives $$ \begin{align} \begin{vmatrix} e_{\vec{x}} & e_{\vec{y}} & e_{\vec{z}} \\ 1 & 0 & \sqrt{3} \\ 1 & \sqrt{3} & 0 \end{vmatrix} = \left(\begin{vmatrix} 0 & \sqrt{3} \\ \sqrt{3} & 0\end{vmatrix}, -\begin{vmatrix} 1 & \sqrt{3} \\ 1 & 0\end{vmatrix}, \begin{vmatrix} 1 & 0 \\ 1 & \sqrt{3}\end{vmatrix}\right) = \left(-3, \sqrt{3}, \sqrt{3} \right) \end{align} $$

Thus, $\vec{p} = t(-3,\sqrt{3},\sqrt{3}), \enspace t\neq 0$ is a orthogonal vector to vector $\vec{u}$ and $\vec{v}$ and the plane they make. If $(x,y,z)$ and $(x_0,y_0,z_0)$ are points on the plane, then $\vec{w} = (x-x_0, y-y_0,z-z_0)$ is a vector on the plane, and $\vec{w}\cdot\vec{p} = 0$:

$$ \begin{align} t(-3,\sqrt{3},\sqrt{3})\cdot(x-x_0, y-y_0,z-z_0) &= 0 \\ t(-3(x-x_0)+\sqrt{3}(y-y_0)+\sqrt{3}(z-z_0) &= 0 \end{align} $$

We can arbitrarily choose $t=1$ (because a vector is orthogonal regardless of its length) and $(x_0,y_0,z_0) = (0,0,0)$ (because the plane passes through the origin, because both vectors go through the origin). This yields:

$$-3x+\sqrt{3}y+\sqrt{3}z = 0$$

BSplitter
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Frank Vel
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Well, since $u$ and $v$ are not parallel, the plane $P$ is

$$ P = \{a \cdot u + b \cdot v \} $$

for $a, b \in \mathbb{R}$. So

$$ P = \{(a, 0, \sqrt{3}a) ~ + ~ (b, \sqrt{3}b, 0) \} = \{(a + b, \sqrt{3}b, \sqrt{3}a)\} $$

To obtain the equation of the plane in Euclidean / Cartesian / xyz-form let

$$ (a + b, \sqrt{3}b, \sqrt{3}a) = (x, y, z)$$

and observe that $z - \sqrt{3}x + y = 0$.

Daniel Pietrobon
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For both $u=(1,0,\sqrt3)$ and $v=(1,\sqrt3,0)$, we observe that $x=\frac{1}{\sqrt3}(y+z)$ and this is the equation of the plane containing these vectors, since it is passing through $(0,0,0)$.

Bob Dobbs
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What if you subtract v from u and then use that as the normal (a,b,c) in the equation ax + by + cz = 0

That will give you the equation.

MAT1503 student
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  • Did you try this? It gives a wrong answer: $\sqrt3 y - \sqrt3 z = 0$. This is clearly not satisfied by the vectors $u$ and $v$, let alone their linear hull (apart from origin). – The Vee Nov 14 '13 at 16:01