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Is there a formula for the angle between a line and a plane in $R^4$? More precisely, let $E_1$,$E_2$ and $E_3$ be three (non-colinear) vectors in $R^4$. We can always define the angle between two vectors by using "dot product". The angle between the vector $E_1$ and the plane spanned by $E_2$ and $E_3$ can be defined as the minimum angle between $E_1$ and a vector in $span(E_2, E_3)$. Can we explicitly write out this angle in terms of coordinates of $E_1$, $E_2$ and $E_3$?

PS: There is a formula for the angle between two planes in $R^4$. See

Angle between two planes in four dimensions

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  • $\begingroup$ Think about the projection of $E_1$ onto the span of $E_2$ and $E_3$. $\endgroup$ – Michael Biro Oct 17 '15 at 1:03
  • $\begingroup$ @Bye_World What is the definition of left contraction? Is there a formula for it? I just need the four dimensional case. $\endgroup$ – Tony B Oct 17 '15 at 1:25
  • $\begingroup$ @Bye_World No, I don't. It seems quite complicated and not easy to understand for undergrad. It seems to be in a new subject called "Clifford Algebras" $\endgroup$ – Tony B Oct 17 '15 at 1:31
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There's a simple formula for the angle between $A$ (a $j$-blade representing some subspace of $\Bbb R^n$) and $B$ (a $k$-blade representing some other subspace of $\Bbb R^n$), where $j\le k$, given by $$\cos(\theta) = \dfrac{\|A\ \raise .2em{\lrcorner}\ B\|}{\|A\|\|B\|}$$ where $\raise .2em{\lrcorner}$ is the left contraction product.


But because you're not familiar with geometric algebra, I'll try to give you just enough (non-rigorous) definitions so that you can calculate the angle between any two subspaces of $\Bbb R^n$.

Definitions:

Wedge Product: We define the wedge product, denoted $a\wedge b$, of two vectors $a, b\in \Bbb R^n$ as an object that is neither a vector nor a scalar and obeys all of the following: $$\begin{align}a \wedge b &= -b \wedge a \\ a\wedge(b\wedge c) &= (a\wedge b)\wedge c \\ a\wedge a &= 0 \\ k(a\wedge b) &= (ka)\wedge b = a\wedge (kb) \tag{$k\in \Bbb R$}\\ a\wedge (b+c) &= a\wedge b + a\wedge c\end{align}$$

Blade: An object formed by the wedge product of $k$ vectors is called a $k$-blade. We define scalars as $0$-blades, vectors as $1$-blades, objects that can be written as $a\wedge b$ as $2$-blades, objects that can be written as $a\wedge b\wedge c$ as $3$-blades, etc.

$k$-vector: A $k$-vector is a linear combination of $k$-blades.

Multivector: A multivector is a linear combination of $k$-vectors.

Grade: Objects in the algebra we're building have "grades". All $k$-blades/ $k$-vectors have grade $k$. But multivectors are in general multigraded objects. For instance, $B = b_0 + b_1e_1 + b_2 e_2 + b_{12}e_1\wedge e_2$, where $b_i$ are scalars and $e_i$ are vectors, is a multigraded object.

Clifford product: The Clifford product is an associative product of multivectors satisfying: $$ab = a\cdot b + a\wedge b \\ (AB)C = A(BC) \\ k(AB) = (kA)B = A(kB) \\ A(B+C) = AB+AC$$ for all vectors $a,b$, scalars $k$, and multivectors $A,B,C$.

  • A consequence of this is that if $\{e_1, \dots, e_n\}$ is an orthonormal basis of $\Bbb R^n$ then $e_ie_i = e_i\cdot e_i = 1$ for all $i$ and $e_ie_j = e_i \wedge e_j$ for $i\ne j$.

Grade projection: The grade projection operator, denoted $\langle A \rangle_i$, returns the grade $i$ parts of the multivector $A$. For instance, $\langle ab\rangle_0 = \langle a\cdot b + a\wedge b\rangle_0 = a\cdot b$.

Norm: The norm of a multivector can be determined in the standard way after decomposing it into an orthonormal basis. For instance, if $A = a_0 + a_1e_1 + a_2e_2 + a_{12}e_1 \wedge e_2$ then $$\|A\| = \sqrt{a_0^2 + a_1^2 + a_2^2 + a_{12}^2}$$

Left contraction: The left contraction product of a $j$-blade $A$ and a $k$-blade $B$ is defined as $$A\ \raise .2em{\lrcorner}\ B = \langle AB\rangle_{k-j}$$


Now for an example. Consider the plane spanned by $a=2e_1 +3e_3$ and $b=e_2+e_3$ and the line spanned by $c=2e_1$.

Then the angle between that line and that plane is given by $$\cos(\theta) = \frac{\|c\ \raise .2em{\lrcorner}\ (a\wedge b)\|}{\|c\|\|a\wedge b\|}$$

So let's calculate it:

$$a\wedge b = (2e_1 +3e_3)\wedge (e_2+e_3) = 2e_1\wedge e_2 + 2e_1\wedge e_3 + 3e_3\wedge e_2 = 2e_1e_2 + 2e_1e_3 - 3e_2e_3 \\ \|a\wedge b\| = \sqrt{4+4+9} = \sqrt{17} \\ \|c\| = \sqrt{4} =2 \\ c\ \raise .2em{\lrcorner}\ (a\wedge b) = \langle (2e_1)(2e_1e_2 + 2e_1e_3 - 3e_2e_3)\rangle_1 = \langle 4e_2 + 4e_3 - 6e_1e_2e_3\rangle_1 = 4e_2 + 4e_3 \\ \|c\ \raise .2em{\lrcorner}\ (a\wedge b)\| = \sqrt{16+16} = \sqrt{32} \\ \implies \theta = \arccos\left(\frac{\sqrt{32}}{2\sqrt{17}}\right) \approx 46.7°$$

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Let $\alpha$ be your plane. $X \in \alpha \Leftrightarrow \exists a, b \in \mathbb{R}$ such as $X = a Y_1 + b Y_2$, where $Y_1 = \frac{X_2 - X_1}{|X_2 - X_1|}$, $Y_2 = \frac{X_3 - X_1}{|X_3 - X_1|}$, and $X_1, X_2, X_3 \in \alpha$.

Let $L$ be your line. $X \in L \Leftrightarrow X = P + tD$, where $P \in L$ and $D$ is the normalized director vector ($|D| = 1$).

The distance between then is $d = \min_{a, b, t} |aY_1 + bY_2 - P - tD|$.

Let $f(a, b, t) = |aY_1 + bY_2 - P - tD|$. To find the minimum, we need to solve $\nabla f = 0$.

Doing the math, we need to solve the following linear system:

$$ \pmatrix{ 1 & Y_1 \cdot Y_2 & -Y_1 \cdot D \\ Y_1 \cdot Y_2 & 1 & -Y_2 \cdot D \\ -Y_1 \cdot D & -Y_2 \cdot D & 1 } \cdot \pmatrix{a \\ b \\ t} = \pmatrix{P \cdot Y_1 \\ P \cdot Y_2 \\ -P \cdot D} $$

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Here's a straightforward approach: find the unique (up to scalar multiples) non-zero vector $E = c_1 E_1 + c_2 E_2 + c_3 E_3$ that is normal to the plane spanned by $E_2$ and $E_3$. (This comes down to solving a $2 \times 3$ system of equations of rank $2$.) Then compute the angle $\theta$ between $E_1$ and $E$. The angle you seek is $\pi/2 - \theta$.

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