formula for the angle between a line and a plane in $R^4$ 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
 A: 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:
$$a\wedge b = (2e_1 +3e_3)\wedge (e_2+e_3) = 2e_1\wedge e_2 + 2e_1\wedge e_3 + 3e_3\wedge e_2 = 2e_1e_2 + 2e_1e_3 - 3e_2e_3 \\ \|a\wedge b\| = \sqrt{4+4+9} = \sqrt{17} \\ \|c\| = \sqrt{4} =2 \\ c\ \raise .2em{\lrcorner}\ (a\wedge b) = \langle (2e_1)(2e_1e_2 + 2e_1e_3 - 3e_2e_3)\rangle_1 = \langle 4e_2 + 4e_3 - 6e_1e_2e_3\rangle_1 = 4e_2 + 4e_3 \\ \|c\ \raise .2em{\lrcorner}\ (a\wedge b)\| = \sqrt{16+16} = \sqrt{32} \\ \implies \theta = \arccos\left(\frac{\sqrt{32}}{2\sqrt{17}}\right) \approx 46.7°$$
A: 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:
$$
\pmatrix{
1 & Y_1 \cdot Y_2 & -Y_1 \cdot D \\
Y_1 \cdot Y_2 & 1 & -Y_2 \cdot D \\
-Y_1 \cdot D & -Y_2 \cdot D & 1
}
\cdot
\pmatrix{a \\ b \\ t}
=
\pmatrix{P \cdot Y_1 \\ P \cdot Y_2 \\ -P \cdot D}
$$
A: 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 $\pi/2 - \theta$.
