docs: maths: fix issues

src/content/docs/maths/matrices/14-eigenvalues-eigenvectors.md

@@ -21,8 +21,11 @@ Roots of the equation $p(\lambda) = 0$ are the eigenvalues of $A$.
 
 :::note
 
-- [Determinant of a matrix](/maths/matrices/determinant/#in-relation-with-eigenvalues)
-  can be written in terms of all of its eigenvalues.
+- Product of the eigenvalues is equal to the
+  [determinant](/maths/matrices/determinant/#in-relation-with-eigenvalues) of
+  the matrix
+- Sum of the eigenvalues is equal to the [trace](/maths/matrices/trace) of a
+  matrix
 - If $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of
   $A^2$
 - $A$ and $A^T$ have the same eigenvalues.

src/content/docs/maths/riemann-integration/17-gamma-function.md

@@ -138,12 +138,14 @@ $\ln$, it's better to try this substitution.
 
 ### Form 4
 
+For $k\in \mathbb{R}$:
+
 ```math
-\Gamma(n) = 2 \int_0^\infty e^{-x^2} x^{2n-1}\,\text{d}x
+\Gamma(n) = k \int_0^\infty e^{-x^k} x^{kn-1}\,\text{d}x
 ```
 
 :::note[Proof Hint]
 
-Use $x=t^2$. $ $
+Use $x=t^k$. $ $
 
 :::

src/content/docs/maths/vectors/03-straight-lines.md

@@ -95,15 +95,21 @@ Using the $\alpha,\beta$ lines mentioned above: $ $
 
 Here $v_1, v_2$ are $2$ vectors parallel to $\alpha, \beta$ respectively.
 
-## Shortest distance to a point
+## Shortest distance from a point
 
-Suppose $x_1$ and $x_2$ lie on a line. Shortest distance to the point $P$ is:
+The distance can be calculated using Pythogoras' theorem.
 
 ```math
 d^2 =
-\frac{
-\bigg|(\underline{x_2} - \overrightarrow{OP}) \times (\underline{x_1}-\overrightarrow{OP})\bigg|^2
-}{
-\big|\underline{x_2} - \underline{x_1}\big|^2
-}
+{\big\lvert\underline{r} - \underline{p}\big\rvert}^2 -
+\bigg[
+\frac{\underline{n}\cdot (\underline{r} - \underline{p})}{\lvert \underline{n} \rvert}
+\bigg]^2
 ```
+
+Here:
+
+- $P$ is the arbitrary point
+- $\underline{p}$ is the position vector of $P$
+- $\underline{r}$ is the position vector of a point on the line
+- $\underline{n}$ is parallel to the line

src/content/docs/maths/vectors/04-planes.md

@@ -77,21 +77,46 @@ The angle between the planes $\phi$ is given by: $ $
 
 Here $\underline{n_A},\underline{n_B}$ are normal to the planes $A,B$.
 
-## Shortest distance to a point
+## Shortest distance from a point
 
-Considering a plane $ax+by+cz=d$.$ $
+Consider the plane $ax+by+cz=d$.$ $
 
 ```math
 \text{distance}=
 \frac{
-\lvert
+\big\lvert
 (\underline{r_1}-\underline{r_0})\cdot\underline{n}
-\rvert
+\big\rvert
 }{
 \lvert{\underline{n}}\rvert
 }
 ```
 
 - $\underline{n}$ is a normal to the plane
 - $\underline{r_0}$ is the position vector of any known point on the plane
-- $\underline{r_1}$ is the position vector to the arbitrary point
+- $\overline{r_1}$ is the position vector to the arbitrary point
+
+## Intersection
+
+In 3D, to prove 2 planes intersect, it has to be proven that there is a point
+satisfiying both of the planes.
+
+### Of 2 planes
+
+Can either be a:
+
+- Plane - when the planes coincicde
+- Line - otherwise
+
+Equation of the line of intersection can be found by:
+
+- Solving $y,z$ with respect to $x$
+- Subject $x$ and symmetric form can be found
+
+### Of 3 planes
+
+Can either be a:
+
+- Plane - when the planes coincide
+- Line
+- Point

src/content/docs/maths/vectors/05-skew-lines.md

@@ -59,5 +59,5 @@ a_1,b_1,c_1
 
 Here
 
-- $\underline{n}$ is the normal to both $l_1,l_2$
+- $\underline{n}$ is the unit normal to both $l_1,l_2$
 - $A$ and $B$ are points lying on each line