Geometry and Trigonometry - Distances (HL)

Find the distance between the points $A(1, -2, 4)$ and $B(3, 0, -1)$.

$d = \sqrt{(3 - 1)^2 + (0 - (-2))^2 + (-1 - 4)^2}$ $d = \sqrt{2^2 + 2^2 + (-5)^2}$ $d = \sqrt{4 + 4 + 25} = \sqrt{33}$

Apply the 3D distance formula by taking the square root of the sum of the squared differences of the $x$, $y$, and $z$ coordinates.

Calculate the shortest distance from the point $P(2, 1, 3)$ to the plane with equation $3x - 2y + 6z = 5$.

Rewrite the plane equation as $3x - 2y + 6z - 5 = 0$. Here $a=3, b=-2, c=6, d=-5$. $D = \frac{|3(2) + (-2)(1) + 6(3) - 5|}{\sqrt{3^2 + (-2)^2 + 6^2}}$ $D = \frac{|6 - 2 + 18 - 5|}{\sqrt{9 + 4 + 36}}$ $D = \frac{17}{\sqrt{49}} = \frac{17}{7}$

Substitute the point coordinates into the numerator of the point-to-plane distance formula and divide by the magnitude of the normal vector $(a, b, c)$.

Find the shortest distance between the skew lines $L_1: \mathbf{r}$ = $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ + $\lambda \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $L_2: \mathbf{r}$ = $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ + $\mu \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$.

Step 1: Find the common normal $\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$: $\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \mathbf{i}(1-0) - \mathbf{j}(1-0) + \mathbf{k}(1-0)$ = $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ Step 2: Vector between points on the lines $\mathbf{a}_2 - \mathbf{a}_1$ = $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ - $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ = $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$. Step 3: Apply the formula: $d = \frac{\left|\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\right|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{|-1 - 1 + 0|}{\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

The shortest distance between skew lines is the scalar projection of the vector joining any two points on the lines onto the cross product of the direction vectors.

1. Identify the plane equation in the form $ax + by + cz + d = 0$: $x + 2y - 2z - 10 = 0$.
2. Identify coefficients: $a=1, b=2, c=-2, d=-10$.
3. Point coordinates: $x_0=4, y_0=-1, z_0=2$.
4. Apply formula: $D = \frac{|1(4) + 2(-1) - 2(2) - 10|}{\sqrt{1^2 + 2^2 + (-2)^2}}$
5. Calculate numerator: $|4 - 2 - 4 - 10| = |-12| = 12$.
6. Calculate denominator: $\sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
7. Result: $D = \frac{12}{3} = 4$.

The point-to-plane distance formula calculates the projection of a vector from any point on the plane to $Q$ onto the plane's normal vector $\mathbf{n} = [1, 2, -2]$.

1. Find a point $P$ on $\Pi_1$. Let $x=0, y=0$, then $2z=6 \Rightarrow z=3$. So $P(0, 0, 3)$.
2. Use the point-to-plane distance formula from $P(0, 0, 3)$ to $\Pi_2: 2x - y + 2z - 15 = 0$.
3. $D = \frac{|2(0) - 1(0) + 2(3) - 15|}{\sqrt{2^2 + (-1)^2 + 2^2}}$
4. Numerator: $|6 - 15| = |-9| = 9$.
5. Denominator: $\sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
6. Result: $D = \frac{9}{3} = 3$.

Since the planes are parallel (they share the same normal vector $[2, -1, 2]$), the distance between them is constant. We can pick any point on one plane and find its distance to the other.