Geometry and Trigonometry - Vector equation of a line in 3D (HL)

Find the vector equation of the line passing through points $A(1, -2, 3)$ and $B(4, 0, -1)$.

1. Find the direction vector $\mathbf{b} = \vec{AB} = \mathbf{b} - \mathbf{a}$ = $\begin{pmatrix} 4 - 1 \\ 0 - (-2) \\ -1 - 3 \end{pmatrix}$ = $\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}$.
2. Use point $A$ as the position vector $\mathbf{a} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$.
3. The equation is $\mathbf{r} =$ $\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$ + $\lambda \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}$.

To find the equation of a line, we need a point on the line and a direction vector. The direction vector is found by subtracting the coordinates of the two given points.

Find the acute angle between the lines $L_1: \mathbf{r}_1$ = $\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ + $\lambda$ $\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $L_2: \mathbf{r}_2$ = $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ + $\mu \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$.

1. Identify direction vectors: $\mathbf{b}_1 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b}_2 = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$.
2. Calculate dot product: $\mathbf{b}_1 \cdot \mathbf{b}_2 = (1)(3) + (2)(0) + (2)(4) = 3 + 0 + 8 = 11$.
3. Calculate magnitudes: $|\mathbf{b}_1| = \sqrt{1^2 + 2^2 + 2^2} = 3$ and $|\mathbf{b}_2| = \sqrt{3^2 + 0^2 + 4^2} = 5$.
4. Calculate $\cos \theta = \frac{11}{3 \times 5} = \frac{11}{15}$.
5. $\theta = \arccos(\frac{11}{15}) \approx 42.8^\circ$.

The angle between two lines is determined solely by the dot product of their direction vectors.

Determine if the following two lines intersect: $L_1: \mathbf{r} =$ $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ $+ \lambda$ $\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ $L_2: \mathbf{r}$ = $\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ + $\mu \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$

1. Equate components: $1 + 2\lambda = 2 + \mu$ (i) $1 + \lambda = 1 - \mu$ (ii) $2\lambda = 2$ (iii)
2. From (iii), $2\lambda = 2 \implies \lambda = 1$.
3. Substitute $\lambda = 1$ into (ii): $1 + 1 = 1 - \mu \implies \mu = -1$.
4. Check if these values satisfy (i): $1 + 2(1) = 3$; $2 + (-1) = 1$. Since $3 \neq 1$, the lines do not intersect.
5. Since direction vectors are not proportional, the lines are skew.

To check for intersection, we set the $x, y, z$ components of both lines equal and solve for the parameters. If a consistent set of parameters exists for all three equations, they intersect; otherwise, they are skew or parallel.

$\begin{pmatrix} 7 \\ 1 \\ 10 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$

Equate the components to find $\lambda$:

1. $7 = 1 + 2\lambda \implies 6 = 2\lambda \implies \lambda = 3$
2. $1 = -2 + \lambda \implies 3 = \lambda$
3. $10 = 1 + 3\lambda \implies 9 = 3\lambda \implies \lambda = 3$

Since the value of $\lambda = 3$ is consistent for all three coordinates, the point $P$ lies on the line.

To check if a point lies on a vector line, substitute the point's coordinates for $\mathbf{r}$ and solve for the parameter $\lambda$. If all three components yield the same $\lambda$, the point is on the line.

The vector equation is $\mathbf{r} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix}$.

This gives parametric equations: $x = 2 - \lambda \implies \lambda = \frac{x - 2}{-1}$ $y = 5 + 4\lambda \implies \lambda = \frac{y - 5}{4}$ $z = -1 + 2\lambda \implies \lambda = \frac{z + 1}{2}$

Equating the expressions for $\lambda$: $\frac{x - 2}{-1} = \frac{y - 5}{4} = \frac{z + 1}{2}$

The Cartesian equation is derived by expressing the parameter $\lambda$ in terms of $x, y,$ and $z$ and setting the resulting expressions equal to each other.