Geometry and Trigonometry

Formula Booklet:

Vector equation of a line	\( r = a + \lambda b\)
Parametric form of the equation of a line	\( x=x_0+\lambda l, y=y_0+\lambda m, z=z_0+\lambda n.\)
Cartesian equations of a line	\( \dfrac{x-x_0}{l} = \dfrac{y-y_0}{m} = \dfrac{z-z_0}{n} \)

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In two dimensions, the vector equation of a line is given by:

$$\vec{r} = \vec{a} + t\vec{b}$$

where \(\vec{r}\) is the position vector of a point on the line, \(\vec{a}\) is the position vector of a known point on the line, \(\vec{b}\) is the direction vector of the line, and \(t\) is a parameter. In three dimensions, the vector equation of a line is given by:

$$\vec{r} = \vec{a} + t\vec{b} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} + t\begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix}$$

where \(\vec{r}\), \(\vec{a}\), and \(\vec{b}\) are all position vectors in three dimensions. One simple application of vector equations of lines is in kinematics, where they can be used to model the motion of an object in space.

Here's an example:

$$\text{Find the vector equation of the line passing through the points } A(1, 2, 3) \text{ and } B(4, 5, 6).$$

We can find the direction vector of the line by subtracting the position vector of point \(A\) from the position vector of point \(B\):

$$\vec{b} = \vec{OB} - \vec{OA} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$$

So the vector equation of the line passing through \(A\) and \(B\) is:

$$\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$$

This is the vector equation of a line in three dimensions passing through points \(A\) and \(B\), where \(t\) is a parameter that can be used to find any point on the line.

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Suppose a particle moves along a straight line in space, and its position vector at time \(t\) is given by:

$$\vec{r}(t) = \begin{pmatrix} 3t \\ 2t + 1 \\ t \end{pmatrix}$$

We can find the velocity vector of the particle by differentiating the position vector with respect to time:

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$$

This tells us that the particle is moving in the direction of the vector \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\). We can also see that the particle is moving with a constant speed, since the magnitude of \(\vec{v}\) is constant:

$$|\vec{v}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}$$

We can use the vector equation of a line to describe the path of the particle. Since the particle is moving in the direction of the vector \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\), we can use this as the direction vector of the line. We can also use the initial position vector of the particle, \(\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\), as a point on the line. Thus, the vector equation of the line describing the path of the particle is:

$$\vec{r}(t) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3t \\ 2t + 1 \\ t \end{pmatrix}$$

So the particle is moving along a line in space, with a constant speed of \(\sqrt{14}\) units per time unit, in the direction of the vector \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\).

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