Question

Show that the given lines intersect and find the acute angle between them. $$\mathbf{r}=2 \mathbf{j}+\mathbf{k}+(3 \mathbf{i}-\mathbf{k}) t_{1} \quad \text { and } \quad \mathbf{r}=7 \mathbf{i}+2 \mathbf{k}+(2 \mathbf{i}-\mathbf{j}+\mathbf{k}) t_{2}$$

Short answer

The lines intersect at (3, 2, 0) and the acute angle between them is \(\cos^{-1}\frac{\root15}{6}\).

Step-by-step solution

Parameterize the Lines

First, express both lines in parametric form. For the first line: \[ \textbf{r}_1 = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} + t_1 \begin{pmatrix} 3 \ 0 \ -1 \end{pmatrix} \] For the second line: \[ \textbf{r}_2 = \begin{pmatrix} 7 \ 0 \ 2 \end{pmatrix} + t_2 \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} \]

Set the Parametric Equations Equal

To find the intersection, set the parametric equations of the lines equal: \[ \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} + t_1 \begin{pmatrix} 3 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} 7 \ 0 \ 2 \end{pmatrix} + t_2 \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} \]

Equate Components

Equate the components of the vectors: \[ \begin{cases} 3t_1 = 7 + 2t_2 \ 2 = -t_2 \ 1 - t_1 = 2 + t_2 \ \text{Solve for } t_2 \text{ and } t_1 \end{cases} \]

Solve for t2 and t1

From the second equation, solve for \( t_2 \): \[ t_2 = -2 \] Substitute \( t_2 = -2 \) in the first and third equations: \[ 3t_1 = 7 + 2(-2) \ 3t_1 = 3 \ t_1 = 1 \]

Substitute t1 and t2 back into Equations

Substitute \( t_1 = 1 \) and \( t_2 = -2 \) back to check intersection: \[ \textbf{r}_1 (t_1=1) = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} + 1 \begin{pmatrix} 3 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \ 0 \end{pmatrix} \]\[ \textbf{r}_2 (t_2=-2) = \begin{pmatrix} 7 \ 0 \ 2 \end{pmatrix} + (-2) \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \ 0 \end{pmatrix} \]

Find Direction Vectors

Compute the direction vectors: \[ \textbf{d}_1 = \begin{pmatrix} 3 \ 0 \ -1 \end{pmatrix} \ \textbf{d}_2 = \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} \]

Compute Dot Product and Magnitudes

Use the dot product formula and magnitudes to find the angle: \[ \textbf{d}_1 \times \textbf{d}_2 = 3\times 2 + 0 \times -1 + (-1) \times 1 = 5 \] The magnitudes are: \[ |\textbf{d}_1| = \begin{Vmatrix} 3 \ 0 \ -1 \end{Vmatrix} = \ \begin{Vmatrix} 3^2 + 0^2 + (-1)^2 \end{Vmatrix} = \ \begin{Vmatrix} 10 \end{Vmatrix} = \ \sqrt{10} \] \[ |\textbf{d}_2| = \begin{Vmatrix} 2 \ -1 \ 1 \end{Vmatrix} = \ \begin{Vmatrix} 2^2 + (-1)^2 + 1^2 \end{Vmatrix} = \ \begin{Vmatrix} 6 \end{Vmatrix} = \ \sqrt{6} \]

Calculate Angle Using Dot Product

Use the dot product to find the cosine of the angle: \[ \text{cos} \theta = \frac{\textbf{d}_1 \times \textbf{d}_2}{|\textbf{d}_1| \times |\textbf{d}_2|} = \ \frac{5}{\sqrt{10} \times \sqrt{6}} = \ \frac{5}{\sqrt{60}} = \ \frac{5}{2\sqrt{15}} = \ \frac{\sqrt{15}}{6} \] \[ \theta = \cos^{-1}\frac{\sqrt{15}}{6} \]

Key concepts

vector mathematics

Vector mathematics is a branch of mathematics that deals with quantities that have both magnitude and direction. Vectors are typically represented as arrows in a coordinate system with their length indicating the magnitude and the arrowhead indicating the direction.
In everyday use, vectors can represent various physical quantities like velocity, force, and displacement.

• Scalar: A quantity with magnitude only.
• Vector: A quantity with both magnitude and direction.

In our example, the vectors \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) are used to represent lines in parametric form, which helps find the intersection point and understand their direction in space.

dot product

The dot product (or scalar product) is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This product reveals how much one vector extends in the direction of another. Mathematically, the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\) is defined as:
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \ldots + a_nb_n\]
In the exercise, the dot product helps determine the angle between the two vectors. For vectors \(\mathbf{d}_1 = \begin{pmatrix} 3 \ 0 \ -1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix}\), the dot product is calculated as:
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = 3 \times 2 + 0 \times -1 + (-1) \times 1 = 5\), which is a critical step in finding the angle between the lines.

parametric equations

Parametric equations define a group of quantities as functions of one or more independent variables called parameters. These equations are often used to describe the motion of a point along a plane or space.
In our exercise, we use parametric equations to represent lines, which include:
\[\mathbf{r}_1 = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} + t_1 \begin{pmatrix} 3 \ 0 \ -1 \end{pmatrix}\] and
\[\mathbf{r}_2 = \begin{pmatrix} 7 \ 0 \ 2 \end{pmatrix} + t_2 \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix}\]
By setting the parametric equations equal, students can find common points, thus determining their intersection. The parameters \(t_1\) and \(t_2\) within these equations are solved to establish at what points the vectors intersect.

angle calculation

Calculating the angle between two vectors involves the dot product and magnitudes of the vectors. The cosine of the angle \(\theta\) between vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:
\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| \, ||\mathbf{b}||}\]
In the exercise, we find:
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = 5\)
The magnitudes are:
\[||\mathbf{d}_1|| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{10}\]
and
\[||\mathbf{d}_2|| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}\]
Thus, the angle calculation is:
\[\cos \theta = \frac{5}{\sqrt{10} \sqrt{6}} = \frac{5}{\sqrt{60}} = \frac{5}{2 \sqrt{15}} = \frac{\sqrt{15}}{6}\]
Finally, to find the angle, use the inverse cosine:
\(\theta = \cos^{-1}\left(\frac{\sqrt{15}}{6}\right)\).