Problem 16
Question
Determine if the described lines are the same line, parallel lines, intersecting or skew lines. If intersecting, give the point of intersection. $$ \begin{array}{l} \overrightarrow{\ell_{1}}(t)=\langle 2,1,1\rangle+t\langle 5,1,3\rangle \\ \vec{\ell}_{2}(t)=\langle 14,5,9\rangle+t\langle 1,1,1\rangle \end{array} $$
Step-by-Step Solution
Verified Answer
The lines intersect at the point (12, 3, 7).
1Step 1: Compare Direction Vectors
The direction vector of \( \ell_1 \) is \( \langle 5, 1, 3 \rangle \), and the direction vector of \( \ell_2 \) is \( \langle 1, 1, 1 \rangle \). Since these vectors are not proportional, the two lines are not parallel. Therefore, we must determine if they intersect or are skew.
2Step 2: Set Parametric Equations Equal
Set the parametric equations of \( \ell_1 \) and \( \ell_2 \) equal to find the point of intersection: \( \langle 2,1,1 \rangle + t \langle 5,1,3 \rangle = \langle 14,5,9 \rangle + s \langle 1,1,1 \rangle \). This provides three equations: \( 2 + 5t = 14 + s \), \( 1 + t = 5 + s \), and \( 1 + 3t = 9 + s \).
3Step 3: Solve the System of Equations
From the second equation, solve for \( t \): \( t = 4 + s \). Substitute \( t = 4 + s \) into the first equation: \( 2 + 5(4 + s) = 14 + s \), simplifying to \( 22 + 5s = 14 + s \), so \( 4s = -8 \), which gives \( s = -2 \). Substitute \( s = -2 \) back to find \( t \): \( t = 4 - 2 = 2 \).
4Step 4: Verify the Solution in the Third Equation
Substitute \( t = 2 \) and \( s = -2 \) into the third equation \( 1 + 3t = 9 + s \): \( 1 + 3(2) = 9 - 2 \), which simplifies to \( 7 = 7 \). This equality confirms that \( t = 2 \) and \( s = -2 \) is a solution, indicating the lines intersect.
5Step 5: Find the Intersection Point
Substitute \( t = 2 \) into the equation of \( \ell_1 \) to find the intersection point: \( \langle 2 + 5(2), 1 + 2, 1 + 3(2) \rangle = \langle 12, 3, 7 \rangle \). Therefore, the point of intersection is \( (12, 3, 7) \).
Key Concepts
Direction VectorsParametric EquationsSystem of EquationsPoint of Intersection
Direction Vectors
Direction vectors are crucial in understanding the orientation of a line in space. They define the direction in which a line extends. For example, in the lines provided in the exercise, the direction vector of line \( \ell_1 \) is \( \langle 5, 1, 3 \rangle \), while for line \( \ell_2 \), it is \( \langle 1, 1, 1 \rangle \). These vectors help us understand how the lines behave relative to each other.
They help determine if lines are parallel or intersect. Parallel lines have proportional direction vectors, meaning one can be expressed as a scalar multiple of the other. Since the direction vectors of \( \ell_1 \) and \( \ell_2 \) are not proportional, the lines are not parallel, which leads us to explore other relationships like intersection or skewing.
They help determine if lines are parallel or intersect. Parallel lines have proportional direction vectors, meaning one can be expressed as a scalar multiple of the other. Since the direction vectors of \( \ell_1 \) and \( \ell_2 \) are not proportional, the lines are not parallel, which leads us to explore other relationships like intersection or skewing.
Parametric Equations
Parametric equations express a line using vector notation, typically involving a direction vector and a point on the line. In this exercise, the lines' parametric equations are given as:
By equating their parametric forms, you can discover if the lines intersect, as shown in the step-by-step solution. This is further examined through a system of equations derived from these parametric expressions.
- \( \ell_1(t) = \langle 2, 1, 1 \rangle + t \langle 5, 1, 3 \rangle \)
- \( \ell_2(t) = \langle 14, 5, 9 \rangle + t \langle 1, 1, 1 \rangle \)
By equating their parametric forms, you can discover if the lines intersect, as shown in the step-by-step solution. This is further examined through a system of equations derived from these parametric expressions.
System of Equations
Setting up a system of equations is crucial for finding where two lines intersect. By comparing parametric equations, such as those given for \( \ell_1 \) and \( \ell_2 \), you create a system that represents where the two lines can potentially meet.
In the example, equating the parametric equations of \( \ell_1 \) and \( \ell_2 \) resulted in three separate equations:
This technique is fundamental when verifying relationships between lines in space.
In the example, equating the parametric equations of \( \ell_1 \) and \( \ell_2 \) resulted in three separate equations:
- \( 2 + 5t = 14 + s \)
- \( 1 + t = 5 + s \)
- \( 1 + 3t = 9 + s \)
This technique is fundamental when verifying relationships between lines in space.
Point of Intersection
Finding the point of intersection is essential when determining if and where two lines meet in space. Once you solve the system of equations obtained from equating parametric equations, you use the resulting \( t \) (or \( s \)) values to find the exact location.
After determining \( t = 2 \) for \( \ell_1 \) and \( s = -2 \) satisfies all equations, substitute \( t \) into the parametric equation of one line, such as \( \ell_1 \) to find this point:
After determining \( t = 2 \) for \( \ell_1 \) and \( s = -2 \) satisfies all equations, substitute \( t \) into the parametric equation of one line, such as \( \ell_1 \) to find this point:
- \( \langle 2 + 5(2), 1 + 2, 1 + 3(2) \rangle \)
Other exercises in this chapter
Problem 15
Find the measure of the angle between the two vectors in both radians and degrees. \(\vec{u}=\langle 8,1,-4\rangle, \vec{v}=\langle 2,2,0\rangle\)
View solution Problem 16
Give the equation of the described plane in standard and general forms. Contains the point (5,7,3) and the line $$ \ell(t)=\left\\{\begin{array}{l} x=t \\ y=t \
View solution Problem 16
Vectors \(\vec{u}\) and \(\vec{v}\) are given. Compute \(\vec{u} \times \vec{v}\) and show this is orthogonal to both \(\vec{u}\) and \(\vec{v}\). \(\vec{u}=\ve
View solution Problem 16
Find the measure of the angle between the two vectors in both radians and degrees. \(\vec{u}=\langle 1,7,2\rangle, \vec{v}=\langle 4,-2,5\rangle\)
View solution