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In the following exercises, points \(P\) and \(Q\) are given. Let \(L\) be the line passing through points \(P\) and \(Q .\) a. Find the vector equation of line \(L\) b. Find parametric equations of line \(L\) c. Find symmetric equations of line \(L\) d. Find parametric equations of the line segment determined by \(P\) and \(Q\) . \(P(7,-2,6), \quad Q(-3,0,6)\)

Short Answer

Expert verified
The vector equation of line \( L \): \( (7, -2, 6) + t(-10, 2, 0) \), parametric: \( x = 7 - 10t, y = -2 + 2t, z = 6 \), symmetric: \( \frac{x - 7}{-10} = \frac{y + 2}{2}, z = 6 \), line segment: \( 0 \leq t \leq 1 \).

Step by step solution

01

Find Direction Vector

To find the vector equation of a line passing through two points, we should first determine the direction vector. Given points are \( P(7, -2, 6) \) and \( Q(-3, 0, 6) \). The direction vector \( \vec{PQ} \) can be found by subtracting the coordinates of \( P \) from \( Q \):\[ \vec{d} = Q - P = (-3 - 7, 0 - (-2), 6 - 6) = (-10, 2, 0). \]
02

Vector Equation of Line L

The vector equation of the line can be formed using a point on the line (take \( P \)) and the direction vector \( \vec{d} \): \[ \vec{r}(t) = \vec{r}_0 + t \cdot \vec{d} = (7, -2, 6) + t(-10, 2, 0). \] This equation represents any point on the line as \( (7 - 10t, -2 + 2t, 6) \).
03

Parametric Equations of Line L

Parametric equations are obtained from the vector equation by setting each component equal to the corresponding component of the position vector: \[ x = 7 - 10t, \] \[ y = -2 + 2t, \] \[ z = 6. \] These equations describe the line in terms of parameter \( t \).
04

Symmetric Equations of Line L

To find the symmetric equations, solve each parametric equation for \( t \) and set them equal: \[ t = \frac{x - 7}{-10} = \frac{y + 2}{2}. \] Since \( z = 6 \) is constant, we include \( z = 6 \) directly. These symmetric equations describe all points on the line without a parameter: \[ \frac{x - 7}{-10} = \frac{y + 2}{2}, \quad z = 6. \]
05

Parametric Equations for Line Segment PQ

To find the parametric equations for the line segment itself, limit \( t \) from 0 to 1, giving parametric equations with restricted \( t \): \[ x = 7 - 10t, \] \[ y = -2 + 2t, \] \[ z = 6. \] These hold for \( 0 \leq t \leq 1 \), thus defining the segment between \( P \) and \( Q \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector
In the world of vector equations, the direction vector is like the compass that guides us. To better understand the concept, consider two points: point \( P(7, -2, 6) \) and point \( Q(-3, 0, 6) \), which are on a line. The direction vector essentially describes how you travel from one point to the other in three-dimensional space.

To find the direction vector \( \vec{PQ} \), subtract the coordinates of point \( P \) from those of point \( Q \). This gives us:
  • \( x: -3 - 7 = -10 \)
  • \( y: 0 - (-2) = 2 \)
  • \( z: 6 - 6 = 0 \)
Thus, the direction vector is \( \vec{d} = (-10, 2, 0) \).

It tells us the line's overall direction, which we use to find further equations that describe the line itself.
Parametric Equations
Once we know our direction vector, we can move on to creating parametric equations. These equations break down the components of a line into individual expressions that depend on a parameter, usually \( t \). Parametric equations provide a way to express the line's coordinates in a controlled manner as \( t \) varies.

From the earlier found vector equation \( \vec{r}(t) = (7, -2, 6) + t(-10, 2, 0) \), we identify the parametric equations:
  • \( x = 7 - 10t \)
  • \( y = -2 + 2t \)
  • \( z = 6 \)
Parametric equations are fantastic because they allow us to "walk" along the line. By plugging in different values for \( t \), we find different points on the line. The parameter \( t \) serves as a slider that moves us from one point to another along the line.
Symmetric Equations
Symmetric equations offer another way to represent the same line. They eliminate the parameter \( t \), providing a format that directly agrees with each coordinate relation without the need for a parameter. This is useful when you want to see how coordinates relate directly to one another.

For the line defined by parametric equations \( x = 7 - 10t \) and \( y = -2 + 2t \), solving for \( t \) and equating the expressions results in:
  • \( t = \frac{x - 7}{-10} \)
  • \( t = \frac{y + 2}{2} \)
This leads to the symmetric equation:\[ \frac{x - 7}{-10} = \frac{y + 2}{2} \]Since \( z \) remains constant as \( z = 6 \), it joins our equation set without alteration. Symmetric equations give us a consistent format to view the line relation without delving into parameterized values.

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Most popular questions from this chapter

Consider line \(L\) of symmetric equations \(x-2=-y=\frac{z}{2}\) and point \(A(1,1,1) .\) a. Find parametric equations for a line parallel to L that passes through point A. b. Find symmetric equations of a line skew to L and that passes through point A. c. Find symmetric equations of a line that intersects L and passes through point A.

\([T]\) The force vector \(F\) acting on a proton with anelectric charge of \(1.6 \times 10^{-19} \mathrm{C}\) moving in a magnetic field \(B\) where the velocity vector \(V\) is given by\(\mathbf{F}=1.6 \times 10^{-19}(\mathbf{v} \times \mathbf{B})\) (here, v is expressed in meters per second, \(B\) in \(T\), and \(F\) in \(N\)). If the magnitude of force \(F\) acting on a proton is\(5.9 \times 10^{-17} \mathrm{~N}\) and the proton is moving at the speed of 300 m/sec in magnetic field \(B\) of magnitude \(2.4 T\), find the angle between velocity vector \(V\) of the proton and magnetic field \(B\). Express the answer in degrees rounded to the nearest integer.

Show that the line passing through points \(P(3,1,0)\) and \(Q(1,4,-3)\) is perpendicular to the line with equation \(x=3 t, y=3+8 t, z=-7+6 t, \quad t \in \mathbb{R}\)

For the following exercises, find the unit vectors. Find the unit vector that has the same direction as vector \(\mathbf{v}\) that begins at \((0,-3)\) and ends at \((4,10)\) .

For the following exercises, the equations of two planes are given. a. Determine whether the planes are parallel, orthogonal, or neither. b. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer. \(5 x-3 y+z=4, \quad x+4 y+7 z=1\)

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