Vectors - Questions On Equation Of Line

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Topic: Vectors - Questions on Equation of Line

Slide 2: Introduction to Vectors

A vector is a quantity that has both magnitude and direction.
It is represented by an arrow.
Examples of vectors: displacement, velocity, force.
Vectors can be added or subtracted.

Slide 3: Equation of a Line

The equation of a line is typically written in the form y = mx + c.
In vector form, the equation of a line is given by r = a + λu, where r is a point on the line, a is a fixed point on the line, u is the direction vector of the line, and λ is a scalar parameter.

Slide 4: Example 1

Find the equation of the line passing through the points A(2, 3) and B(-1, 5).

Step 1: Find the direction vector u by subtracting the coordinates of point A from the coordinates of point B.
- u = B - A = (-1, 5) - (2, 3) = (-3, 2)
Step 2: Choose a point on the line, for example, A(2, 3).
Step 3: Substitute the values of a, u, and r into the equation r = a + λu.
- r = (2, 3) + λ(-3, 2)
- r = (2 - 3λ, 3 + 2λ)

Slide 5: Example 2

Find the equation of the line parallel to the line r = (1, -2) + λ(2, 3).

Step 1: The direction vector of the line is given by u = (2, 3).
Step 2: Choose a point on the line, for example, (1, -2).
Step 3: Substitute the values of a, u, and r into the equation r = a + λu.
- r = (1, -2) + λ(2, 3)

Slide 6: Example 3

Find the equation of the line passing through the point (4, -3) and perpendicular to the line r = (2, 1) + λ(3, -1).

Step 1: The direction vector of the given line is u = (3, -1).
Step 2: The direction vector of the line perpendicular to the given line is u' = (-1, -3).
Step 3: Substitute the values of a, u', and r into the equation r = a + λu.
- r = (4, -3) + λ(-1, -3)

Slide 7: Equation of a Line in 3D

In three-dimensional space, the equation of a line is given by r = a + λu + μv, where r is a point on the line, a is a fixed point on the line, u and v are direction vectors of the line, and λ and μ are scalar parameters.

Slide 8: Example 4

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

Step 1: Find the direction vectors u and v.
- u = B - A = (3, 1, 4) - (1, 2, -3) = (2, -1, 7)
- v = (1, 1, 1) (arbitrarily chosen)
Step 2: Choose a point on the line, for example, A(1, 2, -3).
Step 3: Substitute the values of a, u, v, and r into the equation r = a + λu + μv.
- r = (1, 2, -3) + λ(2, -1, 7) + μ(1, 1, 1)

Slide 9: Example 5

Find the equation of the line passing through the point (2, -1, 3) and parallel to the line r = (4, 5, -1) + λ(3, -2, 1).

Step 1: The direction vector of the given line is u = (3, -2, 1).
Step 2: Choose a point on the line, for example, (2, -1, 3).
Step 3: Substitute the values of a, u, v, and r into the equation r = a + λu + μv.
- r = (2, -1, 3) + λ(3, -2, 1)

Slide 10: Summary

The equation of a line in two dimensions can be written as r = a + λu.
The equation of a line in three dimensions can be written as r = a + λu + μv.
The direction vector of a line determines its slope.
The equation of a line parallel to another line can be written with the same direction vector.

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Example 6: Find the equation of the line passing through the points A(1, -2, 3) and B(2, 4, -1).
- Step 1: Find the direction vectors u and v.
  - u = B - A = (2, 4, -1) - (1, -2, 3) = (1, 6, -4)
  - v = (0, 1, 0) (arbitrarily chosen)
- Step 2: Choose a point on the line, for example, A(1, -2, 3).
- Step 3: Substitute the values of a, u, v, and r into the equation r = a + λu + μv.
  - r = (1, -2, 3) + λ(1, 6, -4) + μ(0, 1, 0)

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Properties of Lines:
1. Two lines are parallel if their direction vectors are scalar multiples of each other.
2. Two lines are perpendicular if the dot product of their direction vectors is zero.
3. The distance between two parallel lines is given by the magnitude of the cross product of their direction vectors divided by the magnitude of one of the direction vectors.
4. The shortest distance between two skew lines is given by the magnitude of the vector connecting a point on one line to the other line.

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Example 7: Determine if the lines r = (1, 2) + λ(2, 3) and s = (3, 4) + μ(1, 2) are parallel or perpendicular.
- Step 1: Calculate the direction vectors u and v.
  - u = (2, 3) and v = (1, 2)
- Step 2: Calculate the dot product of u and v.
  - u · v = (2)(1) + (3)(2) = 8
- Step 3: Determine if u · v = 0 (perpendicular) or not (parallel).
  - Since u · v ≠ 0, the lines are parallel.

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Example 8: Find the distance between the parallel lines r = (1, 2) + λ(2, 3) and s = (3, 4) + μ(2, 3).
- Step 1: Calculate the magnitude of the cross product of the direction vectors u and v.
  - |u × v| = |(2)(3) - (3)(2)| = 0
- Step 2: Choose one of the direction vectors, say u, and calculate its magnitude.
  - |u| = √((2)^2 + (3)^2) = √13
- Step 3: Calculate the distance between the parallel lines.
  - distance = |u × v| / |u| = 0 / √13 = 0

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Example 9: Find the shortest distance between the skew lines r = (1, 2, 3) + λ(2, 1, -1) and s = (3, -1, 4) + μ(1, -2, 3).
- Step 1: Choose a point on the line s, for example, (3, -1, 4).
- Step 2: Calculate the vector connecting the point (3, -1, 4) to the skew line r.
  - direction vector = (1, 2, 3) - (3, -1, 4) = (-2, 3, -1)
- Step 3: Calculate the magnitude of the vector connecting the point (3, -1, 4) to the skew line r.
  - |direction vector| = √((-2)^2 + (3)^2 + (-1)^2) = √14

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Example 10: Find the angle between the lines r = (1, 2) + λ(2, 3) and s = (-1, 4) + μ(3, -2).
- Step 1: Calculate the direction vectors u and v.
  - u = (2, 3) and v = (3, -2)
- Step 2: Calculate the dot product of u and v.
  - u · v = (2)(3) + (3)(-2) = 0
- Step 3: Calculate the magnitudes of u and v.
  - |u| = √((2)^2 + (3)^2) = √13
  - |v| = √((3)^2 + (-2)^2) = √13
- Step 4: Calculate the angle between the lines using the dot product.
  - θ = cos^(-1)(u · v / (|u| |v|))

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Example 11: Find the length of the line segment joining the points A(1, 2, 3) and B(4, 5, 6).
- Step 1: Find the direction vector u by subtracting the coordinates of point A from the coordinates of point B.
  - u = B - A = (4, 5, 6) - (1, 2, 3) = (3, 3, 3)
- Step 2: Calculate the magnitude of the direction vector u.
  - |u| = √((3)^2 + (3)^2 + (3)^2) = 3√3

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Example 12: Section formula.
- If P divides the line joining A and B in the ratio m:n, then the position vector of P is given by
  - P = (mA + nB) / (m + n)
Example 13: Find the coordinates of the point which divides the line segment joining the points A(-3, -2) and B(7, 6) in the ratio 2:3.
- Step 1: Find the position vector of P using the section formula.
  - P = (2A + 3B) / (2 + 3)

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Example 14: Midpoint formula.
- The midpoint M of the line segment joining A and B is given by
  - M = (A + B) / 2
Example 15: Find the midpoint of the line segment joining the points A(4, -2) and B(-2, 6).
- Step 1: Find the midpoint M using the midpoint formula.
  - M = (A + B) / 2

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Recap:
- Vectors have magnitude and direction.
- The equation of a line can be written as r = a + λu in 2D and r = a + λu + μv in 3D.
- Parallel lines have the same direction vectors.
- Perpendicular lines have dot product of direction vectors equal to zero.
- Distance between parallel lines can be calculated using the cross product.
- Shortest distance between skew lines can be calculated using a connecting vector.
- Angle between lines can be calculated using the dot product.
- Length of line segments can be calculated using the distance formula.

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Example 16: Find the ratio in which the line segment joining the points A(1, -2, 3) and B(2, 4, -1) is divided by the yz-plane.
- Step 1: The yz-plane is parallel to the x-axis and intersects the line segment at point P.
- Step 2: Determine the coordinates of point P.
- Step 3: Calculate the ratio using the distance formula.

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Example 17: Find the ratio in which the line segment joining the points A(3, 4, -2) and B(5, 6, 1) is divided by the plane passing through the points P(1, -1, 2), Q(2, 0, -1), and R(3, 2, 1).
- Step 1: Find a normal vector to the plane by taking the cross product of two direction vectors.
- Step 2: Write the equation of the plane.
- Step 3: Calculate the point of intersection of the line segment and the plane.
- Step 4: Determine the ratio using the distance formula.

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Example 18: Find the equation of a plane passing through three non-collinear points.
- Step 1: Determine two direction vectors by subtracting the coordinates of two points from the coordinates of the third point.
- Step 2: Take the cross product of the two direction vectors to find a normal vector to the plane.
- Step 3: Write the equation of the plane using one of the points and the normal vector.
- Step 4: Simplify the equation to obtain the final form.

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Example 19: Find the distance between the point P(1, -2, 3) and the plane 2x - 3y + 4z = 12.
- Step 1: Calculate the perpendicular distance between the plane and the point.
- Step 2: Substitute the coordinates of the point into the equation of the plane to find the distance.

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Example 20: Find the equation of a sphere given the center and radius.
- Step 1: Write the equation of a sphere using the center and radius.
- Step 2: Simplify the equation to obtain the final form.
Example 21: Find the equation of a sphere passing through four non-coplanar points.
- Step 1: Determine the center of the sphere by finding the intersection of the perpendicular bisectors of the line segments joining pairs of points.
- Step 2: Calculate the radius of the sphere using the distance formula.
- Step 3: Write the equation of the sphere using the center and radius.

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Example 22: Find the equation of a cone given the vertex, axis, and semi-vertical angle.
- Step 1: Determine the direction vector of the axis.
- Step 2: Write the equation of the cone using the vertex, axis, and semi-vertical angle.
- Step 3: Simplify the equation to obtain the final form.

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Example 23: Find the equation of a paraboloid given the vertex, axis, and focal length.
- Step 1: Determine the direction vector of the axis.
- Step 2: Write the equation of the paraboloid using the vertex, axis, and focal length.
- Step 3: Simplify the equation to obtain the final form.

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Example 24: Find the equation of a hyperboloid of one sheet given the center, axis, and distance between the foci.
- Step 1: Determine the direction vector of the axis.
- Step 2: Write the equation of the hyperboloid using the center, axis, and distance between the foci.
- Step 3: Simplify the equation to obtain the final form.

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Example 25: Find the equation of a hyperboloid of two sheets given the center, axis, and distance between the foci.
- Step 1: Determine the direction vector of the axis.
- Step 2: Write the equation of the hyperboloid using the center, axis, and distance between the foci.
- Step 3: Simplify the equation to obtain the final form.

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Summary:
- Equations of geometric objects such as lines, planes, spheres, cones, paraboloids, and hyperboloids can be determined using specific information about their shapes, positions, and dimensions.
- Spanning vectors play an important role in the equations of lines and planes.
- The distance formula is used to calculate distances between points, lines, and planes.
- The cross product and dot product are used to determine the relative positions of geometric objects.
- The study of 3D geometry extends the concepts of 2D geometry and provides a deeper understanding of the spatial world around us.