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    Lec 4 Equation of Plane

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    Mahad Elahi
    The document discusses parameterizing a line segment between two points P and Q. It shows the steps to: 1) Find the equation of the line through P and Q using the vector PQ. 2) Determine the parameter t values that correspond to points P and Q to restrict the line to a line segment. 3) Write the parametric equations of the line segment in terms of t between 0 and 1. The example parameterizes the line segment between points P(-3,2,-3) and Q(1,-1,4). It finds the vector PQ, writes the equations of the line, and restricts t from 0 to 1 to write the parametric equations of the

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    Lec 4 Equation of Plane

    Uploaded by

    Mahad Elahi
    51 views6 pages
    The document discusses parameterizing a line segment between two points P and Q. It shows the steps to: 1) Find the equation of the line through P and Q using the vector PQ. 2) Determine the parameter t values that correspond to points P and Q to restrict the line to a line segment. 3) Write the parametric equations of the line segment in terms of t between 0 and 1. The example parameterizes the line segment between points P(-3,2,-3) and Q(1,-1,4). It finds the vector PQ, writes the equations of the line, and restricts t from 0 to 1 to write the parametric equations of the

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    The document discusses parameterizing a line segment between two points P and Q. It shows the steps to: 1) Find the equation of the line through P and Q using the vector PQ. 2) Determine the parameter t values that correspond to points P and Q to restrict the line to a line segment. 3) Write the parametric equations of the line segment in terms of t between 0 and 1. The example parameterizes the line segment between points P(-3,2,-3) and Q(1,-1,4). It finds the vector PQ, writes the equations of the line, and restricts t from 0 to 1 to write the parametric equations of the

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    51 views6 pages

    Lec 4 Equation of Plane

    Uploaded by

    Mahad Elahi
    The document discusses parameterizing a line segment between two points P and Q. It shows the steps to: 1) Find the equation of the line through P and Q using the vector PQ. 2) Determine the parameter t values that correspond to points P and Q to restrict the line to a line segment. 3) Write the parametric equations of the line segment in terms of t between 0 and 1. The example parameterizes the line segment between points P(-3,2,-3) and Q(1,-1,4). It finds the vector PQ, writes the equations of the line, and restricts t from 0 to 1 to write the parametric equations of the

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 6

    Lecture no.

    Parameterization of a Line segment

    Example: Parameterize the line segment joining the points P (-3, 2, -3) and Q (1, -1, 4).

    Solution:

    First of all, we will find the parametric equation of the line through the points P (-3, 2, -3)
    and Q (1, -1, 4) and then restrict the domain of parameter t to obtain the parametric equation of
    the line segment from P to Q.

    Step-1 (equation of line)

    ⃗⃗⃗⃗⃗⃗ = (1 + 3)𝑖 + (−1 − 2)𝑗 + (4 + 3)𝑘


    𝑣⃗ = 𝑃𝑄

    𝑣⃗ = 4𝑖 − 3𝑗 + 7𝑘

    P (-3, 2, -3)

    Equation of line

    𝑥 = −3 + 4𝑡
    𝑦 = 2 − 3𝑡
    𝑧 = −3 + 7𝑡

    Step-2 (line segment)

    In order to find the value of t for which an arbitrary point (x, y, z) of the line is at P (-3, 2, -3) we
    solve the equation

    −3 = −3 + 4𝑡
    2 = 2 − 3𝑡 }⇒𝑡=0
    −3 = −3 + 7𝑡

    Similarly, when (x, y, z) is at Q (1, -1, 4) we solve

    1 = −3 + 4𝑡
    −1 = 2 − 3𝑡} ⇒ 𝑡 = 1
    4 = −3 + 7𝑡
    So, the parametric equation of the line segment is

    𝑥 = −3 + 4𝑡

    𝑦 = 2 − 3𝑡

    𝑧 = −3 + 7𝑡; 0≤𝑡≤1

    Question 19: Find the parametric equations of the line segment joining the points P(-2,0,2) and
    Q(0,2,0).

    Ex. 12.5: 13-20

    The Distance from a Point to a Line in Space


    The distance from a point S to a line L that passes through a point P

    and is parallel to a vector 𝑣⃗ is the absolute value of the scalar

    component of ⃗⃗⃗⃗⃗⃗
    𝑷𝑺 in the direction of the vector normal to the line.

    ⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑠𝑖𝑛 𝜃
    𝑑 = |𝑃𝑆| (1)

    As we know that

    𝑃𝑆 × 𝑣⃗ = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
    ⃗⃗⃗⃗⃗ |𝑃𝑆||𝑣⃗| 𝑠𝑖𝑛 𝜃 𝑛̂

    ⃗⃗⃗⃗⃗ × 𝑣⃗| = ⃗⃗⃗⃗⃗⃗⃗⃗⃗


    |𝑃𝑆 |𝑃𝑆||𝑣⃗| 𝑠𝑖𝑛 𝜃 ⋅ 1

    ⃗⃗⃗⃗⃗ × 𝑣⃗|
    |𝑃𝑆
    = ⃗⃗⃗⃗⃗⃗⃗⃗⃗
    |𝑃𝑆| 𝑠𝑖𝑛 𝜃.
    |𝑣⃗|

    So, equation (1) becomes:

    ⃗⃗⃗⃗⃗ × 𝑣⃗|
    |𝑃𝑆
    𝑑=
    |𝑣⃗|
    Example: Find the distance from the point S (1, 1, 5) to the line:

    𝑥 = 1+𝑡
    𝐿: {𝑦 = 3 − 𝑡
    𝑧 = 2𝑡
    Solution:

    The vector parallel to the line L is

    𝑣⃗ = 𝑖 − 𝑗 + 2𝑘

    The Line passes through the point P (1, 3, 0)

    ⃗⃗⃗⃗⃗ = (1 − 1)𝑖 + (1 − 3)𝑗 + (5 − 0)𝑘


    𝑃𝑆

    ⃗⃗⃗⃗⃗ = 0𝑖 − 2𝑗 + 5𝑘.
    𝑃𝑆

    𝑖 𝑗 𝑘
    ⃗⃗⃗⃗⃗
    𝑃𝑆 × 𝑣⃗ = |0 −2 5|
    1 −1 2

    ⃗⃗⃗⃗⃗ −2 5| 0 5| 0 −2
    𝑃𝑆 × 𝑣⃗ = 𝑖 | −𝑗| +𝑘| |
    −1 2 1 2 1 −1
    ⃗⃗⃗⃗⃗ × 𝑣⃗ = 𝑖(−4 + 5) − 𝑗(0 − 5) + 𝑘(0 + 2)
    𝑃𝑆

    ⃗⃗⃗⃗⃗ × 𝑣⃗ = 𝑖 + 5𝑗 + 2𝑘
    𝑃𝑆

    ⃗⃗⃗⃗⃗ × 𝑣⃗| = √(1)2 + (5)2 + (2)2


    |𝑃𝑆

    ⃗⃗⃗⃗⃗ × 𝑣⃗| = √30


    |𝑃𝑆

    |𝑣⃗| = √(1)2 + (−1)2 + (2)2 = √6

    ⃗⃗⃗⃗⃗ × 𝑣⃗|
    |𝑃𝑆
    𝑑=
    |𝑣⃗|

    √30
    𝑑=
    √6

    𝑑 = √5
    Ex. 12.5: 33-38

    Equation of a Plane in Space:


    A plane in space is determined by knowing a point on the plane and its “tilt” or orientation. This
    “tilt” is defined by specifying a vector that is perpendicular or normal to the plane.

    Suppose that a plane M passes through a point 𝑃𝑜 (𝑥𝑜, 𝑦𝑜, 𝑧𝑜 ) and is normal to the non-zero
    vector 𝑛⃗⃗ = 𝐴𝑖⃗ + 𝐵𝑗⃗ + 𝐶𝑘⃗⃗. Then M is the set of all points 𝑃(𝑥, 𝑦, 𝑧) for which ⃗⃗⃗⃗⃗⃗⃗⃗
    𝑃𝑂 𝑃 is orthogonal
    to 𝑛⃗⃗. Thus, the dot product 𝑛⃗⃗. ⃗⃗⃗⃗⃗⃗⃗⃗
    𝑃𝑂 𝑃 = 0.

    This equation is equivalent to

    (𝐴𝑖⃗ + 𝐵𝑗⃗ + 𝐶𝑘⃗⃗). [(𝑥 − 𝑥𝑜 )𝑖⃗ + (𝑦 − 𝑦𝑜 )𝑗⃗ + (𝑧 − 𝑧𝑜 )𝑘⃗⃗] = 0

    𝐴(𝑥 − 𝑥𝑜 ) + 𝐵(𝑦 − 𝑦𝑜 ) + 𝐶( 𝑧 − 𝑧𝑜 ) = 0

    Remark:

    Another form of the equation of plane.

    𝐴𝑥 − 𝐴𝑥𝑜 + 𝐵𝑦 − 𝐵𝑦𝑜 + 𝐶𝑧 − 𝐶𝑧𝑜 = 0

    𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 − (𝐴𝑥𝑜 + 𝐵𝑦𝑜 + 𝐶𝑧𝑜 ) = 0

    𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 𝐴𝑥𝑜 + 𝐵𝑦𝑜 + 𝐶𝑧𝑜

    𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 𝐷 where 𝐷 = 𝐴𝑥𝑜 + 𝐵𝑦𝑜 + 𝐶𝑧𝑜


    Example 1: Find an equation for the plane through 𝑃(−3,0,7) perpendicular to 𝑛⃗⃗ = 5𝑖⃗ + 2𝑗⃗ −
    𝑘⃗⃗.

    Solution: The equation of plane is

    𝐴(𝑥 − 𝑥𝑜 ) + 𝐵(𝑦 − 𝑦𝑜 ) + 𝐶( 𝑧 − 𝑧𝑜 ) = 0

    5(𝑥 − (−3)) + 2(𝑦 − 0) + (−1)( 𝑧 − 7) = 0

    5(𝑥 + 3) + 2𝑦 − 𝑧 + 7 = 0

    5𝑥 + 15 + 2𝑦 − 𝑧 + 7 = 0

    5𝑥 + 2𝑦 − 𝑧 + 22 = 0

    5𝑥 + 2𝑦 − 𝑧 = −22

    Example 2: Find an equation for the plane passing through three points A (0, 0, 1), B (2, 0, 0)
    and C (0, 3, 0).

    Solution: A vector normal to the plane is:

    𝑛⃗⃗ = ⃗⃗⃗⃗⃗⃗
    𝐴𝐵 × 𝐴𝐶⃗⃗⃗⃗⃗⃗

    𝑖⃗ 𝑗⃗ 𝑘⃗⃗ 0 −1 2 −1 2 0
    = |2 0 −1|= 𝑖⃗ |3 |-𝑗⃗ | | + 𝑘⃗⃗ | |
    −1 0 −1 0 3
    0 3 −1

    =𝑖⃗[(0 − (−3)] − 𝑗⃗(−2 − 0) + 𝑘⃗⃗(6 − 0)

    𝑛⃗⃗ = 3𝑖⃗ + 2𝑗⃗+ 6𝑘⃗⃗

    Now the equation of plane is

    𝐴(𝑥 − 𝑥𝑜 ) + 𝐵(𝑦 − 𝑦𝑜 ) + 𝐶( 𝑧 − 𝑧𝑜 ) = 0

    3(𝑥 − 0) + 2(𝑦 − 0) + 6( 𝑧 − 1) = 0

    3𝑥 + 2𝑦 + 6𝑧 = 6
    Ex. 12.5: 21-26

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