MCV4U CCT
Intersection and Distance in and .
1 Intersection of Lines & Planes [/episode/9-1]
    1.1 Introduction

    In this section, we explore how lines and planes can interact in three-dimensional space ( ), and how two lines can intersect in both and . Understanding these intersections is fundamental to solving geometric problems.

    1.2 Intersection of a Line with a Plane

    A line and a plane in can have three possible types of intersections:

    1. No Points of Intersection: The line is parallel to the plane and distinct from it.
    2. One Point of Intersection: The line passes through the plane.
    3. Infinite Points of Intersection: The line lies entirely within the plane.

    To find the intersection, we typically use the parametric equation of the line and the Cartesian equation of the plane.

    • Line

    • Plane : .

    Substitute the parametric equations of the line into the Cartesian equation of the plane and solve for the parameter .

    1.2.1 Example : Line Intersecting a Plane at a Point

    Find the point of intersection of the plane and the line

    Solution

    Substitute the parametric equations of into :

    Since we found a unique value for , there is one point of intersection. Substitute back into the line's equations:

    The point of intersection is .

    1.2.2 Example : Line Parallel to a Plane

    Determine if the line intersects the plane

    Solution

    Parametric equations for

    Substitute into :

    This was a mistake in setting up the example for "parallel". Let's adjust the plane or line. Consider the line

    and plane

    • Direction vector of line: .
    • Normal vector of plane: .
    • If , the line is parallel to the plane.

    So, the line is parallel to the plane. Now we check if it lies on the plane or is distinct.

    Substitute

    into

    :

    This is a contradiction. Thus, there is no solution for .
    The line is parallel to the plane and does not intersect it.

    1.2.3 Example : Line Lying in a Plane

    Determine if the line intersects the plane

    Solution
    • Direction vector of line: .
    • Normal vector of plane: .

    Thus, the line is parallel to the plane.

    Parametric equations for

    , substitute into :

    This statement is true for all values of .

    Therefore, the line lies entirely within the plane.

    1.3 Intersection of Two Lines

    Two lines in or can:

    1. Intersect at a Single Point.
    2. Be Parallel and Distinct: No intersection.
    3. Be Coincident: Infinite intersection points (they are the same line).
    4. Be Skew (in only): Not parallel and do not intersect.

    To find the intersection, we use parametric equations for both lines and equate the corresponding coordinates.

    • Line :
    • Line :

    Note the use of different parameters ( and ).

    Equating components:

    Solve the system for and .

    • In : Solve the two equations for and . If a unique solution exists, they intersect.

    • In : Use two equations (e.g., (1) and (2)) to find and . Substitute these values into the third equation (3).

      • If (3) is satisfied: Single point of intersection.
      • If (3) is not satisfied: Lines are skew (if not parallel).

      • If the system for from (1) and (2) yields no solution (e.g. ): check if direction vectors are parallel.

        • If yes, parallel lines.
        • If no, this case shouldn't happen if handled correctly (usually leads to skew if (3) then fails).
      • If the system for yields infinite solutions (e.g. ): lines are coincident (if direction vectors are parallel).
    1.3.1 Example : Intersecting Lines in

    Find the intersection of and

    Solution

    Parametric equations:

    Equate components:

    Solve (1) and (2) for and :
    Add (1) and (2):

    Substitute into (1):

    Check these values in (3): . The left side is . The right side of (3) is . Since , the values are consistent.

    Substitute into (or into ):

    The point of intersection is .

    1.3.2 Example : Skew Lines in

    Determine if and intersect.

    Solution

    Direction vectors: , . These are not scalar multiples, so the lines are not parallel or coincident.

    Parametric equations:

    Equate components:

    Solve (1) and (2):
    Add (1) and (2):

    Substitute into (1):

    Check these values in (3): Left side: . Right side: . Since , the values are not consistent.
    The lines do not intersect. Since they are not parallel, they are skew lines.

    1.3.3 Example : Parallel and Distinct Lines in

    Consider

    Determine their intersection.

    Solution

    Direction vectors: , . Since , the lines are parallel.

    Equate components: (1) (2)

    From (2), . Substitute into (1):

    This is a contradiction. There is no solution for (and ).
    Thus, the lines are parallel and distinct.

    1.4 Homework
    1.4.1 Question 1.1 (Easy)

    Find the point of intersection, if any, between the line and the plane .

    Solution

    Substitute the parametric equations of the line into the equation of the plane :

    Combine terms with and constant terms:

    Since we found a unique value for , there is a single point of intersection.
    Substitute back into the parametric equations of the line :

    Answer: The point of intersection is .

    1.4.2 Question 1.2 (Medium)

    Determine whether the lines and intersect, are parallel, or are skew. If they intersect, find the point of intersection.

    Solution

    Direction vectors: ,

    Check if direction vectors are parallel:
    Is for some scalar ?

    Since for all components, .
    The lines are parallel.

    Now, check if they are coincident or distinct.
    To be coincident, a point from must lie on .
    Let be a point on .
    If is on , then for some :

    Equate components:

    Since the same value of satisfies all three equations, the point from is also on .
    Since the lines are parallel and share at least one point, they are coincident.

    Answer: The lines are coincident (and therefore intersect at infinitely many points, they are the same line).

    1.4.3 Question 1.3 (Harder)

    Consider the line and the plane .
    For what value of does the line lie entirely within the plane ?

    Solution

    For the line to lie entirely within the plane , two conditions must be met:

    1. The direction vector of the line, , must be perpendicular to the normal vector of the plane, . This ensures the line is parallel to the plane.

    2. A point from the line must lie on the plane .
      Let's use the point on when : .
      Substitute into the plane equation:

      This is a contradiction. This means that with the current plane equation , the line can never lie in the plane if its direction vector is and passes through .

    Let's re-read the question. It asks for what value of the line lies in the plane. The method is generally to substitute the parametric equations of the line into the plane equation. If the resulting equation in becomes , then the line lies in the plane.

    Substitute into :

    Collect terms with and constant terms:

    • For the line to lie entirely within the plane, this equation must be true for all values of . This can only happen if the form is .
    • For our equation , if the coefficient of is zero, we get , which is a contradiction.

    This implies that for this specific plane and a line of the form , the line cannot lie within the plane, regardless of . It will either intersect at one point or be parallel and distinct.

    Let's check my interpretation.
    Perhaps the question implies that we first satisfy the parallel condition, then check a point.
    If (from ), the line is .
    Direction vector is . Normal vector is

    So the line is parallel to the plane.
    A point on the line (for ) is . Substitute into plane:

    Since , the point is not on the plane. So the line is parallel and distinct.

    There might be an issue with how the question is posed relative to the expected outcome, or I'm missing a nuance.
    If the question meant for "what value of (and possibly adjusting the constant term of the plane or a point on the line) does it lie in the plane?", then is necessary for parallelism.
    However, as strictly stated, for this line and this plane: For this to be true for all (line in plane), we need AND .
    The part is impossible.
    This means there is NO value of for which this specific line lies in this specific plane.

    If the question intended to ask for the line to be parallel to the plane (but not necessarily in it), then makes , leading to , or , which means no solution for , hence parallel and distinct.

    Let's assume the question has a solvable setup, and the equation should become .
    The constant terms must cancel.
    From , we got .
    For to occur, we need:

    1. Coefficient of to be zero: .
    2. The constant part of the equation to be zero: . This is impossible.

    Conclusion based on strict interpretation: There is no value of for which the given line lies entirely within the given plane. The line will always be parallel and distinct if , or intersect at a single point if .

    Self-correction

    (Self-correction: Perhaps the problem intended a different constant for the plane, or a point that forces the line to be in the plane. Given the phrasing, the above is the correct deduction. If a student got this, they should state "no such k exists".)

    Revised approach if a solutionmustexist: If the question had been, for instance, , and the line must lie in it:

    For this to be , we need: and .
    So if the plane was , and , then the line would lie in it.

    Final Answer (as the question is written): For the line to lie in the plane , the equation derived from substituting the line into the plane must be true for all . This requires (so ) AND . Since is false, there is no such value of .

    2 Systems of Equations [/episode/9-2]
      2.1 Introduction

      A system of linear equations is a collection of one or more linear equations involving the same set of variables. We will focus on solving these systems using elementary operations (substitution and elimination).

      Geometrically:

      • A system of two linear equations in two variables ( ) represents two lines in .
      • A system of three linear equations in three variables ( ) represents three planes in .

      The solutions to a system correspond to the points of intersection of these geometric objects.

      2.2 Types of Solutions

      A system of linear equations can have:

      1. Exactly One Solution: The lines/planes intersect at a single point.
      2. Infinitely Many Solutions: The lines are coincident, or the planes intersect in a line or are coincident.
      3. No Solution: The lines are parallel and distinct, or the planes are parallel or intersect in a way that leaves no common point for all three.
      2.3 Elementary Operations on Equations
      1. Multiply an equation by a non-zero constant.
      2. Add a multiple of one equation to another equation.
      3. Swap the positions of two equations.

      These operations produce an equivalent system, meaning they have the same solution set.

      2.4 Solving Systems of Two Equations in Two Variables (Lines in )
      2.4.1 Example Unique Solution (Two Lines Intersecting)

      Solve the system:

      Solution using Elimination

      Multiply equation (2) by −2:

      Add equation (1) and (2'):

      Substitute into equation (1):

      Solution:

      Geometrically, these two lines intersect at the point .

      2.4.2 Example: No Solution (Parallel Lines)

      Solve the system:

      Solution using Substitution

      From equation (1), .
      Substitute this into equation (2):

      This is a contradiction. Therefore, the system has no solution.

      Geometrically, these are parallel and distinct lines.

      2.4.3 Example: Infinitely Many Solutions (Coincident Lines)

      Solve the system:

      Solution using Elimination

      Multiply equation (1) by −3:

      Add equation (1') and (2):

      This is always true. The system has infinitely many solutions.
      To express the solution, let , where is any real number (a parameter).
      From equation (1), .
      Solution:

      Geometrically, these are coincident lines.

      2.5 Solving Systems of Three Equations in Three Variables (Planes in )

      The general strategy is to eliminate one variable from two pairs of equations.

      This reduces the system to two equations in two variables, which can then be solved.

      2.5.1 Example : Unique Solution (Three Planes Intersecting at a Point)

      Solve the system:

      Solution using Elimination

      Step 1: Eliminate using (1) and (2).
      Add (1) and (2):

      Step 2: Eliminate using (1) and (3).
      Multiply (1) by −2:

      Add (1') and (3):

      Step 3: Solve the 2x2 system (4) and (5).

      From (5), . Substitute into (4):

      Substitute into :

      Step 4: Substitute into an original equation (e.g., (1)) to find .

      Solution:

      Geometrically, these three planes intersect at the point .

      2.5.2 Example: Considering Consistency (Thinking Question)

      A system of three linear equations in three variables leads to the equation after some elementary operations. What can you conclude about the geometric configuration of the three planes?

      Solution

      The equation is a contradiction. This means there is no set of values that can satisfy all three original equations simultaneously. Therefore, the system has no solution.

      Geometrically, this means the three planes do not have any common point of intersection. Possible configurations include:

      1. All three planes are parallel and distinct.
      2. Two planes are parallel and distinct, and the third plane intersects them.

      +. The planes intersect in pairs forming parallel lines (a "triangular prism" shape with no common interior).

      1. Two planes are coincident, and the third plane is parallel and distinct to them.
      2.6 Homework
      2.6.1 Question 2.1 (Easy)

      Solve the following system of linear equations:

      Solution

      We can use elimination or substitution. Let's use elimination.
      Multiply equation (1) by 3 to make the y-coefficients opposites:

      Add equation (1') and equation (2):

      Substitute into equation (1):

      Check in equation (2): . This is correct.

      Answer: The solution is .

      2.6.2 Question 2.2 (Medium)

      Solve the following system of linear equations. Describe the geometric interpretation of the solution.

      Solution

      Use elimination.
      Step 1: Eliminate using equations (1) and (2).
      Add (1) and (2):

      Step 2: Eliminate using equations (1) and (3).
      Multiply equation (1) by 3:

      Add (1') and (3):

      Step 3: Solve the system of equations (4) and (5).

      Notice that equation (5) is equation (4): .
      Since they are multiples, they represent the same line in the -plane (if we were thinking in 2D). This means there are infinitely many solutions for and that satisfy both.

      To express the solution, let , where is a parameter.
      Substitute into equation (4):

      Now, substitute and back into one of the original equations (e.g., (1)) to find .

      The solution is:

      This can be written in vector form as .

      Geometric Interpretation: The three planes intersect in a line. The equation found is the parametric equation of this line of intersection.

      Answer: Infinitely many solutions, representing a line: .

      2.6.3 Question 2.3 (Harder)

      Consider the system of equations:

      For what value(s) of will this system have:

      • No solution?
      • Infinitely many solutions?
      • A unique solution? (If possible, find it for a specific that gives a unique solution).
      Solution

      Step 1: Eliminate using (1) and (3).
      Add (1) and (3):

      Step 2: Substitute into equations (1) and (2).

      • Into (1): (4)
      • Into (2): (5)
      • Into (3): (6)

      Notice that equation (4) is , and equation (6) is . These are equivalent: multiply (6) by −1 to get . So equations (4) and (6) are dependent. We essentially have the system:

      • (from (1) and (3) combined effectively)
      • (from (2) after subbing )

      We need to solve this system for and : .
      Substitute this into :

      Case 1: , i.e., .
      Then .
      If , then .
      And we already found .
      So, if , the system has a unique solution: .

      Case 2: , i.e., .
      Substitute into :

      This equation is , which is true for any value of . This indicates infinitely many solutions when .
      If , we have . The equations for and become dependent:

      • (from (4) or (6))

      • (from (5) with ).

        • This is , so it's the same as .

      Let . Then . So, if , the solutions are . This is a line.

      Summary of findings:

      • No solution?
        From :

        • If (i.e. ), we get , which is infinitely many solutions.
        • If , we get , a unique solution.

        It seems there is no case for "no solution" based on this reduction. Let's double check.
        The initial reduction came from (1) and (3). This part is always true.
        The system then reduced to:

        If these two lines (in variables ) are parallel and distinct, we'd get no solution.
        Their "slopes" (if is like function of ):
        From (slope 1)
        From .

        • If , . (slope )
          For them to be parallel, .
          If , then . Both equations become .
          They are coincident, leading to infinite solutions.
        • What if ?

        The system is:

        Substitute into

        And .
        So if (which is ), we get unique solution .
        This fits case.

      It appears there are no values of for which the system has no solution.

      b) Infinitely many solutions? This occurs when . The solutions are .

      c) A unique solution? This occurs when . The unique solution is .
      For example, if (which is ), the unique solution is .

      Answer:

      • No solution: Never.
      • Infinitely many solutions: When . The solution set is the line for .
      • A unique solution: When . The unique solution is .
      3 Intersection of 2 Planes [/episode/9-3]
        3.1 Introduction

        Two distinct planes in can interact in one of two ways:

        • They intersect in a line.
        • They are parallel and distinct (no intersection).

        A third possibility is that the equations represent the same plane (coincident), in which case they "intersect" everywhere on the plane.

        Let the two planes be:

        3.2 Analyzing Normals
        • If and are not parallel (i.e., for any scalar ), the planes must intersect in a line. The direction vector of this line of intersection will be orthogonal to both and , so .

        • If and are parallel (i.e., ), the planes are parallel.

          • To check if they are distinct or coincident:
            If and .
            This means !
            If , the planes are coincident.
            If , the planes are parallel and distinct.
        3.3 Finding the Line of Intersection

        To find the equation of the line of intersection, we solve the system of two equations with three variables. Since there are more variables than equations (when they intersect in a line), there will be infinitely many solutions. We express these solutions in terms of a parameter.

        3.3.1 Example : Planes Intersecting in a Line

        Find the equation of the line of intersection of the planes:

        Solution

        Normals: , . These are not parallel, so the planes intersect in a line.

        We need to solve the system. Eliminate one variable, say .

        Subtract (1) from (2):

        Let (parameter). Then .

        Substitute and into equation (1) to find :

        So, the parametric equations of the line of intersection are:

        This can be written as

        To avoid fractions, we can choose (so ).

        Then

        Line:

        This is an equivalent representation.

        Alternative method for direction vector:

        This is , so it's a valid direction vector. A point on the line can be found by setting one variable to 0 (e.g., ) in the original system:

        Subtracting: . Then . Point .

        So

        This matches.

        3.3.2 Example : Parallel and Distinct Planes

        Show that the planes and are parallel and distinct.

        Solution

        Normals:

        ,

        We see that

        so the normals are parallel, the planes are therefore parallel.

        Now check if they are distinct or coincident.

        The equation for is .

        • Compare with

          If and were coincident, then would be a scalar multiple of .

          If we multiply by 2, we get

        • Comparing

          with

          The coefficients of match (after scaling ), but the constant terms and do not satisfy the same scaling ( , here ). Therefore, the planes are parallel and distinct.

        Alternatively, try to solve the system:

        Multiply (1) by 2:

        Subtract (1') from (2):

        This is a contradiction, so there is no solution.
        The planes do not intersect. Since their normals are parallel, they must be parallel and distinct.

        3.3.3 Example : Coincident Planes

        Determine the nature of intersection of and

        Solution

        Normals:

        so the normals are parallel, the planes are parallel.

        Check for coincidence:

        Multiply by 3:

        This is identical to the equation for . Therefore, the planes are coincident. They intersect at every point on the plane.

        The "solution" is any point satisfying .
        We can parameterize this: Let . Then .

        So

        This describes all points on the plane.

        3.4 Homework
        3.4.1 Question 3.1 (Easy)

        Consider the planes and .
        Determine if these planes are parallel, coincident, or intersect in a line. If they intersect in a line, find a direction vector for the line of intersection.

        Solution

        Normal vectors: ,

        Check if normal vectors are parallel:
        Is ?

        Since for all components, .
        The normal vectors are parallel, so the planes are parallel.

        Now, check if they are distinct or coincident.

        • Plane 1: .
        • Plane 2: .

        Multiply Plane 1 by :

        Comparing this with Plane 2 ( ), the coefficients of match, but the constant terms are different ( ).
        Therefore, the planes are parallel and distinct.

        Answer: The planes are parallel and distinct. They do not intersect.

        3.4.2 Question 3.2 (Medium)

        Find the parametric equations of the line of intersection of the planes and .

        Solution

        The normal vectors are and . They are not parallel, so the planes intersect in a line.

        We need to solve the system of equations:

        Eliminate one variable, say .
        Add (1) and (2):

        Now express one variable in terms of another, or introduce a parameter.
        From (3), .
        Let (parameter).
        Then .

        Substitute and into one of the original plane equations (e.g., (1)) to find :

        The parametric equations of the line of intersection are:

        This can be written in vector form as .

        Alternative method for direction vector:

        This matches the direction vector found above.
        To find a point on the line, set in parametric equations: .
        Check if is on both planes:

        • . (Correct)
        • . (Correct)

        Answer: The parametric equations are .

        3.4.3 Question 3.3 (Harder)

        Find the Cartesian equation of a plane that passes through the point and is perpendicular to the line of intersection of the planes and .

        Solution

        The line of intersection, , of and will have a direction vector that is perpendicular to both and , where and

        The plane is perpendicular to the line . This means the normal vector of , let's call it , must be parallel to the direction vector of , .
        So, we can choose .

        The Cartesian equation of a plane with normal is , so:

        The plane passes through the point . Substitute these coordinates into the equation to find :

        So, the Cartesian equation of plane is .

        Answer: The Cartesian equation of the plane is .

        4 Intersection of 3 Planes [/episode/9-4]
          4.1 Introduction

          The intersection of three planes in can result in several geometric configurations. We solve the system of three linear equations in three variables,

          The possible outcomes are:

          1. Unique Point: The three planes intersect at a single common point. (Normals are not coplanar).
          2. Line of Intersection: The three planes intersect along a common line. (This happens if, for example, one plane contains the line of intersection of the other two).

          3. No Intersection:

            1. All three planes are parallel and distinct.
            2. Two planes are parallel and distinct, intersected by a third.
            3. The planes intersect in pairs forming three parallel lines (a "triangular prism" configuration).
            4. Two planes are coincident, and the third is parallel and distinct.
          4. Plane of Intersection: All three planes are coincident. (Or two are coincident, and the third intersects them in their common plane).

          The method is to use elimination (as in 9.2) to reduce the 3x3 system.

          • If (where ) arises, there is no solution.
          • If arises, there are infinitely many solutions (either a line or a plane).
          • If a unique is found, that's the single point of intersection.
          4.1.1 Example : Unique Point of Intersection

          This was solved as an Example:

          Solution

          4.1.2 Example: Intersection in a Line

          Find the intersection of the planes:

          Solution

          Step 1: Eliminate using (1) and (2). Add (1) and (2):

          Step 2: Eliminate using (1) and (3).

          Multiply (1) by 5:

          Add (1') and (3):

          Step 3: Analyze system (4) and (5).

          These equations are identical—which means we effectively have only two distinct equations (e.g., (1) and (4), or (2) and (4)).
          This signifies an infinite number of solutions, forming a line.

          Let . From (4),

          Substitute and into one of the original plane equations (e.g., (1)) to find :

          The line of intersection is:

          or

          Check normals (optional but insightful):

          Are they coplanar?

          Check scalar triple product:

          Since the scalar triple product is zero, the normal vectors are coplanar. This is consistent with intersection in a line (or parallel configurations, but we found a line).

          4.1.3 Example : No Common Intersection (Triangular Prism)

          Determine the intersection of:

          Solution

          Step 1: Eliminate using (1) and (2). Subtract (1) from (2):

          Step 2: System of (3) and (4).

          Subtract (3) from (4):

          This is a contradiction. Therefore, the system has no solution.

          Geometric Interpretation:

          Let's check the normals:

          No two normals are parallel. So no two planes are parallel.

          • and intersect in a line .
          • and intersect in a line .
          • and intersect in a line .

          The fact that (a contradiction) arose from combining information from all three planes means these three lines of intersection ( ) must be parallel and distinct, forming a triangular prism shape. There is no point common to all three planes.

          4.1.4 Example : All Three Planes Parallel and Distinct

          Describe the intersection of:

          Solution

          The normal vector for all three planes is

          So all three planes are parallel.

          Since the values ( ) are different and not proportional in a way that would make them coincident (e.g. is ), the planes are distinct.

          Thus, there is no intersection.

          Attempting to solve this system, e.g., subtracting from :

          This contradiction confirms no solution.

          4.2 Homework
          4.2.1 Question 4.1 (Easy)

          Find the point of intersection of the following three planes:

          Solution

          We need to solve the system of three linear equations:

          Step 1: Eliminate using (1) and (2).
          Add (1) and (2):

          Step 2: Eliminate using (1) and (3).
          This requires careful choice or modification. Let's use (2) and (3).
          Add (2) and (3):

          (5)

          Step 3: Solve the 2x2 system (4) and (5).

          From (5), . Substitute this into (4):

          Substitute into :

          Step 4: Substitute and into an original equation (e.g., (1)) to find .

          Answer: The point of intersection is .

          4.2.2 Question 4.2 (Medium)

          Analyze the intersection of the following three planes. Describe the geometric configuration.

          Solution

          Observe and :

          The normal vector for is .
          The normal vector for is .
          We see that , so and are parallel.
          If we multiply by 2, we get .
          Comparing this with , the constant terms are different ( ).
          Thus, and are parallel and distinct.

          The third plane is . Its normal vector is not parallel to or . So is not parallel to or .
          Since and are parallel and distinct, they never intersect. Therefore, there can be no point common to all three planes. will intersect in a line, and will intersect in another line, and these two lines of intersection will be parallel.

          Geometric Configuration: Two planes ( ) are parallel and distinct. The third plane ( ) intersects both of these planes, creating two parallel lines of intersection. There is no common point of intersection for all three planes.

          Algebraic verification: Try to solve the system:

          Multiply (1) by 2: (1')
          Subtract (1') from (2):

          This is a contradiction.
          Therefore, the system has no solution.

          Answer: The system has no solution. Geometrically, planes and are parallel and distinct, and plane intersects them in two parallel lines.

          4.2.3 Question 4.3 (Harder)

          Consider the planes:

          The planes and intersect in a line . Determine the relationship that must exist between and such that the plane also contains the line (i.e., the three planes intersect in the line ).

          Solution

          • Step 1: Find the parametric equation of the line of intersection of and .

            Add (1) and (2) to eliminate and :

            This is interesting; is constant. This means the line of intersection is in a plane parallel to the -plane.
            Substitute into (1):

            Let (parameter).
            Then

            .
            So, the line of intersection is .
            In vector form: .
            A point on is (when ).
            The direction vector of is .

          Step 2: For to contain the line , two conditions must be met:

          • The normal vector of , , must be perpendicular to the direction vector of , .

          • Any point on line must also be on plane .
            Let's use . Substitute into :

          So, the conditions are and .
          We can express in terms of fewer parameters. Let and .
          Then and .
          The equation of would be .
          This is .

          Verification by substituting parametric equations of L into Pi3:

          Substitute

          For this to be true for all (meaning the line is in the plane), we need:

          • Coefficient of to be zero: .
          • Constant terms to be equal: .

          These are the same conditions found earlier.

          Answer: The relationship between must be:

          Any plane of the form (where are not both zero to define a plane) will contain the line .

          5 Distance from a Point to a Line [/episode/9-5]

            The distance from a point to a line is defined as the shortest distance, which is the length of the perpendicular line segment from the point to the line.

            5.1 Distance from a Point to a Line in

            Two common methods:

            5.1.1 Method 1: Using the Cartesian Equation of the Line

            If the line is given by and the point is , the distance is:

            5.1.2 Example : Distance in using Cartesian Form

            Find the distance from point to the line .

            Solution

            Here , and .

            The distance is 3 units.

            5.1.3 Method 2: Using Vector Projection (More general, also applies to )

            Let the line be given by

            where:

            • is the position vector of a point on the line
            • is the direction vector of the line

            Let be the external point.

            1. Form the vector .

            2. The distance is the magnitude of the component of that is perpendicular to .
              This can be found using the cross product (in ) or by finding the projection. An easier way for (and ) is related to area of a parallelogram.

              The area of the parallelogram formed by and is (for ).

              This area is also .
              So,

              This formula is primarily for .

              For , we can "embed" into by adding a z-component of 0.

            5.1.4 Example : Distance in using Vector Method

            Find the distance from to the line

            Solution
            • Point on the line , so .
            • Direction vector .
            • Point , so .
            • Vector .

            Using the specific formula derived from the cross product concept:

            5.2 Distance from a Point to a Line in

            Let the line be , and the point be . is the point on the line corresponding to . Vector .

            The distance is given by the formula:

            This formula arises from the fact that , where is the angle between and .
            Also, (from right-triangle trigonometry where is opposite to and is hypotenuse).

            5.2.1 Example : Distance in

            Find the distance from the point to the line .

            Solution

            From the line equation: is a point on the line, so . The direction vector is . The external point is , so .

            Vector

            Now calculate the cross product :

            Now find the magnitudes:

            The distance :

            The distance is units.

            5.2.2 Thinking Question : Zero Distance

            What does it mean if the calculated distance from a point to a line is zero?
            How would look in this case?

            Solution

            If the distance , it means the point lies on the line . The formula for distance is

            For (and assuming is not the zero vector, which it can't be for a line),
            we must have .
            This implies (the zero vector).
            The cross product of two non-zero vectors is the zero vector if and only if the vectors are collinear (parallel).
            So, if is on the line, the vector (connecting two points on the line) must be parallel to the direction vector of the line.

            5.3 Distance Between Two Skew Lines in

            Skew lines are lines in that are not parallel and do not intersect. The shortest distance between them is the length of a unique line segment that is perpendicular to both lines.

            Let the two skew lines be:

            Here, (with position vector ) is a point on , and (with position vector ) is a point on . The direction vectors are and .

            1. The vector is a common normal, i.e., it is perpendicular to both and . This vector gives the direction of the shortest distance segment.
            2. Form a vector connecting a point on to a point on , for example, .
            3. The distance between the skew lines is the absolute value of the scalar projection of onto the common normal .

            The numerator is the absolute value of the scalar triple product, which represents the volume of the parallelepiped formed by the vectors , , and . The denominator is the area of the base parallelogram formed by and . Volume / Base Area = Height, which is the distance.

            5.3.1 Example : Distance Between Skew Lines

            Find the distance between the skew lines: ,

            Solution

            From the line equations:

            , so . Direction vector . , so . Direction vector .

            First, check if they are skew. and are not parallel. Try to find intersection:

            This is false. So lines do not intersect, and are not parallel, thus they are skew.

            Vector connecting points on the lines:

            Common normal vector :

            Magnitude of the common normal:

            Dot product for the numerator of the distance formula:

            The distance :

            The distance between the skew lines is units.

            5.3.2 Thinking Question : Parallel Lines Distance as Skew Lines?

            What happens if you try to use the skew lines distance formula for two parallel lines? , (note same direction vector )

            Solution

            If the lines are parallel, their direction vectors and are scalar multiples, i.e., . For simplicity, assume . The common normal vector in the formula is . If and are parallel, then (the zero vector). The formula for distance is

            This would lead to

            This form is indeterminate, meaning the formula for skew lines is not directly applicable to parallel lines. For parallel lines, you would use the distance from a point on one line to the other line (using the formula from Section 9.5, Distance from a Point to a Line in ).

            5.4 Homework
            5.4.1 Question 5.1 (Easy)

            Find the distance from the point to the line in .

            Solution

            The formula for the distance from a point to a line is:

            Here, .
            The line is , so .

            To rationalize the denominator:

            Answer: The distance is units.

            5.4.2 Question 5.2 (Medium)

            Find the distance from the point to the line in .

            Solution

            The formula for the distance from a point to a line is:

            where is a point on the line (from ) and .

            Here, , so .
            From the line : , so .
            Direction vector .

            Vector :

            Cross product :

            Magnitude of the cross product:

            Magnitude of the direction vector :

            The distance :

            Answer: The distance is units.

            5.4.3 Question 5.3 (Harder)

            Find the shortest distance between the skew lines:

            Solution

            The formula for the distance between skew lines and is:

            From the line equations:

            • , so .
              Direction vector .
            • , so .
              Direction vector .

            Vector connecting points on the lines:

            Common normal vector :

            Magnitude of the common normal:

            Dot product for the numerator:

            The distance :

            Answer: The shortest distance between the skew lines is units.

            6 Distance from a Point to a Plane [/episode/9-6]

              The distance from a point to a plane is the shortest distance, which is the length of the perpendicular line segment from the point to the plane.

              6.1 Formula for Distance

              Let the plane be given by its Cartesian equation .
              Let the point be . The normal vector to the plane is .

              The distance from point to plane is given by:

              6.1.1 Derivation Sketch (Using Scalar Projection)
              1. Let be any point on the plane.
                So, , which means .
              2. Consider the vector .

              3. The distance is the absolute value of the scalar projection of onto the normal vector .

              4. Calculate the dot product:

              5. Substitute :

              6. So,

              6.1.2 Example : Calculating Distance from Point to Plane

              Find the distance from the point to the plane .

              Solution

              Here, . The plane equation gives . The normal vector is .

              Distance

              The distance is units.

              6.1.3 Example : Point on the Plane

              Find the distance from to the plane .

              Solution

              . .
              .
              .
              Distance .

              Since the distance is 0, the point lies on the plane .

              6.1.4 Thinking Question : Distance Between Parallel Planes

              How would you find the distance between two parallel planes: (Note: They must have the same coefficients, or be reducible to such, for their normals to be identical, ensuring they are parallel).

              Solution

              1. Pick any convenient point on one of the planes, say .
                To do this, you can set two variables to simple values (e.g., ) and solve for the third ( , assuming ).
                So, is a point on .

              2. Calculate the distance from this point to the other plane, , using the standard formula:

                Substitute :

              This gives a direct formula for the distance between two parallel planes.

              6.1.5 Example : Distance between parallel planes.

              Find the distance between and .

              Solution using the derived formula


              The distance is 2 units.

              Solution by picking a point

              Let's find a point on .
              Set . Then .
              So, is a point on .

              Now find the distance from to .

              The distance is 2 units. The results match.

              6.2 Homework
              6.2.1 Question 6.1 (Easy)

              Find the distance from the point to the plane .

              Solution

              The formula for the distance from a point to a plane is:

              Here, .
              The plane is , so .
              The normal vector is .

              Distance .

              Answer: The distance is units.

              6.2.2 Question 6.2 (Medium)

              Find the distance between the parallel planes and .

              Solution

              First, confirm the planes are parallel.

              , so they are parallel.
              To use the formula where is .

              • Rewrite to have the same normal as :
                Divide by 2:

                So now we compare

              The common normal (for this form) is .

              Distance (using from and from normalized )

              Rationalize: .

              Alternative Method: Pick a point on and find its distance to .

              Let in :

              So, is a point on .
              Now find distance from to :

              • .

              The results match.

              Answer: The distance between the parallel planes is units.

              6.2.3 Question 6.3 (Harder)

              Find the coordinates of the point on the plane that is closest to the external point . What is this shortest distance?

              Solution

              The point on the plane closest to is the foot of the perpendicular from to the plane. The line segment is thus along the normal to the plane.

              • Step 1: Define the line passing through and perpendicular to the plane .
                The normal vector to the plane is .
                This normal vector is the direction vector for the line .
                The line passes through and has direction .
                Parametric equations for :

              • Step 2: Find the intersection of line with plane . This intersection point is .
                Substitute the parametric equations of into the equation of :

                Combine terms:

              • Step 3: Substitute back into the parametric equations of to find the coordinates of .

                So, the point is .

              • Step 4: Calculate the shortest distance, which is the distance between and .
                This can be found using the distance formula for two points, or by using the formula for distance from a point to a plane.
                Using distance formula for and :

              Alternatively, using the point-to-plane distance formula for P(3,1,2) and :

              • .

              The distances match.

              Answer: The point on the plane closest to is . The shortest distance is units.

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