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10. Suppose that II1 and II2 are parallel planes in R’, given by: Il : ajx + bịy + C1z = di, and II2 : ajx + b¡y+ c1Z = d2, where di + d2. Now, suppose that II3 is another plane given by: II3 : a3x + b3y + c3z = d3, which is not parallel to II, or II2. Show that the line of intersection between II, and II3 is parallel to the line of intersection of II2 and II3. Consult Section 1.2 for the definition of parallel linęs in R 3,

10. Suppose that II1 and II2 are parallel planes in R’, given by: Il : ajx + bịy + C1z = di, and II2 : ajx + b¡y+ c1Z = d2, where di + d2. Now, suppose that II3 is another plane given by: II3 : a3x + b3y + c3z = d3, which is not parallel to II, or II2. Show that the line of intersection between II, and II3 is parallel to the line of intersection of II2 and II3. Consult Section 1.2 for the definition of parallel linęs in R 3,

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter5: Inner Product Spaces
Section5.CM: Cumulative Review
Problem 4CM: Use a software program or a graphing utility to write v as a linear combination of u1, u2, u3, u4,...
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Show that the line of intersection between 1 and 3 is parallel to the line of intersection of2 and 3. Consult Section 1.2 for the definition of parallel lines in 3.

1.2 Section Summary The equation that we obtained, 7x+ 17y- 2z = 0, is called a Cartesian equation for this plane. Thus, we can summarize this Example by saying: The Span of a non-empty set of vectors S = {v1, v2, ..., v} from R" is the set of all possible linear combinations of the vectors in the set. We write: If i = (3,-1, 2), and v = (-4, 2, 3), then: Span(S) = Span({vi, v2, ..., va} ) Span({i, v}) is the plane with Cartesian equation: 7x + 17y – 2z = 0.0 = {xiv1 +x2v2 + ...+xVk |X1, x2, ..., x e R}. We note that the individual vectors v1, v2, ..., V are all members of Span(S), where we let X; = 1 and all the other coefficients 0 in order to produce v. Similarly, the zero vector 0, is also a member of Span(S), where we make all the coefficients zero to produce 0,. We generalize this Example in the following: Definition – Axiom for a Plane in Cartesian Space: If i and v are vectors in R3 that are not parallel to each other, then Span({ủ, v}) is geometrically a plane II in Cartesian space that passes through the origin. It has a Cartesian equation, of the form: 1.2.1: In any R": Span ({0,}) = {0,}. 1.2.2: For all vi, v2, ..., Vk e R": Span ({0n, V1, v2, ..., V}) = Span ({v1, v2, .., va }). ax + by + cz = 0, If v e R" is a non-zero vector, then Span({v}) is geometrically a line L in R" passing through the origin. for some constants, a, b and c, where at least one coefficient is non-zero. In particular, if we write v = (a, b, c) e R³, then L = Span({v}) has vector equation: The letter II is the capital form of the lowercase Greek letter t. In other words, there exist scalars a, b, and c, not all zero, such that: (x, y, z) = t(a, b, c), where t e R. (x, y, z) e Span({i, v}) if and only if ax + by + cz = 0. This can be rewritten into parametric equations: x = at, y = bt, and z = ct, where t e R. 1.2.3: If i andv are non-zero and parallel vectors from some R", then: Span({i, v}) = Span({v}) = Span({ū}), which is still a line L passing through the origin. 1.2.4: If i, v e R² are non-parallel vectors, then: Span({ủ, v}) = R². In other words, any vector w = (x, y) e R2 can be expressed as a linear combination: w = rủ + sv, for some scalarsr and s. The Span of Two Non-Parallel Vectors in R3 is a Plane Through the Origin Axiom for a Plane in Cartesian Space: If i and v are vectors in R3 that are not parallel to each other, then Span({i, }) is geometrically a plane II in Cartesian space that passes through the origin. It has a Cartesian equation, in the form: The Cartesian equation of a plane in R3 is not unique. In our Example, the Cartesian equation: - 21x – 5ly + 6z = 0 ax + by + cz = 0, will represent exactly the same plane 7x+ 17y- 2z = 0, since we multiply the original equation by -3 to get the new equation. Since we can do this with any non-zero scalar, there are an infinite number of Cartesian equations for this plane. for some constants, a, b and c, where at least one coefficient is non-zero. In other words, there exist scalars a, b, and c, not all zero, such that a vector (x, y, z) e Span({ū,v}) if and only if ax + by + cz = 0.
Transcribed Image Text:1.2 Section Summary The equation that we obtained, 7x+ 17y- 2z = 0, is called a Cartesian equation for this plane. Thus, we can summarize this Example by saying: The Span of a non-empty set of vectors S = {v1, v2, ..., v} from R" is the set of all possible linear combinations of the vectors in the set. We write: If i = (3,-1, 2), and v = (-4, 2, 3), then: Span(S) = Span({vi, v2, ..., va} ) Span({i, v}) is the plane with Cartesian equation: 7x + 17y – 2z = 0.0 = {xiv1 +x2v2 + ...+xVk |X1, x2, ..., x e R}. We note that the individual vectors v1, v2, ..., V are all members of Span(S), where we let X; = 1 and all the other coefficients 0 in order to produce v. Similarly, the zero vector 0, is also a member of Span(S), where we make all the coefficients zero to produce 0,. We generalize this Example in the following: Definition – Axiom for a Plane in Cartesian Space: If i and v are vectors in R3 that are not parallel to each other, then Span({ủ, v}) is geometrically a plane II in Cartesian space that passes through the origin. It has a Cartesian equation, of the form: 1.2.1: In any R": Span ({0,}) = {0,}. 1.2.2: For all vi, v2, ..., Vk e R": Span ({0n, V1, v2, ..., V}) = Span ({v1, v2, .., va }). ax + by + cz = 0, If v e R" is a non-zero vector, then Span({v}) is geometrically a line L in R" passing through the origin. for some constants, a, b and c, where at least one coefficient is non-zero. In particular, if we write v = (a, b, c) e R³, then L = Span({v}) has vector equation: The letter II is the capital form of the lowercase Greek letter t. In other words, there exist scalars a, b, and c, not all zero, such that: (x, y, z) = t(a, b, c), where t e R. (x, y, z) e Span({i, v}) if and only if ax + by + cz = 0. This can be rewritten into parametric equations: x = at, y = bt, and z = ct, where t e R. 1.2.3: If i andv are non-zero and parallel vectors from some R", then: Span({i, v}) = Span({v}) = Span({ū}), which is still a line L passing through the origin. 1.2.4: If i, v e R² are non-parallel vectors, then: Span({ủ, v}) = R². In other words, any vector w = (x, y) e R2 can be expressed as a linear combination: w = rủ + sv, for some scalarsr and s. The Span of Two Non-Parallel Vectors in R3 is a Plane Through the Origin Axiom for a Plane in Cartesian Space: If i and v are vectors in R3 that are not parallel to each other, then Span({i, }) is geometrically a plane II in Cartesian space that passes through the origin. It has a Cartesian equation, in the form: The Cartesian equation of a plane in R3 is not unique. In our Example, the Cartesian equation: - 21x – 5ly + 6z = 0 ax + by + cz = 0, will represent exactly the same plane 7x+ 17y- 2z = 0, since we multiply the original equation by -3 to get the new equation. Since we can do this with any non-zero scalar, there are an infinite number of Cartesian equations for this plane. for some constants, a, b and c, where at least one coefficient is non-zero. In other words, there exist scalars a, b, and c, not all zero, such that a vector (x, y, z) e Span({ū,v}) if and only if ax + by + cz = 0.
10. Suppose that II1 and II2 are parallel planes in R³, given by: II1 : ajx + b¡y+cjz = d1, and II2 : ajx + b¡y+cjz = d2, where di + d2. Now, suppose that II3 is another plane given by: d3, II3 : a3x + b3y + C3z || which is not parallel to II1 or II2. Show that the line of intersection between II, and II3 is parallel to the line of intersection of II2 and II3. Consult Section 1.2 for the definition of parallel lines in R³.
Transcribed Image Text:10. Suppose that II1 and II2 are parallel planes in R³, given by: II1 : ajx + b¡y+cjz = d1, and II2 : ajx + b¡y+cjz = d2, where di + d2. Now, suppose that II3 is another plane given by: d3, II3 : a3x + b3y + C3z || which is not parallel to II1 or II2. Show that the line of intersection between II, and II3 is parallel to the line of intersection of II2 and II3. Consult Section 1.2 for the definition of parallel lines in R³.
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