Answered: Linear Algebra Determine the line r…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Linear Algebra

Determine the line r that passes through points A(4, 0, 1) and B(5, 1, 3). Then
calculate the intersection of it with the plane π by P(0, 1, 0) and the line s : x−2=y/2=z-1

Branch of mathematics concerned with mathematical structures that are closed under operations like addition and scalar multiplication. It is the study of linear combinations, vector spaces, lines and planes, and some mappings that are used to perform linear transformations. Linear algebra also includes vectors, matrices, and linear functions. It has many applications from mathematical physics to modern algebra and coding theory.

Expert Solution

Step 1

we have to determine the equation of a line r passing through points A(4,0,1) and B(5,1,3).

then we have to calculate the intersection of it with plane $π$ by P(0,1,0) and the line s: $x - 2 = \frac{y}{2} = z - 1$

as we know that the cartesian equation of the line passing through points $A (x_{1}, y_{1}, z_{1}) a n d B (x_{2}, y_{2}, z_{2})$ is given by:

$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}}$

therefore the cartesian equation of the line r passing through points A(4,0,1) and B(5,1,3) is:

$\begin{array}{rcl} \frac{x - 4}{5 - 4} & = & \frac{y - 0}{1 - 0} = \frac{z - 1}{3 - 1} \\ \frac{x - 4}{1} & = & \frac{y}{1} = \frac{z - 1}{2} \end{array}$

Step 2

the cartesian equation of the line r passing through points A(4,0,1) and B(5,1,3) is:

$\frac{x - 4}{1} = \frac{y}{1} = \frac{z - 1}{2}$

now we have to find the equation of the plane $π$ passing through P(0,1,0) and containing the line s: $x - 2 = \frac{y}{2} = z - 1$ .

as we know that the equation of a plane passing through a point $A (x_{1}, y_{1}, z_{1})$ and having direction ratios of the normal vector to the plane as A,B and C is given by:

$A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0$

let the direction ratios of the normal vector to the plane $π$ be A,B and C and the plane $π$ is passing though point P(0,1,0).

therefore,

the equation of the plane $π$ is:

$\begin{array}{rcl} A (x - 0) + B (y - 1) + C (z - 0) & = & 0 \\ A x + B (y - 1) + C z & = & 0 (1) \end{array}$

Step 3

now as we know that the cartesian equation of the line passing through point $A (x_{1}, y_{1}, z_{1})$ and parallel to vector b having direction ratios as a,b and c is given by:

$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$ .

the equation of the line s $x - 2 = \frac{y}{2} = z - 1$ can be written as:

$\frac{x - 2}{1} = \frac{y - 0}{2} = \frac{z - 1}{1}$

therefore it is the equation of a line passing through point (2,0,1) and parallel to the vector having direction ratios 1,2,and 1

as the plane is containing the line s therefore the plane is passing through point (2,0,1) and parallel to the vector having direction ratios 1,2, and 1.

Step 4

therefore, the point (2,0,1) will satisfy the equation of the plane that is equation (1),

therefore,

$A (2) + B (0 - 1) + C (1) = 0 2 A - B + C = 0 (2)$

now as the plane is parallel to the vector having direction ratio of the vector as 1,2, and 1.

therefore the dot product of the normal vector to the plane having direction ratios A,B and C with the vector to which the plane is parallel is zero.

therefore,

$A \times 1 + B \times 2 + C \times 1 = 0 A + 2 B + C = 0 (3)$

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Linear Algebra Determine the line r that passes through points A(4, 0, 1) and B(5, 1, 3). Then calculate the intersection of it with the plane π by P(0, 1, 0) and the line s : x−2=y/2=z-1