Ryabushko A.P. IDZ 3.1 option 6

Download Mirror

No. 1.6. Given four points A1(0;7;1); A2(2;–1;5); A3(1;6;3); A4(3;–9;8). Necessary:

a) create an equation for the plane A1A2A3;

b) draw up an equation of straight line A1A2;

c) draw up an equation of the straight line A4M, which is perpendicular to the plane A1A2A3;

d) compose an equation for straight line A3N, which is parallel to straight line A1A2;

e) create an equation for a plane that passes through point A4 and is perpendicular to straight line A1A2;

f) calculate the sine of the angle between straight line A1A4 and plane A1A2A3;

g) calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.

a) To compile the equation of the plane A1A2A3, we find the vector product of two vectors lying in this plane:

$\overrightarrow{A_1A_2} = \begin{pmatrix}2 \ -8 \ 4\end{pmatrix}$

$\overrightarrow{A_1A_3} = \begin{pmatrix}1 \ -1 \ 2\end{pmatrix}$

$\overrightarrow{n} = \overrightarrow{A_1A_2} \times \overrightarrow{A_1A_3} = \begin{pmatrix}14 \ 2 \ 18\end{pmatrix}$

Thus, the equation of the plane A1A2A3 has the form:

$14x + 2y + 18z - 56 = $0

b) To compile the equation of straight line A1A2, we will use the parametric form of the straight line equation:

$x = 0 + 2t = 2t$

$y = 7 - 8t$

$z = 1 + 4t$

d) To compose the equation of straight line A3N parallel to straight line A1A2, we use its parametric form:

$x = 1 + 2t$

$y = 6 - 7t$

$z = 3 + 2t$

e) To compile the equation of a plane passing through point A4 and perpendicular to line A1A2, we find a vector that is perpendicular to this line:

$\overrightarrow{A_1A_2} = \begin{pmatrix}2 \ -8 \ 4\end{pmatrix}$

Since the desired plane is perpendicular to the vector $\overrightarrow{A_1A_2}$, its equation has the form:

$2x - 8y + 4z + d = 0$

To determine the coefficient d, we substitute the coordinates of point A4 into the equation:

$2\cdot3 - 8\cdot(-9) + 4\cdot8 + d = 0$

$d = -14$

Thus, the equation of the desired plane has the form:

$2x - 8y + 4z - 14 = $0

c) To compile the equation of the straight line A4M perpendicular to the plane A1A2A3, we find the vector that lies in this plane:

$\overrightarrow{A_1A_2} = \begin{pmatrix}2 \ -8 \ 4\end{pmatrix}$

$\overrightarrow{A_1A_3} = \begin{pmatrix}1 \ -1 \ 2\end{pmatrix}$

$\overrightarrow{n} = \overrightarrow{A_1A_2} \times \overrightarrow{A_1A_3} = \begin{pmatrix}14 \ 2 \ 18\end{pmatrix}$

Since the desired straight line is perpendicular to the vector $\overrightarrow{n}$, its direction vector has the form:

$\overrightarrow{AM} = \begin{pmatrix}x_A - x_M \ y_A - y_M \ z_A - z_M\end{pmatrix}$

where point M lies on line A4M. Since the straight line A4M is perpendicular to the plane A1A2A3, the vector $\overrightarrow{AM}$ must be parallel to the vector $\overrightarrow{n}$:

$\overrightarrow{AM} = t \cdot \begin{pmatrix}14 \ 2 \ 18\end{pmatrix}$

Thus, the equation of line A4M has the form:

$x = 3 + 14t$

$y = -9 + 2t$

$z = 8 + 18t$

f) To calculate the sine of the angle between straight line A1A4 and plane A1A2A3, it is necessary to find the scalar product of a vector that is parallel to straight line A1A4 and a vector that is perpendicular to plane A1A2A3:

$\overrightarrow{A_1A_4} = \begin{pmatrix}3 \ -16 \ 7\end{pmatrix}$

$\overrightarrow{n} = \overrightarrow{A_1A_2} \times \overrightarrow{A_1A_3} = \begin{pmatrix}14 \ 2 \ 18\end{pmatrix}$

$|\overrightarrow{A_1A_4}| = \sqrt{3^2 + (-16)^2 + 7^2} = \sqrt{314}$

$|\overrightarrow{n}| = \sqrt{14^2 + 2^2 + 18^2} = \sqrt{380}$

Since the sine of the angle between vectors is defined as the ratio of the scalar product of vectors to the product of their modules, the sine of this angle is equal to:

$\sin{\alpha} = \frac{\overrightarrow{A_1A_4} \cdot \overrightarrow{n}}{|\overrightarrow{A_1A_4}| \cdot |\overrightarrow{n}|} = \frac{28}{\sqrt{314} \cdot \sqrt{380}} \approx 0.425$

g) To calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3, it is necessary to find the scalar product of a vector perpendicular to the plane A1A2A3 and lying in the Oxy plane, and a vector perpendicular to the Oxy plane and lying in the A1A2A3 plane:

$\overrightarrow{n_1} = \begin{pmatrix}0 \ 0 \ 1\end{pmatrix}$

$\overrightarrow{n_2} = \overrightarrow{A_1A_2} \times \begin{pmatrix}0 \ 0 \ 1\end{pmatrix} = \begin{pmatrix}-8 \ 2 \ 0\end{pmatrix}$

$|\overrightarrow{n_1}| = 1$

$|\overrightarrow{n_2}| = \sqrt{(-8)^2 + 2^2} = 2\sqrt{17}$

Since the cosine of the angle between the vectors is

"Ryabushko A.P. IDZ 3.1 version 6" is a digital product that is a solution to an individual homework assignment in mathematics compiled by A.P. Ryabushko. The solution is made with option number 6 of task 3.1 and is intended for use by students and students who are studying this course.

The product is presented in the form of an electronic document that can be downloaded after payment in the digital goods store. The document is designed in a beautiful html format, which allows you to conveniently view and study its contents on a computer, tablet or mobile device.

The solution to the task contains a complete and detailed description of each step, which makes it easy to understand and master the material. The solution was completed by a professional teacher, which guarantees its high quality and compliance with educational standards.

"Ryabushko A.P. IDZ 3.1 option 6" is an indispensable assistant for students who want to successfully cope with individual homework in mathematics.

***

Ryabushko A.P. IDZ 3.1 option 6 is a geometry task that consists of several points.

No. 1.6. Given four points in three-dimensional space, you need to create equations for the plane and lines passing through these points, as well as calculate the sine and cosine of the angles between some of them.

No. 2.6. It is required to create an equation for a plane passing through two given points and parallel to the selected coordinate axis.

No. 3.6. It is required to find the value of the parameter at which the given lines will be parallel.

If you have any questions, you can contact the seller listed in the seller information.

***

Ease of use and intuitive interface.
High quality content (for example, high-resolution images or clear audio).
Availability and convenient delivery method.
Completeness and completeness of content.
Possibility of receiving technical support and updates.
Uniqueness and originality of content.
Fast loading and opening speed of files.
Compatible with various devices and programs.
High degree of protection against viruses and other security threats.
Convenient payment method and the ability to return goods in case of unsatisfactory quality.
Digital goods can be downloaded and used instantly, which saves time and is convenient for users.
Digital goods have a smaller environmental footprint because they do not require the production and delivery of physical copies.
Digital goods can be easily stored and transmitted through electronic media such as email or cloud storage.
Digital products can be easily updated and modified to meet the changing needs of users.
Digital goods can be accessed anytime and anywhere, providing ease of use to users.
Digital goods can be more affordable than their physical counterparts, making them more accessible to a wider audience.
Digital goods can be safer to use as they can be protected with passwords and encryption, reducing the risk of hacker attacks.